Omid Hatami wrote: Find all polynomials $ f\in\mathbb Z[x]$ such that for each $ a,b,x\in\mathbb N$ \[ a + b + c|f(a) + f(b) + f(c) \] Really nice problem . Here is my solution . From $ f(c)\equiv f( - a - b) (\mod a + b + c)$ then we have \[ a + b + c|f(a) + f(b) + f( - a - b) (\mod a + b + c) \] For each pair of $ (a,b)\in \mathbb{N}$ , if $ f(a) + f(b) + f( - a - b)$ is different 0 then when we choose $ c$ is large enough then the condition $ a + b + c|f(a) + f(b) + f( - a - b)$ is not hold . Therefore we must have $ f(a) + f(b) + f( - a - b) = 0,\forall a,b\in \mathbb{N}$ Take $ a = b$ then we have : \[ f( - 2a) + 2f(a) = 0,\forall a\in \mathbb{N} \] By compare leading coefficient we have $ \tex{deg} P = 1$ Then easy to check that $ f(x) = ax$ with a is a integer .

TTsphn wrote: From $ f(c)\equiv f( - a - b) (\mod a + b + c)$ then we have how did you get this?

here is my solution: we put $ a = b = 0$ and we get $ c |2f(0) + f(c) \equiv 3a_0 \pmod {c} \implies c |3a_0$. so the number $ 3a_0$ has infinitely many divisors which means that $ a_0 = 0 \implies f(0) = 0$. now we put $ c = 0,a = kb,b = b \implies (k + 1)b |f(b) + f(kb)$. now let's suppose that $ f(x) = x.q(x),q(x) \in Z[x]$. let's assume $ deg(q(x)) > 0$ and $ q(x) = a_nx^{n - 1} + ... + a_1$.where $ n > 1,a_n \neq 0$. we must have $ (k + 1)b | b.q(b) + kb.q(kb)$. now we choose $ k$ s.t. $ \gcd(k + 1,b) = 1$. so we must have $ k + 1 | q(kb) - q(b) = b^{n - 1}.a_{n}(k^{n - 1} - 1) + ... + b.a_2(k - 1) \rightarrow k + 1 | \sum_{i}b^{2i - 2}.a_{2i}$. where the summation is done over all positive integer $ i$ s.t. $ 1 \le 2i \le n - 1$. then we have just to choose an integer $ b$ for which the summation isn't zero(obviously it exists) and take a sufficient large $ k$ s.t. $ |k + 1| > |\sum_{i}b^{2i - 2}.a_{2i}|$.now we have a contradiction and we must have $ deg(q(x)) = 0$. we r done! earldbest wrote: TTsphn wrote: From $ f(c)\equiv f( - a - b) (\mod a + b + c)$ then we have how did you get this? as we have proven $ f(0)=0$ so we get $ 0+(a+b)+c | f(0)+f(a+b)+f(c) \implies f(c) \equiv- f(a+b) \pmod {a + b + c}$. but we had $ a+b+c | f(a)+f(b)+f(c)$ which implies $ a+b+c | f(a)+f(b)-f(a+b)$. the rest can be solved in the same way as TTsphn's solution. or by getting $ f(a)+f(b)=f(a+b)$ u can consider cauchy's functions because $ f(x)$ is a polynomial and so continious and get directly the result. but i don't know how we get $ f(c)-f(-a-b) \equiv 0 \pmod {a+b+c}$!?

shoki wrote: earldbest wrote: TTsphn wrote: From $ f(c)\equiv f( - a - b) (\mod a + b + c)$ then we have how did you get this? but i don't know how we get $ f(c) - f( - a - b) \equiv 0 \pmod {a + b + c}$!? I understand now, $ f(c)-f(-a-b)$ is divisible by $ (a+b+c)$ since $ f$ is a polynomial

what a stupid ....!! i didn't pay attention to the sign!

$1.$ $a=b=c=n$ $:$ $3n\mid 3f(n) \: \Rightarrow\: n\mid f(n)$ for all positive integer $n$ so $f(n)=ng(n).$ $2.$ $p>f(1)+\cdots +f(n-1)$ and $b=n-a,\, c=p-n$ : $p$ divides $f(a)+f(n-a)+f(p-n).$ $0<f(a)+f(n-a)<p$ and $f(a)+f(n-a)\equiv -f(p-n)\: \: (mod\: \: p)$ so $f(a)+f(n-a)$ is same for all $1\leq a\leq n-1.$ $3.$ By $2,$ $f(n-1)+f(n+1)=f(n)+f(n) \: \Rightarrow\: f(n)-f(n-1)=f(n+1)-f(n)$ so $f(n)=f(1)+(n-1)c$ for constant $c.$ $4.$ By $1$ and $3,$ $ng(n)=g(1)+(n-1)c.$ $n$ divides $g(1)-c$ for all $n$, so $c=g(1)$ and $g(n)=g(1).$ $5.$ So $g(x)$ is constant and $f(x)=kx$ $(k$ is constant, which is positive integer$).$