Solomon Vishkautsan

Solomon Vishkautsan

This is the personal website of Solomon Vishkautsan. I am a senior lecturer at Tel-Hai Academic College, Israel. My research interests are Algebra and Number Theory.

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Arithmetic dynamics

Cycles of polynomials with integer coefficients

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During work on the article ``Scarcity of cycles for rational functions over a number field'', Jung Kyu Canci and I proved as a corollary that a rational function $latex {\phi:\mathbb{P}^1\rightarrow\mathbb{P}^1}&fg=000000$ defined over $latex {\mathbb{Q}}&fg=000000$ with everywhere good reduction has at most $latex {d+5}&fg=000000$ periodic points. As a special case, we remarked that a monic polynomial with integer coefficients has at most $latex {d+5}&fg=000000$ $latex {\mathbb{Q}}&fg=000000$-rational periodic points. It turns out that much better can be said, and that G. Baron proved in 1991 in an unpublished paper that such a polynomial has at most $latex {d+1}&fg=000000$ periodic points, including the point at infinity. I decided to translate the article from the original German, and bring it the mathematical community's attention. I marked in red additions made by me. Any mistakes are probably due to my translation.

About integer polynomials with fixed points and two-cycles

Gerd Baron

Let $latex {P(x)}&fg=000000$ be an element in the polynomial ring $latex {R[x]}&fg=000000$, where $latex {R}&fg=000000$ is some commutative ring. For an element $latex {a\in{R}}&fg=000000$ one can define a sequence $latex {\left<a_k\right>}&fg=000000$ with $latex {a_1 = a}&fg=000000$ and $latex {a_{n + 1} = P (a_n)}&fg=000000$, for $latex {n\geq{2}}&fg=000000$. A cycle is a finite sequence $latex {(a_1,...,a_n)}&fg=000000$ such that $latex {P(a_i)=a_{i+1}}&fg=000000$ for $latex {1\leq{i}\leq{n-1}}&fg=000000$ and $latex {P(a_n)=a_1}&fg=000000$. The length of a cycle $latex {(a_1,...,a_n)}&fg=000000$ is $latex {n}&fg=000000$. Over the ring $latex {R=\mathbb{Z}}&fg=000000$ we say that a cycle $latex {(a_1,...,a_n)}&fg=000000$ is integral if $latex {a_i\in\mathbb{Z}}&fg=000000$ for any $latex {1\leq{i}\leq{n}}&fg=000000$, and rational if $latex {a_i\in\mathbb{Q}}&fg=000000$ for any $latex {1\leq{i}\leq{n}}&fg=000000$.

W. Narkiewicz has shown that a polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ has no [integral] cycle of length more than two and conjectured also that [integral] cycles of length two can occur only if there is no fixed point. [In particular, since monic polynomials in $latex {\mathbb{Z}[x]}&fg=000000$ can only have integral cycles, one concludes that monic polynomials in $latex {\mathbb{Z}[x]}&fg=000000$ have rational cycles of length at most $latex {2}&fg=000000$.]

Question: Does there exist a polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$, with a rational cycle of length $latex {>2}&fg=000000$?

For the sake of completeness, we give an overview of the proof of the result of Narkiewicz.

Theorem 1 Let $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ and $latex {a \in \mathbb{Z}}&fg=000000$, and define the sequence $latex {\left<a_k\right>}&fg=000000$ with $latex {a_1 = a}&fg=000000$ and $latex {a_{n + 1} = P (a_n)}&fg=000000$. Then $latex {\left<a_k\right>}&fg=000000$ is either (1) pairwise distinct, (2) eventually constant [i.e., $latex {a_n}&fg=000000$ = $latex {a_m}&fg=000000$ for all $latex {n>m}&fg=000000$, therefore, cycles of length 1, which are fixed points] or (3) finally alternating [i.e., $latex {a_{2m} \neq a_{2m + 1}}&fg=000000$, $latex {a_{2n + 1} = a_{2m + 1}}&fg=000000$, $latex {a_{2n}=a_{2m}}&fg=000000$ for all $latex {n> m}&fg=000000$, i.e., two-cycles].

Proof: $latex {a_{n+2} - a_{n+1} = P(a_{n + 1})-P(a_{n}) = (a_{n+1}-a_n)R_n}&fg=000000$ with $latex {R_n\in\mathbb{Z}}&fg=000000$. It follows by induction that $latex {a_{n + k + 1}-a_{n+k} = (a_{n + 1}-a_n)\prod{R_j}}&fg=000000$, where the product runs over all $latex {j}&fg=000000$ from $latex {n}&fg=000000$ to $latex {n + k-1}&fg=000000$.

  1. If $latex {a_n = a_{n + k}}&fg=000000$ for a pair $latex {n,k}&fg=000000$, then $latex {\langle{a_n}\rangle}&fg=000000$ is eventually periodic.
  2. In the special case $latex {k = 1}&fg=000000$ and $latex {a_{n+1} = a_n}&fg=000000$ the sequence is eventually constant.
  3. If $latex {k> 1}&fg=000000$, then the following applies: $latex {a_{n+k+1}-a_{n + k} = a_{n+1}-a_n = a_{n+1}-a_n\prod{R_j}}&fg=000000$ where $latex {a_{n + 1}-a_n\neq{0}}&fg=000000$. So $latex {\prod R_j = 1}&fg=000000$ and $latex {R_j = \pm 1}&fg=000000$ for all relevant $latex {j}&fg=000000$. If all $latex {R_j = 1}&fg=000000$, then $latex {\sum (a_{n + j + 1} - a_{n+j}) = a_{n + k} - a_n = k(a_{n + 1}-a_n) = 0}&fg=000000$, a contradiction. So a $latex {R_j = -1}&fg=000000$ for some $latex {j}&fg=000000$, and $latex {a_{j + 2} - a_{j+1} = -(a_{j + 1} - a_{j}) = a_j - a_{j+1}}&fg=000000$, and therefore $latex {a_{j + 2} = a_j}&fg=000000$ and thus $latex {k = 2}&fg=000000$.

$latex \Box&fg=000000$

Corollary 2 A polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ can only have [integral] cycles of length $latex {2}&fg=000000$ (two-cycles) and $latex {1}&fg=000000$ (fixed points).

Next, we want to show that both can occur simultaneously and thus refute the conjecture by Narkiewicz.

Theorem 3 Let $latex {k, n}&fg=000000$ be natural numbers with $latex {2k <n}&fg=000000$ and $latex {a_1, ..., a_k}&fg=000000$ pairwise distinct positive integers and $latex {R(x) \in \mathbb{Z}[x]}&fg=000000$ a (monic) polynomial. The (monic) polynomial $latex {P(x) = xR(x)\prod_{i=1}^k(x^2-a_i^2) -x}&fg=000000$ has the fixed point $latex {0}&fg=000000$, and the two-cycles $latex {(-a_i, a_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.

Corollary 4 For each $latex {k}&fg=000000$ with $latex {2k < n}&fg=000000$ there are monic polynomials in $latex {\mathbb{Z}[x]}&fg=000000$ of degree $latex {n}&fg=000000$ with $latex {k}&fg=000000$ two-cycles and a fixed point.

Theorem 5 Let $latex {k, n}&fg=000000$ be natural numbers with $latex {k < n}&fg=000000$ and $latex {a_1, ..., a_k}&fg=000000$ pairwise distinct integers other than zero and $latex {R(x) \in \mathbb{Z}[x]}&fg=000000$ a (monic) polynomial. Then the (monic) polynomial $latex {P(x) = xR (x) \prod_{i=1}^k (x-a_i) + x}&fg=000000$ has the fixed points $latex {0}&fg=000000$ and $latex {a_i}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.

Corollary 6 For each $latex {k \leq n}&fg=000000$ there are monic polynomials in $latex {\mathbb{Z}[x]}&fg=000000$ of degree $latex {n}&fg=000000$ with $latex {k}&fg=000000$ fixed points.

Theorem 7 If a polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ has an [integral] two-cycle, then $latex {P(x)}&fg=000000$ has at most one fixed point.

Proof: Proof by contradiction. If $latex {P(x)}&fg=000000$ has the two-cycle $latex {(0, d)}&fg=000000$ and two distinct fixed points $latex {a,b}&fg=000000$ (in particular $latex {ab\neq{0}}&fg=000000$). Since $latex {P (a) = a, P (b) = b}&fg=000000$, the polynomial $latex {S(x) = P(x) -x}&fg=000000$ satisfies $latex {S(a) = S(b) = 0}&fg=000000$, and thus there exists a polynomial $latex {R(x)}&fg=000000$ such that $latex {S (x) = R(x) (x-a) (x-b)}&fg=000000$. Therefore $latex {P(x) = S(x) + x = R(x) (x-a) (x-b) + x}&fg=000000$. From $latex {P(0) = d = R(0)ab}&fg=000000$ it follows $latex {e = R (0) \neq 0, d = eab}&fg=000000$ and because $latex {0=P(d) = R(d)(d-a)(d-b) + d}&fg=000000$ also $latex {R(d)(eb-1)(ea-1) + e = 0}&fg=000000$. From the division chain $latex {e | R(d) | e}&fg=000000$ it therefore follows that $latex {ea-1 = \pm{1}}&fg=000000$ and $latex {eb-1 = \pm 1}&fg=000000$. Since $latex {ea\neq{0}}&fg=000000$ and $latex {eb\neq{0}}&fg=000000$ we get $latex {ea=eb=2}&fg=000000$ contradicting $latex {a\neq{b}}&fg=000000$. $latex \Box&fg=000000$

Theorem 8 If $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ has two [integral] two-cycles $latex {(a, b)}&fg=000000$ and $latex {(c, d)}&fg=000000$, then $latex {a + b = c + d}&fg=000000$.

