Find all polynomials $P(x,y)$ with real coefficients such that for all real numbers $x,y$ and $z$: $$P(x,2yz)+P(y,2zx)+P(z,2xy)=P(x+y+z,xy+yz+zx).$$ Proposed by Sina Saleh
Problem
Source: Iranian TST 2019, third exam day 1, problem 1
Tags: algebra, polynomial
15.04.2019 22:49
nice problem
15.04.2019 22:54
Hope I didn't mess up .
Edit: thx Muriatic
19.08.2019 17:59
@above I solved as same as you did! Hope there is another proof not comparing the coefficient By the way, in the proof you dropped the case: the coefficient of $x^n$ is equal to zero. As before, we will assumed $P$ with constant (1,2)-weighted degree. For the case, 1. $n$ is odd, $P(x, 2yz)+P(y, 2xz)+P(z,2xy)$ can be written in forms of $2xyz \cdot Q(x,y,z)$. Since the representation of symmetric function is unique with elementary polynomial, $P(x+y+z, xy+yz+zx) = xyz \cdot Q(x,y,z)$ implies each sides is equal to zero. 2. If $n$ is even and if coefficient of $y^{n/2}$ is zero, again solved by uniqueness of representation. 3. If not zero, assume the coefficient is equal to 1. Let $x^{n-2m}y^m$ be the term with biggest $x$ degree of $P$. Then, $m$ is not zero by assumption and obviously $n\ne 2m$. Then the term $\sum x^{n-m}y^m$ only appears only from the term $(x+y+z)^{n-2m}(xy+yz+zx)^{m}$, so the coefficient does not cancel out in RHS. However, since $P(x, 2yz)+P(y,2xz)+P(z,2xy)=(x^n+y^n+z^n)+xyz \cdot Q(x,y,z)$, there are no such terms in LHS. So this case is also impossible.
12.01.2020 06:46
The answer is all linear combinations of $x$ and $x^2+2y$. The key is to observe that for all monomial terms on the left side, at least two among $x,y,z$ have an equal degree. This is impossible for the right side, with the only exceptions being $P(x,y) =ax+bx^2+cy+dy^2$. Use the condition to check that we must have $d=0$ and $c=2b$.