Sixteenth Macedonian Mathematical Olympiad

Lemma. If $x \in \mathbb{Z}$, then every prime divisor of $x^2+1$ is of the form $4k+1$.

Proof of the lemma. Let $p \mid x^2+1$. So clearly $\gcd(x, p)=1$. Then $x^2+1 \equiv 0 \pmod{p}$, $x^2 \equiv -1 \pmod{p}$. Hence $(x^2)^{\frac{p-1}{2}} \equiv (-1)^{\frac{p-1}{2}} \pmod{p}$ i.e. $x^{\frac{p-1}{2}} \equiv (-1)^{\frac{p-1}{2}} \pmod{p}$. From Fermat's theorem we get $x^{\frac{p-1}{2}} \equiv 1 \pmod{p}$. So $\frac{p-1}{2} = 2k$ i.e. $p=4k+1$. The lemma is proved.

Now the given equation is equivalent to
$$x^{2010} + 1 = 4y^{2009} + 2007y + 4y^{2008} + 2007 \text{ i.e. } x^{2010} + 1 = (4y^{2008} + 2007)(y+1).$$
But $4y^{2008} + 2007 = 4y^{2008} + 2008 - 1$ is of the form $4k-1$ which implies that it must have a prime divisor of the form $4k-1$. Hence $(x^{1005})^2 + 1$ has a prime divisor of the form $4k-1$ which contradicts the above lemma. So the equation doesn't have a solution in the set of integers.