Proof: Since $latex {P(a) = b}&fg=000000$ and $latex {P(b) = a}&fg=000000$ then the polynomial $latex {S(x) = P(x) + x - (a + b)}&fg=000000$ satisfies $latex {S(a) = S(b) = 0}&fg=000000$. Therefore there exists a polynomial $latex {R(x)}&fg=000000$ such that $latex {S(x) = R(x)(x-a)(x-b)}&fg=000000$. So $latex {P(x) = S(x) - x + a + b = R(x) (x-a) (x-b) - x + a + b}&fg=000000$. The polynomial $latex {Q(x) = P (P (x))}&fg=000000$ has the fixed points $latex {a, b, c, d}&fg=000000$. Because of the shape of $latex {P(x)}&fg=000000$, we get

$latex Q(x) = R (P (x)) (P (x)-a) (P (x) -b) - P (x) + a + b = $
$latex = R (P (x)) (R (x) (x-a) (x-b) -x + b) (R(x) (x-a) (x-b) -x + a) $
$latex \quad -R(x) (x-a) (x-b) + x. $

The polynomial $latex {Q_1 (x) = Q (x) -x}&fg=000000$ has the roots $latex {a, b, c, d}&fg=000000$. So $latex {Q_1 (x) = [R(P(x))(R(x)(x-a) -1)(R(x)(x-b) -1) -R (x)](x-a)(x-b)}&fg=000000$. From $latex {Q_1 (c) = 0}&fg=000000$ it follows that $latex {R(d)(R(c)(c-a) -1) (R(c) (c-b) -1) -R(c) = 0}&fg=000000$ and from $latex {Q_1(d) = 0}&fg=000000$ it follows that $latex {R(c)(R(d)(d-a) -1)(R(d)(d-b) -1)-R (d) = 0}&fg=000000$. With $latex {A = (R(c) (c-a) -1) (R(c) (c-b) -1) (R (d) (d-a) -1) (R (d) (d-b) -1)}&fg=000000$ we get $latex {R (d) (A-1) = R (c) (A-1) = 0}&fg=000000$. Thus, either both $latex {R(d)}&fg=000000$ and $latex {R(c)}&fg=000000$ are $latex {0}&fg=000000$, or $latex {A=1}&fg=000000$. From $latex {R (c) = R (d) = 0}&fg=000000$ follows $latex {P(c) = - c + a + b = d}&fg=000000$, and thus $latex {a + b = c + d}&fg=000000$ (remark that we only needed $latex {R(c)=0}&fg=000000$). If $latex {A = 1}&fg=000000$ we get $latex {R(c)(c-a) -1 =  1}&fg=000000$, etc. From $latex {R(c)(c-a) -1 = R(c)(c-b) -1}&fg=000000$ follows $latex {R(d)(d-a) -1 = R(d)(d-b) -1}&fg=000000$ and $latex {R(c)(a-b) = R(d)(a-b) = 0}&fg=000000$, so $latex {R(c) = R(d) = 0}&fg=000000$, a contradiction (as before). From $latex {R(c)(c-a) -1 \neq R(c)(c-b) -1}&fg=000000$ follows $latex {R (d) (d-a) -1 \neq R (d) (d-b) -1}&fg=000000$ and therefore each one of two expressions equal to $latex {-1}&fg=000000$. But it follows again that $latex {R (c) = R (d) = 0}&fg=000000$, a contradiction. $latex \Box&fg=000000$

Corollary 9 If $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ has the [integral] two-cycles $latex {(a_i,b_i)}&fg=000000$ ($latex {1\leq{i}\leq{k}}&fg=000000$) with fixed $latex {c}&fg=000000$ such that $latex {a_i + b_i = c}&fg=000000$ for all $latex {1\leq{i}\leq{k}}&fg=000000$, then $latex {P(x)}&fg=000000$ is of the form $latex { P (x) = R(x) \prod (x-a_i) (x-b_i) -x + c}&fg=000000$ where $latex {R(x) \in \mathbb{Z}[x]}&fg=000000$. $latex {P(x)}&fg=000000$ is monic iff $latex {R(x)}&fg=000000$ is monic.

Definition 10 Two polynomials $latex {P(x), Q(x) \in \mathbb{Z}[x] }&fg=000000$ are called equivalent [(over $latex {\mathbb{Z}}&fg=000000$)] if there exists $latex {f\in\mathbb{Z}}&fg=000000$, so that $latex {Q (x) = P (x + f) - f}&fg=000000$.

Corollary 11 If $latex {c}&fg=000000$ is even, i.e., $latex {(a_i + b_i) / 2 \in \mathbb{Z}[x]}&fg=000000$, then $latex {P(x)}&fg=000000$ is equivalent to $latex {R(x) \prod (x^2-\tilde{a}_i^2) -x}&fg=000000$ with the cycles $latex {(-\tilde{a}_i, \tilde{a}_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.

Proof: Take $latex {f=c/2}&fg=000000$. $latex \Box&fg=000000$

Theorem 12 If $latex {P(x)}&fg=000000$ has the [integral] two-cycle $latex {(a, b)}&fg=000000$ and the fixed point $latex {e}&fg=000000$, then $latex {2e = a + b}&fg=000000$.

Proof: Analogous to the proof of Theorem~8. $latex \Box&fg=000000$

Corollary 13 If $latex {P(x)}&fg=000000$ has the [integral] two-cycle $latex {(a, b)}&fg=000000$ with $latex {a + b}&fg=000000$ odd, then $latex {P(x)}&fg=000000$ has no [integral] fixed point.

Exactly one of the following cases occurs: (where $latex {P(x)}&fg=000000$ is monic exactly when $latex {R(x)}&fg=000000$ is monic).

  • Let $latex {P(x)}&fg=000000$ have the fixed points $latex {a_i}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$, then $latex {P(x) = R(x) \prod (x-a_i) + x}&fg=000000$. If $latex {k>2}&fg=000000$, then $latex {P(x)}&fg=000000$ has no [integral] two-cycle.
  • Let $latex {P(x)}&fg=000000$ have the $latex {k}&fg=000000$ [integral] two-cycles $latex {(a_i, b_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$, then $latex {a_i + b_i = c}&fg=000000$ (fixed) and $latex {P(x) = R(x) \prod (x-a_i) (x-b_i) -x + c}&fg=000000$.
    • If $latex {c}&fg=000000$ is even, then $latex {P(x)}&fg=000000$ is equivalent to $latex {Q(x) = R(x) \prod (x^2-d_i^2) -x}&fg=000000$ with the $latex {k}&fg=000000$ two-cycles $latex {(-d_i, d_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.
    • If $latex {c}&fg=000000$ is odd, then $latex {P(x)}&fg=000000$ is equivalent to $latex {Q(x) = R(x) \prod (x^2 + x-d_i^2+d_i) -x}&fg=000000$ with the $latex {k}&fg=000000$ two-cycles $latex {(-d_i + 1, d_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.
  • Let $latex {P(x)}&fg=000000$ have a fixed point $latex {c\in\mathbb{Z}}&fg=000000$ and $latex {k}&fg=000000$ [integral] two-cycles $latex {(a_i, b_i)}&fg=000000$,$latex {i\in\{1,...,k\}}&fg=000000$, so that $latex {a_i + b_i = 2c}&fg=000000$, for $latex {i\in\{1,...,k\}}&fg=000000$, and $latex {P(x) = R(x)(x-c) \prod (x-a_i) (x-b_i) -x + 2c}&fg=000000$.
    • $latex {P(x)}&fg=000000$ is equivalent to $latex {Q(x) = [R(x) \prod (x^2-d_i^2) -1]x}&fg=000000$ with the fixed point $latex {0}&fg=000000$ and the $latex {k}&fg=000000$ two-cycles $latex {(-di, di)}&fg=000000$.

Corollary 14 A polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ of degree $latex {d}&fg=000000$ has at most $latex {d}&fg=000000$ integral periodic points. A monic polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ of degree $latex {d}&fg=000000$ has at most $latex {d}&fg=000000$ rational periodic points.
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Bibliography of arithmetic dynamics of quadratic rational maps

4 minute read

Published:

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Here is a list of articles that relate to the dynamics of quadratic rational maps. Some are not arithmetic in nature, but are important for arithmetic considerations as well. Let me know if there are of articles on this topic you think should be in the list.

[1] Bjorn Poonen. The classification of rational preperiodic points of quadratic polynomials over Q: a refined conjecture. Math. Z., 228(1):11-29, 1998. [ bib | DOI | http ]
[2] Michael Stoll. Rational 6-cycles under iteration of quadratic polynomials. LMS J. Comput. Math., 11:367-380, 2008. [ bib | DOI | http ]
[3] E. V. Flynn, Bjorn Poonen, and Edward F. Schaefer. Cycles of quadratic polynomials and rational points on a genus-2 curve. Duke Math. J., 90(3):435-463, 1997. [ bib | DOI | http ]
[4] Benjamin Hutz and Patrick Ingram. On Poonen's conjecture concerning rational preperiodic points of quadratic maps. Rocky Mountain J. Math., 43(1):193-204, 2013. [ bib | DOI | http ]
[5] Xander Faber, Benjamin Hutz, Patrick Ingram, Rafe Jones, Michelle Manes, Thomas J. Tucker, and Michael E. Zieve. Uniform bounds on pre-images under quadratic dynamical systems. Math. Res. Lett., 16(1):87-101, 2009. [ bib | DOI | http ]
[6] Patrick Morton. On certain algebraic curves related to polynomial maps. Compositio Math., 103(3):319-350, 1996. [ bib | http ]
[7] Ralph Walde and Paula Russo. Rational periodic points of the quadratic function Qc(x)=x2+c. Amer. Math. Monthly, 101(4):318-331, 1994. [ bib | DOI | http ]
[8] Xander Faber. Benedetto's trick and existence of rational preperiodic structures for quadratic polynomials. Proc. Amer. Math. Soc., 143(2):685-694, 2015. [ bib | DOI | http ]
[9] Robert L. Benedetto, Ruqian Chen, Trevor Hyde, Yordanka Kovacheva, and Colin White. Small dynamical heights for quadratic polynomials and rational functions. Exp. Math., 23(4):433-447, 2014. [ bib | DOI | http ]
[10] Laura DeMarco. The moduli space of quadratic rational maps. J. Amer. Math. Soc., 20(2):321-355, 2007. [ bib | DOI | http ]
[11] Jung Kyu Canci. Rational periodic points for quadratic maps. Ann. Inst. Fourier (Grenoble), 60(3):953-985, 2010. [ bib | http ]
[12] Timo Erkama. Arithmetic progressions in cycles of quadratic polynomials. Rev. Roumaine Math. Pures Appl., 54(5-6):441-450, 2009. [ bib ]
[13] Dustin Gage and Daniel Jackson. Computer generated images for quadratic rational maps with a periodic critical point. Acta Math. Acad. Paedagog. Nyházi. (N.S.), 27(1):77-88, 2011. [ bib ]
[14] Michelle Manes. Moduli spaces for families of rational maps on P1. J. Number Theory, 129(7):1623-1663, 2009. [ bib | DOI | http ]
[15] Michelle Manes. Q-rational cycles for degree-2 rational maps having an automorphism. Proc. Lond. Math. Soc. (3), 96(3):669-696, 2008. [ bib | DOI | http ]
[16] Patrick Morton. Arithmetic properties of periodic points of quadratic maps. II. Acta Arith., 87(2):89-102, 1998. [ bib ]
[17] Patrick Morton. Arithmetic properties of periodic points of quadratic maps. Acta Arith., 62(4):343-372, 1992. [ bib ]
[18] Clayton Petsche and Brian Stout. On quadratic rational maps with prescribed good reduction. Proc. Amer. Math. Soc., 143(3):1145-1158, 2015. [ bib | DOI | http ]
[19] John R. Doyle, Xander Faber, and David Krumm. Preperiodic points for quadratic polynomials over quadratic fields. New York J. Math., 20:507-605, 2014. [ bib | .html ]
[20] Jan Kiwi. Puiseux series dynamics of quadratic rational maps. Israel J. Math., 201(2):631-700, 2014. [ bib | DOI | http ]
[21] Robert L. Devaney, Núria Fagella, Antonio Garijo, and Xavier Jarque. Sierpiński curve Julia sets for quadratic rational maps. Ann. Acad. Sci. Fenn. Math., 39(1):3-22, 2014. [ bib | DOI | http ]
[22] Selim Berker, Adam L. Epstein, and Kevin M. Pilgrim. Remarks on the period three cycles of quadratic rational maps. Nonlinearity, 16(1):93-100, 2003. [ bib | DOI | http ]
[23] John Milnor. Geometry and dynamics of quadratic rational maps. Experiment. Math., 2(1):37-83, 1993. With an appendix by the author and Lei Tan. [ bib | http ]
[24] David Lukas, Michelle Manes, and Diane Yap. A census of quadratic post-critically finite rational functions defined over Q. LMS Journal of Computation and Mathematics, 17:314-329, 2014. [ bib | DOI | http ]
[25] J. Blanc, J.K. Canci, and N. D. Elkies. Moduli spaces of quadratic rational maps with a marked periodic point of small order. 2014. [ bib | http ]
[26] Zhiming Wang and Robin Zhang. On quadratic periodic points of quadratic polynomials. 2015. [ bib | http ]
[27] Thierry Bousch. Sur quelques problèmes de dynamique holomorphe. PhD thesis, Universite de Paris-Sud, 1992. [ bib | .html ]
[28] Chatchawan Panraksa. Arithmetic dynamics of quadratic polynomials and dynamical units. 2011. [ bib | .pdf ]

This file was generated by bibtex2html 1.96.

