, 2010

Let $n$ be the degree of the polynomial $P$, i.e.,
$$P(x, y) = a_n x^n + a_{n-1} x^{n-1} y + \dots + a_1 x y^{n-1} + a_0 y^n,$$
where $n > 0$. Note that $n$ must be even because otherwise the condition $P(\sin t, \cos t) = 1$ for $t = 0$ would imply $a_0 = 1$ while the same condition for $t = \pi$ would imply $a_0 = -1$.
Since $P$ has no constant term, $P(0, 0) = 0$. Now assume that $x \neq 0$ or $y \neq 0$ and take $c = \sqrt{x^2 + y^2}$. Since
$$\left(\frac{x}{\sqrt{x^2 + y^2}}\right)^2 + \left(\frac{y}{\sqrt{x^2 + y^2}}\right)^2 = 1,$$
there exists some real number $t$ such that $\sin t = x/\sqrt{x^2 + y^2}$ and $\cos t = y/\sqrt{x^2 + y^2}$ and therefore $P(\sin t, \cos t) = 1$. By homogeneity, $P(x, y) = c^n \cdot P(x/c, y/c)$, hence
$$P(x, y) = \left(\sqrt{x^2 + y^2}\right)^n \cdot P\left(\frac{x}{\sqrt{x^2 + y^2}}, \frac{y}{\sqrt{x^2 + y^2}}\right) = \left(\sqrt{x^2 + y^2}\right)^n.$$

for all $x, y$. The case $n = 2k$ implies $P(x, y) = (x^2 + y^2)^k$ which satisfies also the condition $P(0, 0) = 0$.

Like in Solution 1, express the polynomial as a sum of $n + 1$ monomials with coefficients $a_0, \dots, a_n$ and show that $n = 2k$.
We prove the claim of the problem by induction on $k$. In case $k = 0$ (omitting the extra assumption that $P$ is non-constant) the claim holds obviously. Assume now that $k > 0$ and the claim holds for $k - 1$. Substituting $t = 0$ and $t = \frac{\pi}{2}$ into $P(\sin t, \cos t) = 1$ gives $a_0 = 1$ and $a_n = 1$, respectively. Hence the polynomial $P(x, y) - (x^2 + y^2)^k$ does not have terms with $x^n$ and $y^n$. Let $Q(x, y)$ be such that $P(x, y) - (x^2 + y^2)^k \equiv x y \cdot Q(x, y)$. Then $\sin t \cos t \cdot Q(\sin t, \cos t) = 0$ for every real number $t$, hence $Q(\sin t, \cos t) = 0$ for every $t$ such that $\sin 2t \neq 0$. By continuity of $Q(\sin t, \cos t)$ as a function of $t$, it follows that $Q(\sin t, \cos t) \equiv 0$. Now define $R(x, y) = Q(x, y) + (x^2 + y^2)^{k-1}$. As both $Q$ and $R$ are homogeneous polynomials of degree $2(k-1)$, the assumptions of the problem hold for polynomial $R$. By the induction hypothesis, $R(x, y) \equiv (x^2 + y^2)^{k-1}$. Hence $Q(x, y) \equiv 0$ and $P(x, y) \equiv (x^2 + y^2)^k$.