Remark that $P(x,y) = (\sqrt{x^2 + y^2})^n$ where $n = \deg P$. Furthermore, as $P(\pm 1,0) = 1$ we know $n$ is even so let $n = 2k$. Now as there does not exist any non-zero two variable real polynomial $Q(x,y)$ such that $Q(x,y) = 0$ for all real $x,y$, we know that $P(x,y) - (x^2+y^2)^k = 0 \implies P(x,y) = (x^2+y^2)^k$ and thus we are done.

Pardon me as this may be my fault, but I can't understand a single line of that solution. What is your justification for the first line? Moreover, doesn't $n = 2k$ just because $P$ is a polynomial?

Yeah I guess I should have elaborated. Write $P(x,y) = \sum_{i=0}^n a_ix^iy^{n-i}$. $P(1,0) = 1 \implies a_0 = 1$. Then $P(-1,0) = 1 \implies a_0(-1)^n = 1 \implies n$ is even. And for the first line, scale $(x,y)$ to $(x_0, y_0)$ where $x_0^2 + y_0^2 = 1$. Then clearly $P(x_0,y_0) = 1$ by the problem. Then as $P$ is homogenous $P(x,y) = \left ( \frac{x}{x_0} \right )^n P(x_0,y_0) = \left ( \frac{x}{x_0} \right )^n = (\sqrt{x^2+y^2})^n$.

chaotic_iak wrote: Suppose $P(x,y)$ is a homogenous non-constant polynomial with real coefficients such that $P(\sin t, \cos t) = 1$ for all real $t$. Prove that $P(x,y) = (x^2+y^2)^k$ for some positive integer $k$. (A polynomial $A(x,y)$ with real coefficients and having a degree of $n$ is homogenous if it is the sum of $a_ix^iy^{n-i}$ for some real number $a_i$, for all integer $0 \le i \le n$.) Let $n=\deg P$, $P(x,y)=a_nx^n+a_{n-1}x^{n-1}y+\cdots +a_1xy^{n-1}+a_0y^n$. From $P(0,1)=1$, $a_0=1$. From $P(0,-1)=1$, $a_0=(-1)^n$. Therefore $n\equiv 0\pmod 2$, let $n=2k(k\in\mathbb Z^+)$. When $x^2+y^2>0$, let $c=\sqrt {x^2+y^2}$. Obviously $\left(\frac xc\right)^2+\left(\frac yc\right)^2=1$. So $\exists \varphi\in\mathbb R$, $\sin\varphi =\frac xc$, and $\cos \varphi =\frac yc$. Then $P(x,y)=c^nP\left(\frac xc,\frac yc\right)=(\sqrt {x^2+y^2})^nP(\sin\varphi,\cos\varphi)=(x^2+y^2)^{\frac{n}{2}}=(x^2+y^2)^k$.