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good reduction

Cycles of polynomials with integer coefficients

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During work on the article ``Scarcity of cycles for rational functions over a number field'', Jung Kyu Canci and I proved as a corollary that a rational function $latex {\phi:\mathbb{P}^1\rightarrow\mathbb{P}^1}&fg=000000$ defined over $latex {\mathbb{Q}}&fg=000000$ with everywhere good reduction has at most $latex {d+5}&fg=000000$ periodic points. As a special case, we remarked that a monic polynomial with integer coefficients has at most $latex {d+5}&fg=000000$ $latex {\mathbb{Q}}&fg=000000$-rational periodic points. It turns out that much better can be said, and that G. Baron proved in 1991 in an unpublished paper that such a polynomial has at most $latex {d+1}&fg=000000$ periodic points, including the point at infinity. I decided to translate the article from the original German, and bring it the mathematical community's attention. I marked in red additions made by me. Any mistakes are probably due to my translation.

About integer polynomials with fixed points and two-cycles

Gerd Baron

Let $latex {P(x)}&fg=000000$ be an element in the polynomial ring $latex {R[x]}&fg=000000$, where $latex {R}&fg=000000$ is some commutative ring. For an element $latex {a\in{R}}&fg=000000$ one can define a sequence $latex {\left<a_k\right>}&fg=000000$ with $latex {a_1 = a}&fg=000000$ and $latex {a_{n + 1} = P (a_n)}&fg=000000$, for $latex {n\geq{2}}&fg=000000$. A cycle is a finite sequence $latex {(a_1,...,a_n)}&fg=000000$ such that $latex {P(a_i)=a_{i+1}}&fg=000000$ for $latex {1\leq{i}\leq{n-1}}&fg=000000$ and $latex {P(a_n)=a_1}&fg=000000$. The length of a cycle $latex {(a_1,...,a_n)}&fg=000000$ is $latex {n}&fg=000000$. Over the ring $latex {R=\mathbb{Z}}&fg=000000$ we say that a cycle $latex {(a_1,...,a_n)}&fg=000000$ is integral if $latex {a_i\in\mathbb{Z}}&fg=000000$ for any $latex {1\leq{i}\leq{n}}&fg=000000$, and rational if $latex {a_i\in\mathbb{Q}}&fg=000000$ for any $latex {1\leq{i}\leq{n}}&fg=000000$.

W. Narkiewicz has shown that a polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ has no [integral] cycle of length more than two and conjectured also that [integral] cycles of length two can occur only if there is no fixed point. [In particular, since monic polynomials in $latex {\mathbb{Z}[x]}&fg=000000$ can only have integral cycles, one concludes that monic polynomials in $latex {\mathbb{Z}[x]}&fg=000000$ have rational cycles of length at most $latex {2}&fg=000000$.]

Question: Does there exist a polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$, with a rational cycle of length $latex {>2}&fg=000000$?

For the sake of completeness, we give an overview of the proof of the result of Narkiewicz.

Theorem 1 Let $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ and $latex {a \in \mathbb{Z}}&fg=000000$, and define the sequence $latex {\left<a_k\right>}&fg=000000$ with $latex {a_1 = a}&fg=000000$ and $latex {a_{n + 1} = P (a_n)}&fg=000000$. Then $latex {\left<a_k\right>}&fg=000000$ is either (1) pairwise distinct, (2) eventually constant [i.e., $latex {a_n}&fg=000000$ = $latex {a_m}&fg=000000$ for all $latex {n>m}&fg=000000$, therefore, cycles of length 1, which are fixed points] or (3) finally alternating [i.e., $latex {a_{2m} \neq a_{2m + 1}}&fg=000000$, $latex {a_{2n + 1} = a_{2m + 1}}&fg=000000$, $latex {a_{2n}=a_{2m}}&fg=000000$ for all $latex {n> m}&fg=000000$, i.e., two-cycles].

Proof: $latex {a_{n+2} - a_{n+1} = P(a_{n + 1})-P(a_{n}) = (a_{n+1}-a_n)R_n}&fg=000000$ with $latex {R_n\in\mathbb{Z}}&fg=000000$. It follows by induction that $latex {a_{n + k + 1}-a_{n+k} = (a_{n + 1}-a_n)\prod{R_j}}&fg=000000$, where the product runs over all $latex {j}&fg=000000$ from $latex {n}&fg=000000$ to $latex {n + k-1}&fg=000000$.

  1. If $latex {a_n = a_{n + k}}&fg=000000$ for a pair $latex {n,k}&fg=000000$, then $latex {\langle{a_n}\rangle}&fg=000000$ is eventually periodic.
  2. In the special case $latex {k = 1}&fg=000000$ and $latex {a_{n+1} = a_n}&fg=000000$ the sequence is eventually constant.
  3. If $latex {k> 1}&fg=000000$, then the following applies: $latex {a_{n+k+1}-a_{n + k} = a_{n+1}-a_n = a_{n+1}-a_n\prod{R_j}}&fg=000000$ where $latex {a_{n + 1}-a_n\neq{0}}&fg=000000$. So $latex {\prod R_j = 1}&fg=000000$ and $latex {R_j = \pm 1}&fg=000000$ for all relevant $latex {j}&fg=000000$. If all $latex {R_j = 1}&fg=000000$, then $latex {\sum (a_{n + j + 1} - a_{n+j}) = a_{n + k} - a_n = k(a_{n + 1}-a_n) = 0}&fg=000000$, a contradiction. So a $latex {R_j = -1}&fg=000000$ for some $latex {j}&fg=000000$, and $latex {a_{j + 2} - a_{j+1} = -(a_{j + 1} - a_{j}) = a_j - a_{j+1}}&fg=000000$, and therefore $latex {a_{j + 2} = a_j}&fg=000000$ and thus $latex {k = 2}&fg=000000$.

$latex \Box&fg=000000$

Corollary 2 A polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ can only have [integral] cycles of length $latex {2}&fg=000000$ (two-cycles) and $latex {1}&fg=000000$ (fixed points).

Next, we want to show that both can occur simultaneously and thus refute the conjecture by Narkiewicz.

Theorem 3 Let $latex {k, n}&fg=000000$ be natural numbers with $latex {2k <n}&fg=000000$ and $latex {a_1, ..., a_k}&fg=000000$ pairwise distinct positive integers and $latex {R(x) \in \mathbb{Z}[x]}&fg=000000$ a (monic) polynomial. The (monic) polynomial $latex {P(x) = xR(x)\prod_{i=1}^k(x^2-a_i^2) -x}&fg=000000$ has the fixed point $latex {0}&fg=000000$, and the two-cycles $latex {(-a_i, a_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.

Corollary 4 For each $latex {k}&fg=000000$ with $latex {2k < n}&fg=000000$ there are monic polynomials in $latex {\mathbb{Z}[x]}&fg=000000$ of degree $latex {n}&fg=000000$ with $latex {k}&fg=000000$ two-cycles and a fixed point.

Theorem 5 Let $latex {k, n}&fg=000000$ be natural numbers with $latex {k < n}&fg=000000$ and $latex {a_1, ..., a_k}&fg=000000$ pairwise distinct integers other than zero and $latex {R(x) \in \mathbb{Z}[x]}&fg=000000$ a (monic) polynomial. Then the (monic) polynomial $latex {P(x) = xR (x) \prod_{i=1}^k (x-a_i) + x}&fg=000000$ has the fixed points $latex {0}&fg=000000$ and $latex {a_i}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.

Corollary 6 For each $latex {k \leq n}&fg=000000$ there are monic polynomials in $latex {\mathbb{Z}[x]}&fg=000000$ of degree $latex {n}&fg=000000$ with $latex {k}&fg=000000$ fixed points.

Theorem 7 If a polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ has an [integral] two-cycle, then $latex {P(x)}&fg=000000$ has at most one fixed point.

Proof: Proof by contradiction. If $latex {P(x)}&fg=000000$ has the two-cycle $latex {(0, d)}&fg=000000$ and two distinct fixed points $latex {a,b}&fg=000000$ (in particular $latex {ab\neq{0}}&fg=000000$). Since $latex {P (a) = a, P (b) = b}&fg=000000$, the polynomial $latex {S(x) = P(x) -x}&fg=000000$ satisfies $latex {S(a) = S(b) = 0}&fg=000000$, and thus there exists a polynomial $latex {R(x)}&fg=000000$ such that $latex {S (x) = R(x) (x-a) (x-b)}&fg=000000$. Therefore $latex {P(x) = S(x) + x = R(x) (x-a) (x-b) + x}&fg=000000$. From $latex {P(0) = d = R(0)ab}&fg=000000$ it follows $latex {e = R (0) \neq 0, d = eab}&fg=000000$ and because $latex {0=P(d) = R(d)(d-a)(d-b) + d}&fg=000000$ also $latex {R(d)(eb-1)(ea-1) + e = 0}&fg=000000$. From the division chain $latex {e | R(d) | e}&fg=000000$ it therefore follows that $latex {ea-1 = \pm{1}}&fg=000000$ and $latex {eb-1 = \pm 1}&fg=000000$. Since $latex {ea\neq{0}}&fg=000000$ and $latex {eb\neq{0}}&fg=000000$ we get $latex {ea=eb=2}&fg=000000$ contradicting $latex {a\neq{b}}&fg=000000$. $latex \Box&fg=000000$

Theorem 8 If $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ has two [integral] two-cycles $latex {(a, b)}&fg=000000$ and $latex {(c, d)}&fg=000000$, then $latex {a + b = c + d}&fg=000000$.

