The problem is from Putnam and Beyond.
If $x + y + z = 0$, prove that $\frac{x^2 + y^2 + z^2}{2}\frac{x^5 + y^5 + z^5}{5} = \frac{x^7 + y^7 + z^7}{7}.$
The solution is as follows.
Consider the polynomial $P(X) = X^3 +pX+q$, whose zeros are $x, y, z$. Then $x^2 + y^2 + z^2 = (x + y + z)^2 − 2(xy + xz + yz) = −2p$. Adding the relations $x^3 = −px − q, y^3 = −py − q,$ and $z^3 = −pz − q$, which hold because x, y,z are zeros of P (X), we obtain $x^3 + y^3 + z^3 = −3q$.
Similarly,
$x^4 + y^4 + z^4 = −p(x^2 + y^2 + z^2) − q(x + y + z) = 2p^2$,
and therefore
$x^5 + y^5 + z^5 = −p(x^3 + y^3 + z^3) − q(x^2 + y^2 + z^2) = 5pq$,
$x^7 + y^7 + z^7 = −p(x^5 + y^5 + z^5) − q(x^4 + y^4 + z^4) = −5p^2q − 2p^2q = −7p^2q.$
The relation from the statement reduces to the obvious $\frac{−2p}{2} \frac{5pq}{5} = \frac{−7p^2q}{7}.$
My problem with the solution is that it assumes the existence of the polynomial $P(X) = X^3 + pX + q$, although by Viete's relations, $q = x+y+z = 0$. This means that writing things in terms of $q$ doesn't really make sense. I'm surely missing something - what?
but with $z=-x-y$ we get $\frac{x^2+y^2+(-x-y)^2}{2}=x^2+y^2+xy$ $\frac{x^5+y^5+(-x-y)^5}{5}=-xy(y+x)(x^2+xy+y^2)$ and for the right hand side $\frac{x^7+y^7+(-x-y)^7}{7}=-xy(x+y)(x^2+xy+y^2)^2$ a nice problem!