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
ali3985
15.04.2019 22:49
nice problem
Pathological
15.04.2019 22:54
Hope I didn't mess up .
All linear combinations of $x$ and $x^2 + 2y$.
For a term $x^ay^b$ in $P(x, y)$, define its $\textbf{degree}$ to be the quantity $a + 2b$. Then, notice that if we partition the terms in $P$ based on their degree, all the terms of degree $k$ must satisfy the equation above, for every $k \ge 0$. It's easy to check for $0 \le n \le 4$ that the only nonzero combinations of degree $n$ terms which work are $cx$ for $n=1$ and $x^2 + 2y$ for $n =2.$ It therefore suffices to show that there is no nonzero combination of degree $n$ terms which satisfies the above, for $n \ge 5.$
Assume the opposite, for contradiction. Suppose that $P(x, y) = x^n + cx^{n-2}y + dx^{n-4}y^2 + \cdots$ satisfied the above. By looking at the $x^{n-1}y$ coefficient on both sides, we obtain that $c = -n.$ By looking at the $x^{n-2}y^2$ coefficient on both sides, we get that $d = \frac{n(n-3)}{2}.$ Finally, looking at the $x^{n-2}yz$ term on both sides, we obtain that $-2c = n(n-1) + c \cdot (2n-3) + 2d,$ which rewrites as:
$$2n = n(n-1) - n(2n-3) + n(n-3) \Leftrightarrow 2n = -n,$$
clearly absurd.
$\square$
Edit: thx Muriatic
bumjoooon
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.
Idio-logy
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 $c=2b$.