Blog Post number 2
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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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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 It turns out that much better can be said, and that G. Baron proved in 1991 in an unpublished paper that such a polynomial has at most $latex {d+1}&fg=000000$ periodic points, including the point at infinity.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 $latex {n + k-1}&fg=000000$.
- 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.
- In the special case $latex {k = 1}&fg=000000$ and $latex {a_{n+1} = a_n}&fg=000000$ the sequence is eventually constant.
- 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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Blog Post number 4
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