Proof: Since $latex {P(a) = b}&fg=000000$ and $latex {P(b) = a}&fg=000000$ then the polynomial $latex {S(x) = P(x) + x - (a + b)}&fg=000000$ satisfies $latex {S(a) = S(b) = 0}&fg=000000$. Therefore there exists a polynomial $latex {R(x)}&fg=000000$ such that $latex {S(x) = R(x)(x-a)(x-b)}&fg=000000$. So $latex {P(x) = S(x) - x + a + b = R(x) (x-a) (x-b) - x + a + b}&fg=000000$. The polynomial $latex {Q(x) = P (P (x))}&fg=000000$ has the fixed points $latex {a, b, c, d}&fg=000000$. Because of the shape of $latex {P(x)}&fg=000000$, we get

$latex Q(x) = R (P (x)) (P (x)-a) (P (x) -b) - P (x) + a + b = $
$latex = R (P (x)) (R (x) (x-a) (x-b) -x + b) (R(x) (x-a) (x-b) -x + a) $
$latex \quad -R(x) (x-a) (x-b) + x. $

The polynomial $latex {Q_1 (x) = Q (x) -x}&fg=000000$ has the roots $latex {a, b, c, d}&fg=000000$. So $latex {Q_1 (x) = [R(P(x))(R(x)(x-a) -1)(R(x)(x-b) -1) -R (x)](x-a)(x-b)}&fg=000000$. From $latex {Q_1 (c) = 0}&fg=000000$ it follows that $latex {R(d)(R(c)(c-a) -1) (R(c) (c-b) -1) -R(c) = 0}&fg=000000$ and from $latex {Q_1(d) = 0}&fg=000000$ it follows that $latex {R(c)(R(d)(d-a) -1)(R(d)(d-b) -1)-R (d) = 0}&fg=000000$. With $latex {A = (R(c) (c-a) -1) (R(c) (c-b) -1) (R (d) (d-a) -1) (R (d) (d-b) -1)}&fg=000000$ we get $latex {R (d) (A-1) = R (c) (A-1) = 0}&fg=000000$. Thus, either both $latex {R(d)}&fg=000000$ and $latex {R(c)}&fg=000000$ are $latex {0}&fg=000000$, or $latex {A=1}&fg=000000$. From $latex {R (c) = R (d) = 0}&fg=000000$ follows $latex {P(c) = - c + a + b = d}&fg=000000$, and thus $latex {a + b = c + d}&fg=000000$ (remark that we only needed $latex {R(c)=0}&fg=000000$). If $latex {A = 1}&fg=000000$ we get $latex {R(c)(c-a) -1 =  1}&fg=000000$, etc. From $latex {R(c)(c-a) -1 = R(c)(c-b) -1}&fg=000000$ follows $latex {R(d)(d-a) -1 = R(d)(d-b) -1}&fg=000000$ and $latex {R(c)(a-b) = R(d)(a-b) = 0}&fg=000000$, so $latex {R(c) = R(d) = 0}&fg=000000$, a contradiction (as before). From $latex {R(c)(c-a) -1 \neq R(c)(c-b) -1}&fg=000000$ follows $latex {R (d) (d-a) -1 \neq R (d) (d-b) -1}&fg=000000$ and therefore each one of two expressions equal to $latex {-1}&fg=000000$. But it follows again that $latex {R (c) = R (d) = 0}&fg=000000$, a contradiction. $latex \Box&fg=000000$

Corollary 9 If $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ has the [integral] two-cycles $latex {(a_i,b_i)}&fg=000000$ ($latex {1\leq{i}\leq{k}}&fg=000000$) with fixed $latex {c}&fg=000000$ such that $latex {a_i + b_i = c}&fg=000000$ for all $latex {1\leq{i}\leq{k}}&fg=000000$, then $latex {P(x)}&fg=000000$ is of the form $latex { P (x) = R(x) \prod (x-a_i) (x-b_i) -x + c}&fg=000000$ where $latex {R(x) \in \mathbb{Z}[x]}&fg=000000$. $latex {P(x)}&fg=000000$ is monic iff $latex {R(x)}&fg=000000$ is monic.

Definition 10 Two polynomials $latex {P(x), Q(x) \in \mathbb{Z}[x] }&fg=000000$ are called equivalent [(over $latex {\mathbb{Z}}&fg=000000$)] if there exists $latex {f\in\mathbb{Z}}&fg=000000$, so that $latex {Q (x) = P (x + f) - f}&fg=000000$.

Corollary 11 If $latex {c}&fg=000000$ is even, i.e., $latex {(a_i + b_i) / 2 \in \mathbb{Z}[x]}&fg=000000$, then $latex {P(x)}&fg=000000$ is equivalent to $latex {R(x) \prod (x^2-\tilde{a}_i^2) -x}&fg=000000$ with the cycles $latex {(-\tilde{a}_i, \tilde{a}_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.

Proof: Take $latex {f=c/2}&fg=000000$. $latex \Box&fg=000000$

Theorem 12 If $latex {P(x)}&fg=000000$ has the [integral] two-cycle $latex {(a, b)}&fg=000000$ and the fixed point $latex {e}&fg=000000$, then $latex {2e = a + b}&fg=000000$.

Proof: Analogous to the proof of Theorem~8. $latex \Box&fg=000000$

Corollary 13 If $latex {P(x)}&fg=000000$ has the [integral] two-cycle $latex {(a, b)}&fg=000000$ with $latex {a + b}&fg=000000$ odd, then $latex {P(x)}&fg=000000$ has no [integral] fixed point.

Exactly one of the following cases occurs: (where $latex {P(x)}&fg=000000$ is monic exactly when $latex {R(x)}&fg=000000$ is monic).

  • Let $latex {P(x)}&fg=000000$ have the fixed points $latex {a_i}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$, then $latex {P(x) = R(x) \prod (x-a_i) + x}&fg=000000$. If $latex {k>2}&fg=000000$, then $latex {P(x)}&fg=000000$ has no [integral] two-cycle.
  • Let $latex {P(x)}&fg=000000$ have the $latex {k}&fg=000000$ [integral] two-cycles $latex {(a_i, b_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$, then $latex {a_i + b_i = c}&fg=000000$ (fixed) and $latex {P(x) = R(x) \prod (x-a_i) (x-b_i) -x + c}&fg=000000$.
    • If $latex {c}&fg=000000$ is even, then $latex {P(x)}&fg=000000$ is equivalent to $latex {Q(x) = R(x) \prod (x^2-d_i^2) -x}&fg=000000$ with the $latex {k}&fg=000000$ two-cycles $latex {(-d_i, d_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.
    • If $latex {c}&fg=000000$ is odd, then $latex {P(x)}&fg=000000$ is equivalent to $latex {Q(x) = R(x) \prod (x^2 + x-d_i^2+d_i) -x}&fg=000000$ with the $latex {k}&fg=000000$ two-cycles $latex {(-d_i + 1, d_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.
  • Let $latex {P(x)}&fg=000000$ have a fixed point $latex {c\in\mathbb{Z}}&fg=000000$ and $latex {k}&fg=000000$ [integral] two-cycles $latex {(a_i, b_i)}&fg=000000$,$latex {i\in\{1,...,k\}}&fg=000000$, so that $latex {a_i + b_i = 2c}&fg=000000$, for $latex {i\in\{1,...,k\}}&fg=000000$, and $latex {P(x) = R(x)(x-c) \prod (x-a_i) (x-b_i) -x + 2c}&fg=000000$.
    • $latex {P(x)}&fg=000000$ is equivalent to $latex {Q(x) = [R(x) \prod (x^2-d_i^2) -1]x}&fg=000000$ with the fixed point $latex {0}&fg=000000$ and the $latex {k}&fg=000000$ two-cycles $latex {(-di, di)}&fg=000000$.

Corollary 14 A polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ of degree $latex {d}&fg=000000$ has at most $latex {d}&fg=000000$ integral periodic points. A monic polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ of degree $latex {d}&fg=000000$ has at most $latex {d}&fg=000000$ rational periodic points.
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periodic points

Cycles of polynomials with integer coefficients

11 minute read

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During work on the article ``Scarcity of cycles for rational functions over a number field'', Jung Kyu Canci and I proved as a corollary that a rational function $latex {\phi:\mathbb{P}^1\rightarrow\mathbb{P}^1}&fg=000000$ defined over $latex {\mathbb{Q}}&fg=000000$ with everywhere good reduction has at most $latex {d+5}&fg=000000$ periodic points. As a special case, we remarked that a monic polynomial with integer coefficients has at most $latex {d+5}&fg=000000$ $latex {\mathbb{Q}}&fg=000000$-rational periodic points. It turns out that much better can be said, and that G. Baron proved in 1991 in an unpublished paper that such a polynomial has at most $latex {d+1}&fg=000000$ periodic points, including the point at infinity. I decided to translate the article from the original German, and bring it the mathematical community's attention. I marked in red additions made by me. Any mistakes are probably due to my translation.

About integer polynomials with fixed points and two-cycles

Gerd Baron

Let $latex {P(x)}&fg=000000$ be an element in the polynomial ring $latex {R[x]}&fg=000000$, where $latex {R}&fg=000000$ is some commutative ring. For an element $latex {a\in{R}}&fg=000000$ one can define a sequence $latex {\left<a_k\right>}&fg=000000$ with $latex {a_1 = a}&fg=000000$ and $latex {a_{n + 1} = P (a_n)}&fg=000000$, for $latex {n\geq{2}}&fg=000000$. A cycle is a finite sequence $latex {(a_1,...,a_n)}&fg=000000$ such that $latex {P(a_i)=a_{i+1}}&fg=000000$ for $latex {1\leq{i}\leq{n-1}}&fg=000000$ and $latex {P(a_n)=a_1}&fg=000000$. The length of a cycle $latex {(a_1,...,a_n)}&fg=000000$ is $latex {n}&fg=000000$. Over the ring $latex {R=\mathbb{Z}}&fg=000000$ we say that a cycle $latex {(a_1,...,a_n)}&fg=000000$ is integral if $latex {a_i\in\mathbb{Z}}&fg=000000$ for any $latex {1\leq{i}\leq{n}}&fg=000000$, and rational if $latex {a_i\in\mathbb{Q}}&fg=000000$ for any $latex {1\leq{i}\leq{n}}&fg=000000$.

W. Narkiewicz has shown that a polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ has no [integral] cycle of length more than two and conjectured also that [integral] cycles of length two can occur only if there is no fixed point. [In particular, since monic polynomials in $latex {\mathbb{Z}[x]}&fg=000000$ can only have integral cycles, one concludes that monic polynomials in $latex {\mathbb{Z}[x]}&fg=000000$ have rational cycles of length at most $latex {2}&fg=000000$.]

Question: Does there exist a polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$, with a rational cycle of length $latex {>2}&fg=000000$?

For the sake of completeness, we give an overview of the proof of the result of Narkiewicz.

