Any Ideas?

Well, $P(X) = X$ is a solution.

amparvardi wrote: Find all polynomials $P(x)$ with real coefficients such that for all pairwise distinct positive integers $x, y, z, t$ with $x^2 + y^2 + z^2 = 2t^2$ and $\gcd(x, y, z, t ) = 1,$ the following equality holds \[2P^2(t ) + 2P(xy + yz + zx) = P^2(x + y + z) .\] Note. $P^2(k)=\left( P(k) \right)^2.$

Ukraine Math Olympiad 2009 Grade 11 P4 wrote: Find all polynomials $P(x)$ with real coefficients such that for all pairwise distinct positive integers $x, y, z, t$ with $x^2 + y^2 + z^2 = 2t^2$ and $\gcd(x, y, z, t ) = 1,$ the following equality holds \[2P^2(t ) + 2P(xy + yz + zx) = P^2(x + y + z) .\] Note. $P^2(k)=\left( P(k) \right)^2.$ As @above we obtain \[2P^2(x) + 2P(x^2) = P^2(2x), \,\, (*)\]for all $x$. Suppose that $n= \deg P \geq 2$. (refer @above for these trivial cases). Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$. To start, let us compare the coefficients of $x^{2n}$ at both sides of $(*)$. We obtain $$(2^n \cdot a_n)^2=2a_n+2a_n^2 \Rightarrow a_n=\frac{1}{2^{2n-1}-1}$$ Now, we will prove that $a_{n-i}=0$, for all $1\leq i \leq n$ in order to conclude. To prove the base case $i=1$, let us compare the coefficients of $x^{2n-1}$ at both sides of $(*)$. We easily obtain $$2a_na_{n-1}=2 \cdot 2^na_n \cdot 2^{n-1}a_{n-1},$$and since $a_n \neq 0$, we obtain $a_{n-1}=0$, as desired. Now we employ strong induction. Suppose that $a_{n-i}=0$ for all $1 \leq i \leq k$. We will prove that $a_{n-k-1}=0$. To that end, we compare the coefficients of $x^{2n-k-1}$ at both sides of $(*)$. We need to examine though the parity of $k$. Case 1: $k$ is even. Then, since in the expansion of $P(x^2)$ only even powers appear, we infer that $2n-k-1$ does not. In addition, in the expansions of $P^2(x)$ and $P^2(2x)$, the power $x^{2n-k-1}$ will emerge from two $p,q \leq n$, which are such that $p+q=2n-k-1$. If though at least one of $p,q$ is $\geq n-k$ and $ \leq n-1$, then $a_pa_q=0$, therefore the coefficient of $x^{p+q}=x^{2n-k-1}$ will be zero. Consequently, we may consider only $p,q$ that are either $n$ or $\leq n-k-1$. If both are $\leq n-k-1$ we get $$2n-k-1=p+q \leq 2n-2k-2,$$a contradiction. Hence one of the two is equal to $n$ and the other to $n-k-1$. That is, the coefficient of $x^{2n-k-1}$ in $2P^2(x)$ will be $4a_na_{n-k-1}$. Similarly we deal with $P^2{2x}$, to obtain the coefficient $2 \cdot 2^n a_n \cdot 2^{n-k-1}a_{n-k-1}=2^{2n-k}a_na_{n-k-1}$. Therefore, $$2^{2n-k}a_na_{n-k-1}=4a_na_{n-k-1},$$and since $2n-k >n \geq 2$ and $a_n \neq 0$, we must have $a_{n-k-1}=0$, as desired. Case 2: $k$ is odd. This case is almost identical with the previous one, although here we may add the coefficient $a_{\frac{2n-k-1}{2}}$ to the LHS, hence we obtain $$(2^{2n-k}-4)a_na_{n-k-1}=a_{\frac{2n-k-1}{2}}$$Since $$\frac{2n-k-1}{2} =n-k+\frac{k-1}{2} >n-k$$we have that $a_{\frac{2n-k-1}{2}}=0$ by the induction hypothesis, hence we instantly obtain that $a_{n-k-1}=0$, as desired. To the problem, we conclude that $P(x)=\frac{x^n}{2^{2n-1}-1}$, which evidently satisfies.

I still don't understand the case of k odd and $a_{\dfrac{2n-k-1}{2}}=0$