Find all polynomials $p$ with integer coefficients such that the number $p(a) - p(b)$ is divisible by $a + b$ for all integers $a, b$, provided that $a + b \neq 0$.
Problem
Source: Baltic Way 2021, Problem 19
Tags: number theory, number theory proposed, Polynomials
sanyalarnab
15.11.2021 21:25
Let $P(a,b): \implies a+b \mid p(a)-p(b)$ We know $a+b \mid p(a)-p(-b)$ So subtracting this result with $P(a,b)$, we get $$a+b \mid p(b)-p(-b)$$Now fixing $b$ and varying $a$ over the integers, we get that $p(b)-p(-b)$ has infinitely many divisors. So $p(b)=p(-b)$ Or $\boxed{p(b)=\sum_{i=0}^{n}a_ix^{2i}}$ where $a_n \neq 0$ and $n \in \mathbb{N}$. Now check that $p(a)-p(b)= \sum_{i=0}^{n}a_i\{(a^2)^i-(b^2)^i\}=(a+b)\sum_{i=0}^{n}a_iF_i$ where $F_i \in \mathbb{Z}$.
Andrew_Dou
15.11.2021 21:44
Let $p(x)=\sum_{k=0}^n a_{n-k} x^{n-k}$ for integers $a_1,a_2,a_3, \dots a_n.$
We have that $p(a)-p(b)=\sum_{k=0}^n a_{n-k} a^{n-k}-\sum_{k=0}^n a_{n-k} b^{n-k}=\sum_{k=0}^n a_{n-k} (a^{n-k}-b^{n-k}).$ Since we must have $a+b \mid \sum_{k=0}^n a_{n-k} (a^{n-k}-b^{n-k}),$ for all integers $a,b$ it is equivalent to having $a+b \mid \sum_{k=0}^n (a^{n-k}-b^{n-k})$ (by choosing sufficiently large $a,b$ so that $a+b>a_k$ and $a+b$ is prime)
For $a+b \mid \sum_{k=0}^n (a^{n-k}+b^{n-k}),$ $a=-b$ must result in $\sum_{k=0}^n (a^{n-k}+b^{n-k})=0.$ The reason we can take $a=-b$ under the condition that $a+b \neq 0$ is due to the following.
Consider $p(a)-p(b)=f(a,b)$ for a polynomial $f(a,b).$ If $a+b \mid p(a)-p(b)=f(a,b),$ then $f(a,b)=(a+b)g(a,b)$ for some polynomial $g(a,b).$
Considering $f(a,b)$ and $g(a,b)$ as "single" variable polynomials in terms of $a,$ then we know that $f(a,b)=(a+b) g(a,b)$ if and only if $-b$ is a root of $f(a,b).$ So when $-b$ is a root of $f(a,b),$ which implies $f(-b,b)=0,$ then $a+b \mid f(a,b).$
This does not force $a+b=0$ because it is only considering that $f(-b,b)=0,$ which implies it is factorable, and does not mean $a$ must be $-b.$
Case 1: If the coefficients of $p$ are nonzero:
n is even: If $n$ is even, then $\sum_{k=0}^n (-b)^{n-k}+b^{n-k}=-2(b^{n-1}+b^{n-3}+ \dots +b).$ Hence, we must have $b^{n-1}+b^{n-3} +\dots+b=0 \implies b(\frac{b^{\frac{n-2}{2}-1}}{b^2-1})=0,$ yielding $b=0$ as integer solutions, thus $p(x)=\sum_{k=0}^n a_{n-k} x^{(n-k)}.$ However, this only holds for $b=0$ and thus no solutions.
n is odd: $n$ being odd yields the same conclusion to $n$ being even.
Case 2: If the coefficients of $p$ can be zero:
Clearly, $\sum_{k=0}^{n} a_{n-k} x^{2n}$ is a solution for integers $a_k.$
There are no other solutions because for all odd powers in $p(x),$ there will be $-2b^j$ terms, where $0\leq j<n,$ requiring $b=0$ as seen in case 1.