Theorem 1 Let $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ and $latex {a \in \mathbb{Z}}&fg=000000$, and define the sequence $latex {\left<a_k\right>}&fg=000000$ with $latex {a_1 = a}&fg=000000$ and $latex {a_{n + 1} = P (a_n)}&fg=000000$. Then $latex {\left<a_k\right>}&fg=000000$ is either (1) pairwise distinct, (2) eventually constant [i.e., $latex {a_n}&fg=000000$ = $latex {a_m}&fg=000000$ for all $latex {n>m}&fg=000000$, therefore, cycles of length 1, which are fixed points] or (3) finally alternating [i.e., $latex {a_{2m} \neq a_{2m + 1}}&fg=000000$, $latex {a_{2n + 1} = a_{2m + 1}}&fg=000000$, $latex {a_{2n}=a_{2m}}&fg=000000$ for all $latex {n> m}&fg=000000$, i.e., two-cycles].

Proof: $latex {a_{n+2} - a_{n+1} = P(a_{n + 1})-P(a_{n}) = (a_{n+1}-a_n)R_n}&fg=000000$ with $latex {R_n\in\mathbb{Z}}&fg=000000$. It follows by induction that $latex {a_{n + k + 1}-a_{n+k} = (a_{n + 1}-a_n)\prod{R_j}}&fg=000000$, where the product runs over all $latex {j}&fg=000000$ from $latex {n}&fg=000000$ to $latex {n + k-1}&fg=000000$.

  1. If $latex {a_n = a_{n + k}}&fg=000000$ for a pair $latex {n,k}&fg=000000$, then $latex {\langle{a_n}\rangle}&fg=000000$ is eventually periodic.
  2. In the special case $latex {k = 1}&fg=000000$ and $latex {a_{n+1} = a_n}&fg=000000$ the sequence is eventually constant.
  3. If $latex {k> 1}&fg=000000$, then the following applies: $latex {a_{n+k+1}-a_{n + k} = a_{n+1}-a_n = a_{n+1}-a_n\prod{R_j}}&fg=000000$ where $latex {a_{n + 1}-a_n\neq{0}}&fg=000000$. So $latex {\prod R_j = 1}&fg=000000$ and $latex {R_j = \pm 1}&fg=000000$ for all relevant $latex {j}&fg=000000$. If all $latex {R_j = 1}&fg=000000$, then $latex {\sum (a_{n + j + 1} - a_{n+j}) = a_{n + k} - a_n = k(a_{n + 1}-a_n) = 0}&fg=000000$, a contradiction. So a $latex {R_j = -1}&fg=000000$ for some $latex {j}&fg=000000$, and $latex {a_{j + 2} - a_{j+1} = -(a_{j + 1} - a_{j}) = a_j - a_{j+1}}&fg=000000$, and therefore $latex {a_{j + 2} = a_j}&fg=000000$ and thus $latex {k = 2}&fg=000000$.

$latex \Box&fg=000000$

Corollary 2 A polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ can only have [integral] cycles of length $latex {2}&fg=000000$ (two-cycles) and $latex {1}&fg=000000$ (fixed points).

Next, we want to show that both can occur simultaneously and thus refute the conjecture by Narkiewicz.

Theorem 3 Let $latex {k, n}&fg=000000$ be natural numbers with $latex {2k <n}&fg=000000$ and $latex {a_1, ..., a_k}&fg=000000$ pairwise distinct positive integers and $latex {R(x) \in \mathbb{Z}[x]}&fg=000000$ a (monic) polynomial. The (monic) polynomial $latex {P(x) = xR(x)\prod_{i=1}^k(x^2-a_i^2) -x}&fg=000000$ has the fixed point $latex {0}&fg=000000$, and the two-cycles $latex {(-a_i, a_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.

Corollary 4 For each $latex {k}&fg=000000$ with $latex {2k < n}&fg=000000$ there are monic polynomials in $latex {\mathbb{Z}[x]}&fg=000000$ of degree $latex {n}&fg=000000$ with $latex {k}&fg=000000$ two-cycles and a fixed point.

Theorem 5 Let $latex {k, n}&fg=000000$ be natural numbers with $latex {k < n}&fg=000000$ and $latex {a_1, ..., a_k}&fg=000000$ pairwise distinct integers other than zero and $latex {R(x) \in \mathbb{Z}[x]}&fg=000000$ a (monic) polynomial. Then the (monic) polynomial $latex {P(x) = xR (x) \prod_{i=1}^k (x-a_i) + x}&fg=000000$ has the fixed points $latex {0}&fg=000000$ and $latex {a_i}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.

Corollary 6 For each $latex {k \leq n}&fg=000000$ there are monic polynomials in $latex {\mathbb{Z}[x]}&fg=000000$ of degree $latex {n}&fg=000000$ with $latex {k}&fg=000000$ fixed points.

Theorem 7 If a polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ has an [integral] two-cycle, then $latex {P(x)}&fg=000000$ has at most one fixed point.

Proof: Proof by contradiction. If $latex {P(x)}&fg=000000$ has the two-cycle $latex {(0, d)}&fg=000000$ and two distinct fixed points $latex {a,b}&fg=000000$ (in particular $latex {ab\neq{0}}&fg=000000$). Since $latex {P (a) = a, P (b) = b}&fg=000000$, the polynomial $latex {S(x) = P(x) -x}&fg=000000$ satisfies $latex {S(a) = S(b) = 0}&fg=000000$, and thus there exists a polynomial $latex {R(x)}&fg=000000$ such that $latex {S (x) = R(x) (x-a) (x-b)}&fg=000000$. Therefore $latex {P(x) = S(x) + x = R(x) (x-a) (x-b) + x}&fg=000000$. From $latex {P(0) = d = R(0)ab}&fg=000000$ it follows $latex {e = R (0) \neq 0, d = eab}&fg=000000$ and because $latex {0=P(d) = R(d)(d-a)(d-b) + d}&fg=000000$ also $latex {R(d)(eb-1)(ea-1) + e = 0}&fg=000000$. From the division chain $latex {e | R(d) | e}&fg=000000$ it therefore follows that $latex {ea-1 = \pm{1}}&fg=000000$ and $latex {eb-1 = \pm 1}&fg=000000$. Since $latex {ea\neq{0}}&fg=000000$ and $latex {eb\neq{0}}&fg=000000$ we get $latex {ea=eb=2}&fg=000000$ contradicting $latex {a\neq{b}}&fg=000000$. $latex \Box&fg=000000$

Theorem 8 If $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ has two [integral] two-cycles $latex {(a, b)}&fg=000000$ and $latex {(c, d)}&fg=000000$, then $latex {a + b = c + d}&fg=000000$.

Proof: Since $latex {P(a) = b}&fg=000000$ and $latex {P(b) = a}&fg=000000$ then the polynomial $latex {S(x) = P(x) + x - (a + b)}&fg=000000$ satisfies $latex {S(a) = S(b) = 0}&fg=000000$. Therefore there exists a polynomial $latex {R(x)}&fg=000000$ such that $latex {S(x) = R(x)(x-a)(x-b)}&fg=000000$. So $latex {P(x) = S(x) - x + a + b = R(x) (x-a) (x-b) - x + a + b}&fg=000000$. The polynomial $latex {Q(x) = P (P (x))}&fg=000000$ has the fixed points $latex {a, b, c, d}&fg=000000$. Because of the shape of $latex {P(x)}&fg=000000$, we get

$latex Q(x) = R (P (x)) (P (x)-a) (P (x) -b) - P (x) + a + b = $
$latex = R (P (x)) (R (x) (x-a) (x-b) -x + b) (R(x) (x-a) (x-b) -x + a) $
$latex \quad -R(x) (x-a) (x-b) + x. $

The polynomial $latex {Q_1 (x) = Q (x) -x}&fg=000000$ has the roots $latex {a, b, c, d}&fg=000000$. So $latex {Q_1 (x) = [R(P(x))(R(x)(x-a) -1)(R(x)(x-b) -1) -R (x)](x-a)(x-b)}&fg=000000$. From $latex {Q_1 (c) = 0}&fg=000000$ it follows that $latex {R(d)(R(c)(c-a) -1) (R(c) (c-b) -1) -R(c) = 0}&fg=000000$ and from $latex {Q_1(d) = 0}&fg=000000$ it follows that $latex {R(c)(R(d)(d-a) -1)(R(d)(d-b) -1)-R (d) = 0}&fg=000000$. With $latex {A = (R(c) (c-a) -1) (R(c) (c-b) -1) (R (d) (d-a) -1) (R (d) (d-b) -1)}&fg=000000$ we get $latex {R (d) (A-1) = R (c) (A-1) = 0}&fg=000000$. Thus, either both $latex {R(d)}&fg=000000$ and $latex {R(c)}&fg=000000$ are $latex {0}&fg=000000$, or $latex {A=1}&fg=000000$. From $latex {R (c) = R (d) = 0}&fg=000000$ follows $latex {P(c) = - c + a + b = d}&fg=000000$, and thus $latex {a + b = c + d}&fg=000000$ (remark that we only needed $latex {R(c)=0}&fg=000000$). If $latex {A = 1}&fg=000000$ we get $latex {R(c)(c-a) -1 =  1}&fg=000000$, etc. From $latex {R(c)(c-a) -1 = R(c)(c-b) -1}&fg=000000$ follows $latex {R(d)(d-a) -1 = R(d)(d-b) -1}&fg=000000$ and $latex {R(c)(a-b) = R(d)(a-b) = 0}&fg=000000$, so $latex {R(c) = R(d) = 0}&fg=000000$, a contradiction (as before). From $latex {R(c)(c-a) -1 \neq R(c)(c-b) -1}&fg=000000$ follows $latex {R (d) (d-a) -1 \neq R (d) (d-b) -1}&fg=000000$ and therefore each one of two expressions equal to $latex {-1}&fg=000000$. But it follows again that $latex {R (c) = R (d) = 0}&fg=000000$, a contradiction. $latex \Box&fg=000000$

Corollary 9 If $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ has the [integral] two-cycles $latex {(a_i,b_i)}&fg=000000$ ($latex {1\leq{i}\leq{k}}&fg=000000$) with fixed $latex {c}&fg=000000$ such that $latex {a_i + b_i = c}&fg=000000$ for all $latex {1\leq{i}\leq{k}}&fg=000000$, then $latex {P(x)}&fg=000000$ is of the form $latex { P (x) = R(x) \prod (x-a_i) (x-b_i) -x + c}&fg=000000$ where $latex {R(x) \in \mathbb{Z}[x]}&fg=000000$. $latex {P(x)}&fg=000000$ is monic iff $latex {R(x)}&fg=000000$ is monic.

Definition 10 Two polynomials $latex {P(x), Q(x) \in \mathbb{Z}[x] }&fg=000000$ are called equivalent [(over $latex {\mathbb{Z}}&fg=000000$)] if there exists $latex {f\in\mathbb{Z}}&fg=000000$, so that $latex {Q (x) = P (x + f) - f}&fg=000000$.

