Here is my solution, can someone check, please?

Anytime we find $x$ in a monomial, we write it as $\left [ (x-\sqrt{2}y)+\sqrt{2}y \right ]$ and expand (if it appears raised at a power greater than one) via the binomial theorem. This way, we obtain a new polynomial $Q$ in variables $x-\sqrt{2}y$ and $y$, i.e. $P(x,y)=Q(x-\sqrt{2}y,y)$. Now, in $Q$, separate all monomials having $x-\sqrt{2}$ (and force a factor of $x-\sqrt{2}y$) from the ones containing only $y$.

, we have arrived at a representation $P(x,y)=Q(x-\sqrt{2}y,y)=(x-\sqrt{2}y)R(x-\sqrt{2}y,y)+S(y)=(x-\sqrt{2}y)T(x,y)+S(y)$. We have now three main observations: $P(\sqrt{2}x,x)=P(\sqrt{2}x+2x,\sqrt{2}x+x) , \dfrac{\sqrt{2}x}{x}=\sqrt{2}=\dfrac{\sqrt{2}(\sqrt{2}x+x)}{\sqrt{2}x+x}=\dfrac{\sqrt{2}x+2x}{\sqrt{2}x+x}$ and $P(\sqrt{2}x,x)=S(x)$ so $$P(\sqrt{2},1)=P(\sqrt{2}+2,\sqrt{2}+1)=P(3\sqrt{2}+4,2\sqrt{2}+3)=....\ \Rightarrow \ S(1)=S(\sqrt{2}+1)=S(2\sqrt{2}+3)=...$$ So the polynomial $S$ takes the same value for an infinite number of times, which means it is constant. As $P(0,0)=S(0)$, we conclude that $S\equiv P(0,0)$. Thereby, $P(x,y)-P(0,0)=(x-\sqrt{2}y)T(x,y)$. Analogously, $P(x,y)-P(0,0)=(x+\sqrt{2}y)U(x,y)$, so therefore $$P(x,y)-P(0,0)=(x^2-2y^2)V(x,y)\ \ \ (*)$$ As $P(x,y)=P(x+2y,x+y)$, subsituting in $(*)$, we get that $V(x,y)=-V(x+2y,x+y)$, from which we have $V(x,y)=V(3x+4y,2x+3y)$. We apply the exact same steps to $V$ to conclude that $V(x,y)-V(0,0)=(x^2-2y^2)W(x,y)$. But $V(0,0)=-V(0,0)$, i.e $V(0,0)=0$, hence $V(x,y)=(x^2-2y^2)W(x,y)$, which substituted back in $(*)$ yields $$P(x,y)-P(0,0)=(x^2-2y^2)^2W(x,y)\ \ \ (**)$$ As $P(x,y)=P(x+2y,x+y)$, substituting back in $(**)$, we get that $W(x,y)=W(x+2y,x+y)$ with $\mathrm{deg}\ W< \mathrm{deg} \ P$. Due to $(**)$, it is easy to see that if $W(x,y)$ can be written as $Z((x^2-2y^2)^2)$, then $P$ can be written too. An induction on the degree of $P$ yields now the desired conclusion.

A good change of polynomial can make the problem trivial in order to see how one can find the transformation just note that you can view things as matrix and we want a matrix to multiply to get some thing simpler and hence the definition of $R$.I will skip the tedious calculations of finding $R$ and will just use it.First we define $P(x,y)=R(x+\sqrt{2} y,x- \sqrt{2} y)$ then by putting $z=x+\sqrt{2} y,t=x- \sqrt{2} y$ the condition is that $R(z,t)=Q((1+\sqrt{2} )z ,(1-\sqrt{2})t)$ This is easy to solve we just need to compare the coefficients Let $R(x,y)=\sum\limits_{i+j<N} a_{ij}z^it^j$ then we should have $(1+\sqrt{2})^i(1-\sqrt{2})^j=1$ for any non zero monomial.This means that any nonzero coefficient of some $z^it^j$ satisfies $i=j \equiv 0 \pmod 2$ meaning that $R(x,y)=Q((xy)^2)$ for some polynomial $Q$ that is $P(x,y)=Q((x^2-2y^2)^2)$.We are done.

