Some details might be missing from this (feel free to comment!), but I think it's OK. $A(a,b,c)\stackrel{\Delta}{\iff}(\forall a,b,c\in\mathbb{R})P(ab,c^2+1)+P(bc,a^2+1)+P(ca,b^2+1)=0$ $A(0,0,0)\implies P(0,1)=0$ $A(0,0,c)\implies P(0,y)=0$ for $y>1$. Therefore $x\mid P(x,y)$ $A(a,b,0)\implies P(x,1)=0$ for $x\in\mathbb{R}$. Therefore $y-1\mid P(x,y)$ So put $P(x,y)=x(y-1)Q(x,y)$. Then $A(a,b,c)\implies cQ(ab,c^2+1)+aQ(bc,a^2+1)+bQ(ca,b^2+1)=0$ Denote the last assertion by $B(a,b,c)$. Then $B(0,0,c)\implies Q(0,y)=0$ for $y>1$. Therefore $x\mid Q(x,y)$. So put $Q(x,y)=xR(x,y)$ to get $R(ab,c^2+1)+R(bc,a^2+1)+R(ca,b^2+1)=0$ We're back to the original equation. Hence we conclude that any solution to it is divisible by $x^{2n}(y-1)^n$ for an arbitrary natural $n$, which means $P(x,y)\equiv 0$.

Farenhajt wrote: $A(a,b,c)\implies cQ(ab,c^2+1)+aQ(bc,a^2+1)+bQ(ca,b^2+1)=0$ Denote the last assertion by $B(a,b,c)$. Then $B(0,0,c)\implies Q(0,y)=0$ for $y>1$. Therefore $x\mid Q(x,y)$. So put $Q(x,y)=xR(x,y)$ to get I don't think your solution is true. At the line above, you just arrive at this conclusion when $abc \neq 0$ So how could you make the assertion $B(0,0,c)$

Actually he can make get that assertion, he didn't explain it because it is quite clear it is true. It is so because if (one variable) polinomial has infinitely many roots than it is zero all the time so here you just fix variables first to non zero values and then one to 0 and one to non zero value and by using that fact for one variable polinomials you get that you can make assertion even when a b or c is zero. (Same thing he used when he got that polynomial is zero for all x larger than 1 so it is as well for every x). Hope this helped you a bit.

$A(0,0,c)\implies P(0,y)=0$ for $y>1$. Therefore $x\mid P(x,y)$ Why? Its also true?

geomath wrote: $A(0,0,c)\implies P(0,y)=0$ for $y>1$. Therefore $x\mid P(x,y)$ Why? Its also true? the first assertion is just true and occours when putting in c into the function, and the second assertion: $x|P(x,y)$ is true because the polynomial P(0,y) is 0 for all y, so then the first variable of the polynomial (x) has to be 0 for P(x,y) to be 0, thus all summands of P(x,y) contain at least one degree of x, so therefore $x|P(x,y)$

I was just wondering if I am allowed to make this claim? Let $q(x,y) = p(x,y+1)$. Then, $q(ab,c^2)+q(bc,a^2)+q(ca,b^2) = 0$. By letting $a=b=c$, $q(a^2, a^2)=0$ i.e. $q(x,x)=0$. Thus $(x-y) | q(x,y)$? Thanks!

Does anyone know if I am allowed to do this? Are there some general rules/identities that I have/can to follow when dealing with two variable polynomials (compared to one variable polynomials)? Thank you!

You can.

Seriously? Let $P(x,y)=\sum a_{ij} x^i y^j$ If $P\not=0$, we can pick one pair of $(i,j)$ that $x^i y^j$ is a term with the largest degree in $P$ (i.e. $i+j$ is maximum) Then in the left side of the equation of the problem, the coefficient of $x^i y^i z^{2j}$ becomes $a_{ij}$, which implies $a_{ij}=0$, but it isn't. So it's a contradiction and $P$ must be 0.