peregrinefalcon88 wrote: Now examining (1) tells us that the degree of any term containing x's and y's is at most 2 since we know that the degree of the x-only-terms is 2. We now know that Why? In fact for any polynomial $P(x,y)=x^2+x^2y^2+y^2$ we have also have leading powers to be $2$ for $P(x,0)=x^2$ and for $P(0,y)=y^2$. Someone please try to solve this problem or post official solutions.

Uh... Did you look at equation 1? Your example doesn't work, because $P(x, x)$ is then $x^4+2x^2$, of degree $4$.

The step hal9v4ik highlighted is definitely suspect, though: for example, the polynomial $P(x, y) = x^3y - xy^3$ satisfies all the numbered equations (1) through (6) that peregrinefalcon88 wrote down, but has terms of degree larger than 2. So certainly the derivation of (7) (and so the solution) is wrong. Edit: okay, and here's a much more direct and less wrong way to get to the same place: we have \begin{align*} P(2x, 2y) & = P((x + y) + (x - y), (x + y) - (x - y)) \\ & = 2P(x + y, x - y) \\ & = 4P(x, y), \end{align*} so $P$ is homogeneous of degree 2. The last steps are routine.

Okay,I also got this but never thought homogenity for polynomial with two variables.

I'm sorry JBL ,but your example is wrong,because if $ P(x,y)=x^3y-y^3x=xy(x-y)(x+y) $ , then $ LHS=4xy(x-y)(x+y) $ ,but $ RHS=2xy(x-y)(x+y) $.

Sardor, that was his point. It of course doesn't satisfy the equation, because as he writes later in his same post, that any solution must be homogeneous of degree 2, which it is not. His point is that peregrinefalcon88's proposed solution method was wrong because it never explicitly ruled out $x^3y-xy^3$ as a solution.

Hello,can someone provide correct solution?