Corollary 11 If $latex {c}&fg=000000$ is even, i.e., $latex {(a_i + b_i) / 2 \in \mathbb{Z}[x]}&fg=000000$, then $latex {P(x)}&fg=000000$ is equivalent to $latex {R(x) \prod (x^2-\tilde{a}_i^2) -x}&fg=000000$ with the cycles $latex {(-\tilde{a}_i, \tilde{a}_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.

Proof: Take $latex {f=c/2}&fg=000000$. $latex \Box&fg=000000$

Theorem 12 If $latex {P(x)}&fg=000000$ has the [integral] two-cycle $latex {(a, b)}&fg=000000$ and the fixed point $latex {e}&fg=000000$, then $latex {2e = a + b}&fg=000000$.

Proof: Analogous to the proof of Theorem~8. $latex \Box&fg=000000$

Corollary 13 If $latex {P(x)}&fg=000000$ has the [integral] two-cycle $latex {(a, b)}&fg=000000$ with $latex {a + b}&fg=000000$ odd, then $latex {P(x)}&fg=000000$ has no [integral] fixed point.

Exactly one of the following cases occurs: (where $latex {P(x)}&fg=000000$ is monic exactly when $latex {R(x)}&fg=000000$ is monic).

  • Let $latex {P(x)}&fg=000000$ have the fixed points $latex {a_i}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$, then $latex {P(x) = R(x) \prod (x-a_i) + x}&fg=000000$. If $latex {k>2}&fg=000000$, then $latex {P(x)}&fg=000000$ has no [integral] two-cycle.
  • Let $latex {P(x)}&fg=000000$ have the $latex {k}&fg=000000$ [integral] two-cycles $latex {(a_i, b_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$, then $latex {a_i + b_i = c}&fg=000000$ (fixed) and $latex {P(x) = R(x) \prod (x-a_i) (x-b_i) -x + c}&fg=000000$.
    • If $latex {c}&fg=000000$ is even, then $latex {P(x)}&fg=000000$ is equivalent to $latex {Q(x) = R(x) \prod (x^2-d_i^2) -x}&fg=000000$ with the $latex {k}&fg=000000$ two-cycles $latex {(-d_i, d_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.
    • If $latex {c}&fg=000000$ is odd, then $latex {P(x)}&fg=000000$ is equivalent to $latex {Q(x) = R(x) \prod (x^2 + x-d_i^2+d_i) -x}&fg=000000$ with the $latex {k}&fg=000000$ two-cycles $latex {(-d_i + 1, d_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.
  • Let $latex {P(x)}&fg=000000$ have a fixed point $latex {c\in\mathbb{Z}}&fg=000000$ and $latex {k}&fg=000000$ [integral] two-cycles $latex {(a_i, b_i)}&fg=000000$,$latex {i\in\{1,...,k\}}&fg=000000$, so that $latex {a_i + b_i = 2c}&fg=000000$, for $latex {i\in\{1,...,k\}}&fg=000000$, and $latex {P(x) = R(x)(x-c) \prod (x-a_i) (x-b_i) -x + 2c}&fg=000000$.
    • $latex {P(x)}&fg=000000$ is equivalent to $latex {Q(x) = [R(x) \prod (x^2-d_i^2) -1]x}&fg=000000$ with the fixed point $latex {0}&fg=000000$ and the $latex {k}&fg=000000$ two-cycles $latex {(-di, di)}&fg=000000$.

Corollary 14 A polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ of degree $latex {d}&fg=000000$ has at most $latex {d}&fg=000000$ integral periodic points. A monic polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ of degree $latex {d}&fg=000000$ has at most $latex {d}&fg=000000$ rational periodic points.
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polynomials

Cycles of polynomials with integer coefficients

11 minute read

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During work on the article ``Scarcity of cycles for rational functions over a number field'', Jung Kyu Canci and I proved as a corollary that a rational function $latex {\phi:\mathbb{P}^1\rightarrow\mathbb{P}^1}&fg=000000$ defined over $latex {\mathbb{Q}}&fg=000000$ with everywhere good reduction has at most $latex {d+5}&fg=000000$ periodic points. As a special case, we remarked that a monic polynomial with integer coefficients has at most $latex {d+5}&fg=000000$ $latex {\mathbb{Q}}&fg=000000$-rational periodic points. It turns out that much better can be said, and that G. Baron proved in 1991 in an unpublished paper that such a polynomial has at most $latex {d+1}&fg=000000$ periodic points, including the point at infinity. I decided to translate the article from the original German, and bring it the mathematical community's attention. I marked in red additions made by me. Any mistakes are probably due to my translation.

About integer polynomials with fixed points and two-cycles

Gerd Baron

Let $latex {P(x)}&fg=000000$ be an element in the polynomial ring $latex {R[x]}&fg=000000$, where $latex {R}&fg=000000$ is some commutative ring. For an element $latex {a\in{R}}&fg=000000$ one can define a sequence $latex {\left<a_k\right>}&fg=000000$ with $latex {a_1 = a}&fg=000000$ and $latex {a_{n + 1} = P (a_n)}&fg=000000$, for $latex {n\geq{2}}&fg=000000$. A cycle is a finite sequence $latex {(a_1,...,a_n)}&fg=000000$ such that $latex {P(a_i)=a_{i+1}}&fg=000000$ for $latex {1\leq{i}\leq{n-1}}&fg=000000$ and $latex {P(a_n)=a_1}&fg=000000$. The length of a cycle $latex {(a_1,...,a_n)}&fg=000000$ is $latex {n}&fg=000000$. Over the ring $latex {R=\mathbb{Z}}&fg=000000$ we say that a cycle $latex {(a_1,...,a_n)}&fg=000000$ is integral if $latex {a_i\in\mathbb{Z}}&fg=000000$ for any $latex {1\leq{i}\leq{n}}&fg=000000$, and rational if $latex {a_i\in\mathbb{Q}}&fg=000000$ for any $latex {1\leq{i}\leq{n}}&fg=000000$.

W. Narkiewicz has shown that a polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ has no [integral] cycle of length more than two and conjectured also that [integral] cycles of length two can occur only if there is no fixed point. [In particular, since monic polynomials in $latex {\mathbb{Z}[x]}&fg=000000$ can only have integral cycles, one concludes that monic polynomials in $latex {\mathbb{Z}[x]}&fg=000000$ have rational cycles of length at most $latex {2}&fg=000000$.]

Question: Does there exist a polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$, with a rational cycle of length $latex {>2}&fg=000000$?

For the sake of completeness, we give an overview of the proof of the result of Narkiewicz.

Theorem 1 Let $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ and $latex {a \in \mathbb{Z}}&fg=000000$, and define the sequence $latex {\left<a_k\right>}&fg=000000$ with $latex {a_1 = a}&fg=000000$ and $latex {a_{n + 1} = P (a_n)}&fg=000000$. Then $latex {\left<a_k\right>}&fg=000000$ is either (1) pairwise distinct, (2) eventually constant [i.e., $latex {a_n}&fg=000000$ = $latex {a_m}&fg=000000$ for all $latex {n>m}&fg=000000$, therefore, cycles of length 1, which are fixed points] or (3) finally alternating [i.e., $latex {a_{2m} \neq a_{2m + 1}}&fg=000000$, $latex {a_{2n + 1} = a_{2m + 1}}&fg=000000$, $latex {a_{2n}=a_{2m}}&fg=000000$ for all $latex {n> m}&fg=000000$, i.e., two-cycles].

Proof: $latex {a_{n+2} - a_{n+1} = P(a_{n + 1})-P(a_{n}) = (a_{n+1}-a_n)R_n}&fg=000000$ with $latex {R_n\in\mathbb{Z}}&fg=000000$. It follows by induction that $latex {a_{n + k + 1}-a_{n+k} = (a_{n + 1}-a_n)\prod{R_j}}&fg=000000$, where the product runs over all $latex {j}&fg=000000$ from $latex {n}&fg=000000$ to $latex {n + k-1}&fg=000000$.

  1. If $latex {a_n = a_{n + k}}&fg=000000$ for a pair $latex {n,k}&fg=000000$, then $latex {\langle{a_n}\rangle}&fg=000000$ is eventually periodic.
  2. In the special case $latex {k = 1}&fg=000000$ and $latex {a_{n+1} = a_n}&fg=000000$ the sequence is eventually constant.
  3. If $latex {k> 1}&fg=000000$, then the following applies: $latex {a_{n+k+1}-a_{n + k} = a_{n+1}-a_n = a_{n+1}-a_n\prod{R_j}}&fg=000000$ where $latex {a_{n + 1}-a_n\neq{0}}&fg=000000$. So $latex {\prod R_j = 1}&fg=000000$ and $latex {R_j = \pm 1}&fg=000000$ for all relevant $latex {j}&fg=000000$. If all $latex {R_j = 1}&fg=000000$, then $latex {\sum (a_{n + j + 1} - a_{n+j}) = a_{n + k} - a_n = k(a_{n + 1}-a_n) = 0}&fg=000000$, a contradiction. So a $latex {R_j = -1}&fg=000000$ for some $latex {j}&fg=000000$, and $latex {a_{j + 2} - a_{j+1} = -(a_{j + 1} - a_{j}) = a_j - a_{j+1}}&fg=000000$, and therefore $latex {a_{j + 2} = a_j}&fg=000000$ and thus $latex {k = 2}&fg=000000$.

$latex \Box&fg=000000$

Corollary 2 A polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ can only have [integral] cycles of length $latex {2}&fg=000000$ (two-cycles) and $latex {1}&fg=000000$ (fixed points).

Next, we want to show that both can occur simultaneously and thus refute the conjecture by Narkiewicz.

Theorem 3 Let $latex {k, n}&fg=000000$ be natural numbers with $latex {2k <n}&fg=000000$ and $latex {a_1, ..., a_k}&fg=000000$ pairwise distinct positive integers and $latex {R(x) \in \mathbb{Z}[x]}&fg=000000$ a (monic) polynomial. The (monic) polynomial $latex {P(x) = xR(x)\prod_{i=1}^k(x^2-a_i^2) -x}&fg=000000$ has the fixed point $latex {0}&fg=000000$, and the two-cycles $latex {(-a_i, a_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.

Corollary 4 For each $latex {k}&fg=000000$ with $latex {2k < n}&fg=000000$ there are monic polynomials in $latex {\mathbb{Z}[x]}&fg=000000$ of degree $latex {n}&fg=000000$ with $latex {k}&fg=000000$ two-cycles and a fixed point.

Theorem 5 Let $latex {k, n}&fg=000000$ be natural numbers with $latex {k < n}&fg=000000$ and $latex {a_1, ..., a_k}&fg=000000$ pairwise distinct integers other than zero and $latex {R(x) \in \mathbb{Z}[x]}&fg=000000$ a (monic) polynomial. Then the (monic) polynomial $latex {P(x) = xR (x) \prod_{i=1}^k (x-a_i) + x}&fg=000000$ has the fixed points $latex {0}&fg=000000$ and $latex {a_i}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.