RagvaloD wrote: $P(x,y)$ is polynomial with real coefficients and $P(x+2y,x+y)=P(x,y)$. Prove that exists polynomial $Q(t)$ such that $P(x,y)=Q((x^2-2y^2)^2)$ A. Golovanov Really instructive problem on polynomials. Here's my solution. First, we can WLOG $P$ is homogenous. This is because we can rewrite $P(x,y) = \sum_{i} f_i (x) y^i$ where $f_i$ is a polynomial with coefficient $x$. Now, as the condition says $P(x,y) = P(x+2y, x + y)$ for all $x,y \in \mathbb{R}$, then notice that the change of variable will only affect terms of the same degree, as it changes linearly. Therefore, we can take $f_i(x) y^i$ instead. $\textbf{Claim 01.}$ There exists a polynomial $R \in \mathbb{R}[x,y]$ such that $P(x,y) = R(x^2, y^2)$. $\textit{Proof.}$ To prove this, we need to prove that $P(x,y) = P(-x,y) = P(x,-y)$. By homogeneity, it is enough to prove that $P(1,x) = P(1,-x)$ and $P(x,1) = P(-x,1)$ for infinitely many values of $x$. Now, notice that $P(x + 2y, x + y) = P(x,y) = P(2y - x, x - y)$. Now, we know that when $x = 1, y = 0$, we get that $P(1,1) = P(1,0) = P(-1,1)$. Therefore, we can define a chain of recursion as follow: \begin{align*} \begin{cases} a_1 = 1, b_1 = 1, \ &a_n = 2a_{n - 1} + b_{n - 1}, b_n = a_{n - 1} + 2b_{n - 1} \\ c_1 = -1, d_1 = 1, \ &c_n = 2d_{n - 1} - c_{n - 1}, d_n = c_{n - 1} - d_{n - 1} \end{cases} \end{align*}One can easily check by induction that $a_{2k} = c_{2k}$, $a_{2k + 1} = -c_{2k + 1}$, $b_{2k} = -d_{2k}$, $b_{2k + 1} = d_{2k + 1}$. By the condition of the problem, we know that $P(a_n, b_n) = P(c_n, d_n)$ for every $n \in \mathbb{N}$. Thus, $P(a_{2k}, b_{2k}) = P(a_{2k}, -b_{2k})$ for every $k \in \mathbb{N}$, which proves that $P (1,x) = P(1,-x)$ for all $x = \frac{b_{2k}}{a_{2k}}$, and it is easy to see that there are infinitely many values of this form. Similarly, $P(x,1) = P(-x,1)$ for all $x = \frac{c_{2k + 1}}{d_{2k + 1}}$. Thus, we have proven what we wanted, and therefore there exists $R \in \mathbb{R}[x,y]$ such that $P(x,y) = R(x^2,y^2)$. $\textbf{Claim 02.}$ There exists a polynomial $S \in \mathbb{R}[x]$ such that $R(x,y) = S(x - 2y)$. $\textit{Proof.}$ To prove this, we'll prove that $x - 2y$ is a factor of $R(x,y)$, that is, $R(2,1) = 0$. To prove this, notice that \[ R(2,1) = P(\sqrt{2},1) = P(\sqrt{2} + 2, \sqrt{2} + 1) = R(6 + 4 \sqrt{2}, 3 + 2 \sqrt{2}) = (3 + 2 \sqrt{2})^{\text{deg} \ R} R(2,1) \]This forces $\text{deg} \ R = 0$ or $R(2,1) = 0$. The former forces $P$ being a constant, and therefore we could just take the same polynomial $P = Q$, which is the wanted constant. Therefore, we must have $R(2,1) = 0$. By homogeneity, we have $R(2x,x) = 0$ for all $x \in \mathbb{R}$. Therefore, $x - 2y$ is a factor of $P(x,y)$. Now, write $R(x,y) = (x - 2y) R_1 (x,y)$. Therefore, we have \[ R(x^2, y^2) = P(x,y) = P(x+2y, x+y) = R(x^2 + 4xy + 4y^2, x^2 + 2xy + y^2) \]\[ (x^2 - 2y^2) R_1 (x^2,y^2) = -(x^2 - 2y^2) R_1 (x^2 + 4xy + 4y^2, x^2 + 2xy + y^2) \]Thus, this is reduced to $R_1 (x^2, y^2) = - R_1 (x^2 + 4xy + 4y^2 ,x^2 + 2xy + y^2)$. We claim that $R_1 (2,1) = 0$. To prove this, notice that we still have $R_1( 2,1) = -R_1 ((\sqrt{2} + 2)^2, (\sqrt{2} + 1)^2) = -(3 + 2 \sqrt{2})^{\text{deg} \ R_1} R_1(2,1)$. By the same reason, this either implies $R_1$ being a constant or $R_1(2,1) = 0$. Therefore, we can keep continue this process, and since the degree of $P$ itself is finite, this process must terminate until we get that $R(x,y) = a_n (x - 2y)^n + a_{n - 1} (x - 2y)^{n - 1} + \dots + a_{0}$, and therefore, there exists a polynomial $S$ such that $R(x,y) = S(x - 2y)$. $\textbf{Claim 03.}$ There exists a polynomial $T \in \mathbb{R}[x]$ such that $S(x) = T(x^2)$. $\textit{Proof.}$ By a famous lemma, it suffices to prove that $S(x) = S(-x)$ for all $x \in \mathbb{R}$. To prove this, notice that \[ S(x^2 - 2y^2) = R(x^2, y^2) = P(x,y) = P(x+2y, x + y) = R((x+2y)^2, (x+y)^2) = S((x+2y)^2 - 2(x+y)^2) = S(-(x^2 - 2y^2)) \]Notice that $x^2 - 2y^2$ is surjective in $\mathbb{R}$, as it covers all positive values when $y = 0$ and it covers all negative values when $x = 0$. Therefore, we have $S(a) = S(-a)$ for all $a \in \mathbb{R}$. This implies that there exists $T \in \mathbb{R}[x]$ such that $S(x) = T(x^2)$. To conclude our journey and by all our lemma, then we have \[ P(x,y) = R(x^2,y^2) = S(x^2 - 2y^2) = T((x^2 - 2y^2)^2) \]Take $Q = T$, we are done.