Ex: $p(x)=4x^6-2x^2.$
sanyalarnab
15.11.2021 21:55
Andrew_Dou wrote: For $a+b \mid \sum_{k=0}^n i_k(a^{n-k}+b^{n-k}),$ $a=-b$ must result in $\sum_{k=0}^n i_k(a^{n-k}+b^{n-k})=0.$ Here I have a question. How can you take $a=-b$ as question says for all $a,b$ s.t. $a+b \neq 0$
Andrew_Dou
16.11.2021 03:35
My logic was consider $p(a)-p(b)=f(a,b)$ for a polynomial $f(a,b).$ If $a+b \mid p(a)-p(b)=f(a,b),$ then $f(a,b)=(a+b)g(a,b)$ for some polynomial $g(a,b).$
Considering $f(a,b)$ and $g(a,b)$ as "single" variable polynomials in terms of $a,$ then we know that $f(a,b)=(a+b) g(a,b)$ if and only if $-b$ is a root of $f(a,b).$ So when $-b$ is a root of $f(a,b),$ which implies $f(-b,b)=0,$ then $a+b \mid f(a,b).$
I think this does not imply $a+b=0$ because it is only considering that $f(-b,b)=0,$ which implies it is factorable, and does not mean $a$ must be $-b.$
grupyorum
16.11.2021 18:43
Let $\textstyle P(x) = \sum_{0\le k\le d}\alpha_k x^k$. We claim $P$ satisfies the given condition iff $\alpha_k =0$ for all $k\equiv 1\pmod{2}$. Observe that trivially $b\equiv -a\pmod{a+b}$, hence $P(b)\equiv P(-a)\pmod{a+b}$, and thus we are left with the condition \[ a+b \mid \sum_{0\le k\le d}\alpha_k\left(a^k-(-a)^k\right). \]We then see straightforwardly that this holds if $\alpha_k =0$ for all $k$ odd. Now, assume $P$ is such that the set, $I=\{i:i\equiv 1\pmod{2},\alpha_i\ne 0\}$, is non-empty. Then we show $P$ does not satisfy the condition. Note that keeping $a$ fixed and ensuring $b\ne -a$, we have \[ a+b \mid \sum_{i\in I}2\alpha_i a^i. \]Note that the polynomial $\textstyle Q(x) = \sum_{i\in I}2\alpha_i x^i$ is clearly not a zero polynomial (as its coefficients are non-zero), and thus for a sufficiently large integral $a$, $\textstyle \sum_{i\in I}2\alpha_i a^i\ne 0$. Having chosen $a$ in this way, and then setting $b$ such that \[ |a+b| > \left|\sum_{i\in I}2\alpha_i a^i\right|, \]we immediately obtain a contradiction.
Tintarn
17.11.2021 19:47
Andrew_Dou wrote:
My logic was consider $p(a)-p(b)=f(a,b)$ for a polynomial $f(a,b).$ If $a+b \mid p(a)-p(b)=f(a,b),$ then $f(a,b)=(a+b)g(a,b)$ for some polynomial $g(a,b).$
This is true, but not at all obvious. Indeed, I think it is essentially equivalent to the original problem (so it it not that helpful to "reduce" to this fact).
Andrew_Dou
17.11.2021 19:55
I see... I will add that to my post then, thank you for the advice.
Orestis_Lignos
18.11.2021 10:44
Note that $(a-(-b)) \mid p(a)-p(-b)$, hence $(a+b) \mid (p(b)-p(-b))$ for all $a \neq -b$, so by fixing $b$ and taking $a$ to be pretty large, we obtain that $p(b)=p(-b)$ for all $b$. Let $p(x)=\sum_{i=0}^{n} a_i x^{i}$. Then, $p(x)=p(-x) \Rightarrow \sum_{i=0}^{n} a_i(x^{i}-(-x)^{i})=0$ Since the polynomial in the LHS is zero, $a_i$ must be equal to zero for all odd $i$, since otherwise we would have at least one non-zero parenthesis, absurd. Thus, $p$ can be any polynomial such that all coefficients of odd powers are zero. It is straightforward to check that all these polynomials indeed satisfy the given condition.