Corollary 6 For each $latex {k \leq n}&fg=000000$ there are monic polynomials in $latex {\mathbb{Z}[x]}&fg=000000$ of degree $latex {n}&fg=000000$ with $latex {k}&fg=000000$ fixed points.

Theorem 7 If a polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ has an [integral] two-cycle, then $latex {P(x)}&fg=000000$ has at most one fixed point.

Proof: Proof by contradiction. If $latex {P(x)}&fg=000000$ has the two-cycle $latex {(0, d)}&fg=000000$ and two distinct fixed points $latex {a,b}&fg=000000$ (in particular $latex {ab\neq{0}}&fg=000000$). Since $latex {P (a) = a, P (b) = b}&fg=000000$, the polynomial $latex {S(x) = P(x) -x}&fg=000000$ satisfies $latex {S(a) = S(b) = 0}&fg=000000$, and thus there exists a polynomial $latex {R(x)}&fg=000000$ such that $latex {S (x) = R(x) (x-a) (x-b)}&fg=000000$. Therefore $latex {P(x) = S(x) + x = R(x) (x-a) (x-b) + x}&fg=000000$. From $latex {P(0) = d = R(0)ab}&fg=000000$ it follows $latex {e = R (0) \neq 0, d = eab}&fg=000000$ and because $latex {0=P(d) = R(d)(d-a)(d-b) + d}&fg=000000$ also $latex {R(d)(eb-1)(ea-1) + e = 0}&fg=000000$. From the division chain $latex {e | R(d) | e}&fg=000000$ it therefore follows that $latex {ea-1 = \pm{1}}&fg=000000$ and $latex {eb-1 = \pm 1}&fg=000000$. Since $latex {ea\neq{0}}&fg=000000$ and $latex {eb\neq{0}}&fg=000000$ we get $latex {ea=eb=2}&fg=000000$ contradicting $latex {a\neq{b}}&fg=000000$. $latex \Box&fg=000000$

Theorem 8 If $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ has two [integral] two-cycles $latex {(a, b)}&fg=000000$ and $latex {(c, d)}&fg=000000$, then $latex {a + b = c + d}&fg=000000$.

Proof: Since $latex {P(a) = b}&fg=000000$ and $latex {P(b) = a}&fg=000000$ then the polynomial $latex {S(x) = P(x) + x - (a + b)}&fg=000000$ satisfies $latex {S(a) = S(b) = 0}&fg=000000$. Therefore there exists a polynomial $latex {R(x)}&fg=000000$ such that $latex {S(x) = R(x)(x-a)(x-b)}&fg=000000$. So $latex {P(x) = S(x) - x + a + b = R(x) (x-a) (x-b) - x + a + b}&fg=000000$. The polynomial $latex {Q(x) = P (P (x))}&fg=000000$ has the fixed points $latex {a, b, c, d}&fg=000000$. Because of the shape of $latex {P(x)}&fg=000000$, we get

$latex Q(x) = R (P (x)) (P (x)-a) (P (x) -b) - P (x) + a + b = $
$latex = R (P (x)) (R (x) (x-a) (x-b) -x + b) (R(x) (x-a) (x-b) -x + a) $
$latex \quad -R(x) (x-a) (x-b) + x. $

The polynomial $latex {Q_1 (x) = Q (x) -x}&fg=000000$ has the roots $latex {a, b, c, d}&fg=000000$. So $latex {Q_1 (x) = [R(P(x))(R(x)(x-a) -1)(R(x)(x-b) -1) -R (x)](x-a)(x-b)}&fg=000000$. From $latex {Q_1 (c) = 0}&fg=000000$ it follows that $latex {R(d)(R(c)(c-a) -1) (R(c) (c-b) -1) -R(c) = 0}&fg=000000$ and from $latex {Q_1(d) = 0}&fg=000000$ it follows that $latex {R(c)(R(d)(d-a) -1)(R(d)(d-b) -1)-R (d) = 0}&fg=000000$. With $latex {A = (R(c) (c-a) -1) (R(c) (c-b) -1) (R (d) (d-a) -1) (R (d) (d-b) -1)}&fg=000000$ we get $latex {R (d) (A-1) = R (c) (A-1) = 0}&fg=000000$. Thus, either both $latex {R(d)}&fg=000000$ and $latex {R(c)}&fg=000000$ are $latex {0}&fg=000000$, or $latex {A=1}&fg=000000$. From $latex {R (c) = R (d) = 0}&fg=000000$ follows $latex {P(c) = - c + a + b = d}&fg=000000$, and thus $latex {a + b = c + d}&fg=000000$ (remark that we only needed $latex {R(c)=0}&fg=000000$). If $latex {A = 1}&fg=000000$ we get $latex {R(c)(c-a) -1 =  1}&fg=000000$, etc. From $latex {R(c)(c-a) -1 = R(c)(c-b) -1}&fg=000000$ follows $latex {R(d)(d-a) -1 = R(d)(d-b) -1}&fg=000000$ and $latex {R(c)(a-b) = R(d)(a-b) = 0}&fg=000000$, so $latex {R(c) = R(d) = 0}&fg=000000$, a contradiction (as before). From $latex {R(c)(c-a) -1 \neq R(c)(c-b) -1}&fg=000000$ follows $latex {R (d) (d-a) -1 \neq R (d) (d-b) -1}&fg=000000$ and therefore each one of two expressions equal to $latex {-1}&fg=000000$. But it follows again that $latex {R (c) = R (d) = 0}&fg=000000$, a contradiction. $latex \Box&fg=000000$

Corollary 9 If $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ has the [integral] two-cycles $latex {(a_i,b_i)}&fg=000000$ ($latex {1\leq{i}\leq{k}}&fg=000000$) with fixed $latex {c}&fg=000000$ such that $latex {a_i + b_i = c}&fg=000000$ for all $latex {1\leq{i}\leq{k}}&fg=000000$, then $latex {P(x)}&fg=000000$ is of the form $latex { P (x) = R(x) \prod (x-a_i) (x-b_i) -x + c}&fg=000000$ where $latex {R(x) \in \mathbb{Z}[x]}&fg=000000$. $latex {P(x)}&fg=000000$ is monic iff $latex {R(x)}&fg=000000$ is monic.

Definition 10 Two polynomials $latex {P(x), Q(x) \in \mathbb{Z}[x] }&fg=000000$ are called equivalent [(over $latex {\mathbb{Z}}&fg=000000$)] if there exists $latex {f\in\mathbb{Z}}&fg=000000$, so that $latex {Q (x) = P (x + f) - f}&fg=000000$.

Corollary 11 If $latex {c}&fg=000000$ is even, i.e., $latex {(a_i + b_i) / 2 \in \mathbb{Z}[x]}&fg=000000$, then $latex {P(x)}&fg=000000$ is equivalent to $latex {R(x) \prod (x^2-\tilde{a}_i^2) -x}&fg=000000$ with the cycles $latex {(-\tilde{a}_i, \tilde{a}_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.

Proof: Take $latex {f=c/2}&fg=000000$. $latex \Box&fg=000000$

Theorem 12 If $latex {P(x)}&fg=000000$ has the [integral] two-cycle $latex {(a, b)}&fg=000000$ and the fixed point $latex {e}&fg=000000$, then $latex {2e = a + b}&fg=000000$.

Proof: Analogous to the proof of Theorem~8. $latex \Box&fg=000000$

Corollary 13 If $latex {P(x)}&fg=000000$ has the [integral] two-cycle $latex {(a, b)}&fg=000000$ with $latex {a + b}&fg=000000$ odd, then $latex {P(x)}&fg=000000$ has no [integral] fixed point.

Exactly one of the following cases occurs: (where $latex {P(x)}&fg=000000$ is monic exactly when $latex {R(x)}&fg=000000$ is monic).

  • Let $latex {P(x)}&fg=000000$ have the fixed points $latex {a_i}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$, then $latex {P(x) = R(x) \prod (x-a_i) + x}&fg=000000$. If $latex {k>2}&fg=000000$, then $latex {P(x)}&fg=000000$ has no [integral] two-cycle.
  • Let $latex {P(x)}&fg=000000$ have the $latex {k}&fg=000000$ [integral] two-cycles $latex {(a_i, b_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$, then $latex {a_i + b_i = c}&fg=000000$ (fixed) and $latex {P(x) = R(x) \prod (x-a_i) (x-b_i) -x + c}&fg=000000$.
    • If $latex {c}&fg=000000$ is even, then $latex {P(x)}&fg=000000$ is equivalent to $latex {Q(x) = R(x) \prod (x^2-d_i^2) -x}&fg=000000$ with the $latex {k}&fg=000000$ two-cycles $latex {(-d_i, d_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.
    • If $latex {c}&fg=000000$ is odd, then $latex {P(x)}&fg=000000$ is equivalent to $latex {Q(x) = R(x) \prod (x^2 + x-d_i^2+d_i) -x}&fg=000000$ with the $latex {k}&fg=000000$ two-cycles $latex {(-d_i + 1, d_i)}&fg=000000$, $latex {i\in\{1,...,k\}}&fg=000000$.
  • Let $latex {P(x)}&fg=000000$ have a fixed point $latex {c\in\mathbb{Z}}&fg=000000$ and $latex {k}&fg=000000$ [integral] two-cycles $latex {(a_i, b_i)}&fg=000000$,$latex {i\in\{1,...,k\}}&fg=000000$, so that $latex {a_i + b_i = 2c}&fg=000000$, for $latex {i\in\{1,...,k\}}&fg=000000$, and $latex {P(x) = R(x)(x-c) \prod (x-a_i) (x-b_i) -x + 2c}&fg=000000$.
    • $latex {P(x)}&fg=000000$ is equivalent to $latex {Q(x) = [R(x) \prod (x^2-d_i^2) -1]x}&fg=000000$ with the fixed point $latex {0}&fg=000000$ and the $latex {k}&fg=000000$ two-cycles $latex {(-di, di)}&fg=000000$.

Corollary 14 A polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ of degree $latex {d}&fg=000000$ has at most $latex {d}&fg=000000$ integral periodic points. A monic polynomial $latex {P(x) \in \mathbb{Z}[x]}&fg=000000$ of degree $latex {d}&fg=000000$ has at most $latex {d}&fg=000000$ rational periodic points.
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quadratic polynomials

Bibliography of arithmetic dynamics of quadratic rational maps

4 minute read

Published:

<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN" "http://www.w3.org/TR/REC-html40/loose.dtd">
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Here is a list of articles that relate to the dynamics of quadratic rational maps. Some are not arithmetic in nature, but are important for arithmetic considerations as well. Let me know if there are of articles on this topic you think should be in the list.

