There are none because nothing would work for $a = 0$, $b = 0$, and $c = 0$.

walnutwaldo20 wrote: There are none because nothing would work for $a = 0$, $b = 0$, and $c = 0$. That was the problem in the contest, because in romanian books, if $0$ divides a number $a$, then $a$ must be $0$. Yet the problem can be solved with the addition of "for all a,b,c with $a^2+b^2$ different than $c^2$".

@above I think there's a difference between this two versions. If "if $0$ divides a number $a$, then $a$ must be $0$" is true, then we can just plug in $(a,b,c)=(3t,4t,5t)$ and get $P(3x)+P(4x)\equiv P(5x)$. But if the problem add an extra condition "for all $a^2+b^2\neq c^2$", we can't conclude that.

ThE-dArK-lOrD wrote: @above I think there's a difference between this two versions. If "if $0$ divides a number $a$, then $a$ must be $0$" is true, then we can just plug in $(a,b,c)=(3t,4t,5t)$ and get $P(3x)+P(4x)\equiv P(5x)$. But if the problem add an extra condition "for all $a^2+b^2\neq c^2$", we can't conclude that. I expressed myself wrong. The original problem was without the extra condition, so your conclusion for the pythagorean triplets is correct. I was saying that the problem still can be solved without the condition, but with different observations.

So, after we get $P(3x)+P(4x)\equiv P(5x)$, the problem is easy (just compare the leading coefficient to get that $\deg (P)=2$ if $P$ is non-constant (otherwise $P\equiv 0$) and then use $a^2\mid P(a)$ for all $a\in \mathbb{Z}$ to get $P(x)\equiv dx^2$ for a constant $d\in \mathbb{Z}_{\neq 0}$). Let's looking for the more interesting problem: Find all polynomials $P\in \mathbb{Z}[x]$ such that $a^2+b^2-c^2 \mid P(a)+P(b)-P(c)$, for all integers $a,b,c$ that $a^2+b^2\neq c^2$.

Let $a=3x+4y,b=3y+4x,c=5x+5y$ $a^2+b^2-c^2=-2xy$ $2xy|P(3x+4y)+P(3y+4x)-P(5x+5y)$ so $y|P(3x)+P(4x)-P(5x)$ for every $x,y$ So $P(3x)+P(4x)=P(5x)$ and so $P(x)=dx^2$ or $P(x)=0$

Nice try!!!!