Found this question about polynomials with integer coefficients in a book about problem solving:
Let $\ p(x) $ be a polynomial with integer coefficients. Let $\ a, b, c $ be distinct integers. Is it possible that $\ p(a) = b, p(b) = c, p(c) = a $?
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2$\begingroup$ Do you mean for all $a,b,c$ you have that property? $\endgroup$– gist076923Commented Jan 4, 2023 at 0:29
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$\begingroup$ No. The question is: does there exist a polynomial $p \in \mathbb{Z}[x]$ and distinct integers $a,b,c$ such that $p(a) = b$, $p(b) = c$, and $p(c) = a$? $\endgroup$– diracdeltafunkCommented Jan 4, 2023 at 0:32
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$\begingroup$ yeah, it doesn't need to be all integers a, b, c. just that for some distinct integers a, b, c there exists a polynomial 𝑝 ∈ ℤ[𝑥] such that the property holds. $\endgroup$– Martin WestinCommented Jan 4, 2023 at 0:34
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1$\begingroup$ That would mean $p(p(p(x)))-x=0$ has $3$ distinct integer solutions. $\endgroup$– Robert ShoreCommented Jan 4, 2023 at 0:35
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1$\begingroup$ @Prism it's a swedish book called "Matematiska Utmaningar - En kurs i problemlösning", by Pual Vaderlind. It contains a lot of competition-like problems in all kinds of areas of math $\endgroup$– Martin WestinCommented Jan 6, 2023 at 18:42
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1 Answer
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Use the fact that $x - y \mid p(x) - p(y)$ for $p(x) \in \mathbb{Z}[x]$, three times; we conclude that
- $a - b$ divides $p(a) - p(b) = b - c$,
- $b - c$ divides $p(b) - p(c) = c - a$, and
- $c - a$ divides $p(c) - p(a) = a - b$.
So each of the differences $a - b, b - c, c - a$ divide each other, and so must be equal up to sign. But this is impossible since one of $a, b, c$ must be in between the other two.
answered Jan 4, 2023 at 0:48
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$\begingroup$ How do we know that divisibility relationship holds? $\endgroup$ Commented Jan 4, 2023 at 0:49
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$\begingroup$ @Robert: it's an exercise. You can do it by first considering the case $p(x) = x^n$ then taking linear combinations, or by working $\bmod x - y$. $\endgroup$ Commented Jan 4, 2023 at 0:50
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1$\begingroup$ This is an excellent answer. I justified the conclusion a bit differently. Sharing it here in case others find it helpful. You have shown that $a-b, b-c, c-a$ must all be equal to each other, up to a sign. But if $a-b=c-b$, then $a=c$, contradiction. Similarly, if $a-b = a-c$, then $b=c$, contradiction. Thus, we may assume that $a-b = b-c = c-a$. Using these equalities, we get $b=\frac{a+c}{2}$, and $a=\frac{b+c}{2}$, and $c=\frac{a+b}{2}$. However, one of these numbers (namely, the smallest one) cannot be average of the other two, a final contradiction. $\endgroup$– PrismCommented Jan 4, 2023 at 1:03