[1] Bjorn Poonen. The classification of rational preperiodic points of quadratic polynomials over Q: a refined conjecture. Math. Z., 228(1):11-29, 1998. [ bib | DOI | http ]
[2] Michael Stoll. Rational 6-cycles under iteration of quadratic polynomials. LMS J. Comput. Math., 11:367-380, 2008. [ bib | DOI | http ]
[3] E. V. Flynn, Bjorn Poonen, and Edward F. Schaefer. Cycles of quadratic polynomials and rational points on a genus-2 curve. Duke Math. J., 90(3):435-463, 1997. [ bib | DOI | http ]
[4] Benjamin Hutz and Patrick Ingram. On Poonen's conjecture concerning rational preperiodic points of quadratic maps. Rocky Mountain J. Math., 43(1):193-204, 2013. [ bib | DOI | http ]
[5] Xander Faber, Benjamin Hutz, Patrick Ingram, Rafe Jones, Michelle Manes, Thomas J. Tucker, and Michael E. Zieve. Uniform bounds on pre-images under quadratic dynamical systems. Math. Res. Lett., 16(1):87-101, 2009. [ bib | DOI | http ]
[6] Patrick Morton. On certain algebraic curves related to polynomial maps. Compositio Math., 103(3):319-350, 1996. [ bib | http ]
[7] Ralph Walde and Paula Russo. Rational periodic points of the quadratic function Qc(x)=x2+c. Amer. Math. Monthly, 101(4):318-331, 1994. [ bib | DOI | http ]
[8] Xander Faber. Benedetto's trick and existence of rational preperiodic structures for quadratic polynomials. Proc. Amer. Math. Soc., 143(2):685-694, 2015. [ bib | DOI | http ]
[9] Robert L. Benedetto, Ruqian Chen, Trevor Hyde, Yordanka Kovacheva, and Colin White. Small dynamical heights for quadratic polynomials and rational functions. Exp. Math., 23(4):433-447, 2014. [ bib | DOI | http ]
[10] Laura DeMarco. The moduli space of quadratic rational maps. J. Amer. Math. Soc., 20(2):321-355, 2007. [ bib | DOI | http ]
[11] Jung Kyu Canci. Rational periodic points for quadratic maps. Ann. Inst. Fourier (Grenoble), 60(3):953-985, 2010. [ bib | http ]
[12] Timo Erkama. Arithmetic progressions in cycles of quadratic polynomials. Rev. Roumaine Math. Pures Appl., 54(5-6):441-450, 2009. [ bib ]
[13] Dustin Gage and Daniel Jackson. Computer generated images for quadratic rational maps with a periodic critical point. Acta Math. Acad. Paedagog. Nyházi. (N.S.), 27(1):77-88, 2011. [ bib ]
[14] Michelle Manes. Moduli spaces for families of rational maps on P1. J. Number Theory, 129(7):1623-1663, 2009. [ bib | DOI | http ]
[15] Michelle Manes. Q-rational cycles for degree-2 rational maps having an automorphism. Proc. Lond. Math. Soc. (3), 96(3):669-696, 2008. [ bib | DOI | http ]
[16] Patrick Morton. Arithmetic properties of periodic points of quadratic maps. II. Acta Arith., 87(2):89-102, 1998. [ bib ]
[17] Patrick Morton. Arithmetic properties of periodic points of quadratic maps. Acta Arith., 62(4):343-372, 1992. [ bib ]
[18] Clayton Petsche and Brian Stout. On quadratic rational maps with prescribed good reduction. Proc. Amer. Math. Soc., 143(3):1145-1158, 2015. [ bib | DOI | http ]
[19] John R. Doyle, Xander Faber, and David Krumm. Preperiodic points for quadratic polynomials over quadratic fields. New York J. Math., 20:507-605, 2014. [ bib | .html ]
[20] Jan Kiwi. Puiseux series dynamics of quadratic rational maps. Israel J. Math., 201(2):631-700, 2014. [ bib | DOI | http ]
[21] Robert L. Devaney, Núria Fagella, Antonio Garijo, and Xavier Jarque. Sierpiński curve Julia sets for quadratic rational maps. Ann. Acad. Sci. Fenn. Math., 39(1):3-22, 2014. [ bib | DOI | http ]
[22] Selim Berker, Adam L. Epstein, and Kevin M. Pilgrim. Remarks on the period three cycles of quadratic rational maps. Nonlinearity, 16(1):93-100, 2003. [ bib | DOI | http ]
[23] John Milnor. Geometry and dynamics of quadratic rational maps. Experiment. Math., 2(1):37-83, 1993. With an appendix by the author and Lei Tan. [ bib | http ]
[24] David Lukas, Michelle Manes, and Diane Yap. A census of quadratic post-critically finite rational functions defined over Q. LMS Journal of Computation and Mathematics, 17:314-329, 2014. [ bib | DOI | http ]
[25] J. Blanc, J.K. Canci, and N. D. Elkies. Moduli spaces of quadratic rational maps with a marked periodic point of small order. 2014. [ bib | http ]
[26] Zhiming Wang and Robin Zhang. On quadratic periodic points of quadratic polynomials. 2015. [ bib | http ]
[27] Thierry Bousch. Sur quelques problèmes de dynamique holomorphe. PhD thesis, Universite de Paris-Sud, 1992. [ bib | .html ]
[28] Chatchawan Panraksa. Arithmetic dynamics of quadratic polynomials and dynamical units. 2011. [ bib | .pdf ]

This file was generated by bibtex2html 1.96.

</body></html>

quadratic rational maps

Bibliography of arithmetic dynamics of quadratic rational maps

4 minute read

Published:

<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN" "http://www.w3.org/TR/REC-html40/loose.dtd">
</p>

Here is a list of articles that relate to the dynamics of quadratic rational maps. Some are not arithmetic in nature, but are important for arithmetic considerations as well. Let me know if there are of articles on this topic you think should be in the list.

[1] Bjorn Poonen. The classification of rational preperiodic points of quadratic polynomials over Q: a refined conjecture. Math. Z., 228(1):11-29, 1998. [ bib | DOI | http ]
[2] Michael Stoll. Rational 6-cycles under iteration of quadratic polynomials. LMS J. Comput. Math., 11:367-380, 2008. [ bib | DOI | http ]
[3] E. V. Flynn, Bjorn Poonen, and Edward F. Schaefer. Cycles of quadratic polynomials and rational points on a genus-2 curve. Duke Math. J., 90(3):435-463, 1997. [ bib | DOI | http ]
[4] Benjamin Hutz and Patrick Ingram. On Poonen's conjecture concerning rational preperiodic points of quadratic maps. Rocky Mountain J. Math., 43(1):193-204, 2013. [ bib | DOI | http ]
[5] Xander Faber, Benjamin Hutz, Patrick Ingram, Rafe Jones, Michelle Manes, Thomas J. Tucker, and Michael E. Zieve. Uniform bounds on pre-images under quadratic dynamical systems. Math. Res. Lett., 16(1):87-101, 2009. [ bib | DOI | http ]
[6] Patrick Morton. On certain algebraic curves related to polynomial maps. Compositio Math., 103(3):319-350, 1996. [ bib | http ]
[7] Ralph Walde and Paula Russo. Rational periodic points of the quadratic function Qc(x)=x2+c. Amer. Math. Monthly, 101(4):318-331, 1994. [ bib | DOI | http ]
[8] Xander Faber. Benedetto's trick and existence of rational preperiodic structures for quadratic polynomials. Proc. Amer. Math. Soc., 143(2):685-694, 2015. [ bib | DOI | http ]
[9] Robert L. Benedetto, Ruqian Chen, Trevor Hyde, Yordanka Kovacheva, and Colin White. Small dynamical heights for quadratic polynomials and rational functions. Exp. Math., 23(4):433-447, 2014. [ bib | DOI | http ]
[10] Laura DeMarco. The moduli space of quadratic rational maps. J. Amer. Math. Soc., 20(2):321-355, 2007. [ bib | DOI | http ]
[11] Jung Kyu Canci. Rational periodic points for quadratic maps. Ann. Inst. Fourier (Grenoble), 60(3):953-985, 2010. [ bib | http ]
[12] Timo Erkama. Arithmetic progressions in cycles of quadratic polynomials. Rev. Roumaine Math. Pures Appl., 54(5-6):441-450, 2009. [ bib ]
[13] Dustin Gage and Daniel Jackson. Computer generated images for quadratic rational maps with a periodic critical point. Acta Math. Acad. Paedagog. Nyházi. (N.S.), 27(1):77-88, 2011. [ bib ]
[14] Michelle Manes. Moduli spaces for families of rational maps on P1. J. Number Theory, 129(7):1623-1663, 2009. [ bib | DOI | http ]
[15] Michelle Manes. Q-rational cycles for degree-2 rational maps having an automorphism. Proc. Lond. Math. Soc. (3), 96(3):669-696, 2008. [ bib | DOI | http ]
[16] Patrick Morton. Arithmetic properties of periodic points of quadratic maps. II. Acta Arith., 87(2):89-102, 1998. [ bib ]
[17] Patrick Morton. Arithmetic properties of periodic points of quadratic maps. Acta Arith., 62(4):343-372, 1992. [ bib ]
[18] Clayton Petsche and Brian Stout. On quadratic rational maps with prescribed good reduction. Proc. Amer. Math. Soc., 143(3):1145-1158, 2015. [ bib | DOI | http ]
[19] John R. Doyle, Xander Faber, and David Krumm. Preperiodic points for quadratic polynomials over quadratic fields. New York J. Math., 20:507-605, 2014. [ bib | .html ]
[20] Jan Kiwi. Puiseux series dynamics of quadratic rational maps. Israel J. Math., 201(2):631-700, 2014. [ bib | DOI | http ]
[21] Robert L. Devaney, Núria Fagella, Antonio Garijo, and Xavier Jarque. Sierpiński curve Julia sets for quadratic rational maps. Ann. Acad. Sci. Fenn. Math., 39(1):3-22, 2014. [ bib | DOI | http ]
[22] Selim Berker, Adam L. Epstein, and Kevin M. Pilgrim. Remarks on the period three cycles of quadratic rational maps. Nonlinearity, 16(1):93-100, 2003. [ bib | DOI | http ]
[23] John Milnor. Geometry and dynamics of quadratic rational maps. Experiment. Math., 2(1):37-83, 1993. With an appendix by the author and Lei Tan. [ bib | http ]
[24] David Lukas, Michelle Manes, and Diane Yap. A census of quadratic post-critically finite rational functions defined over Q. LMS Journal of Computation and Mathematics, 17:314-329, 2014. [ bib | DOI | http ]
[25] J. Blanc, J.K. Canci, and N. D. Elkies. Moduli spaces of quadratic rational maps with a marked periodic point of small order. 2014. [ bib | http ]
[26] Zhiming Wang and Robin Zhang. On quadratic periodic points of quadratic polynomials. 2015. [ bib | http ]
[27] Thierry Bousch. Sur quelques problèmes de dynamique holomorphe. PhD thesis, Universite de Paris-Sud, 1992. [ bib | .html ]
[28] Chatchawan Panraksa. Arithmetic dynamics of quadratic polynomials and dynamical units. 2011. [ bib | .pdf ]

This file was generated by bibtex2html 1.96.

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