11. $x$ and $y$ are two positive integers such that $x^{2}-y^{2}$ is positive, a multiple of 2011, and has exactly 2011 positive divisors. How many ordered pairs $(x, y)$ satisfy these conditions? Note: 2011 is a prime number.
(A) 2010
(B) 2011
(C) 1005
(D) 0
(E) There are infinitely many.
11. The answer is (C). First, we note that a number has exactly 2011 positive divisors if and only if it is of the form $p^{2010}$ with $p$ being a prime. Therefore, we need to solve $x^{2}-y^{2}=2011^{2010}$.

By imposing that $x+y=p^{\alpha}$ and that $x-y=p^{\beta}$, with $\alpha+\beta=2010$, we obtain that $x=\frac{p^{\alpha}+p^{\beta}}{2}$, that $y=\frac{p^{\alpha}-p^{\beta}}{2}$, and thus $\alpha>\beta$ since $y$ must be positive. The equation $\alpha+\beta=2010$ with $\alpha>\beta$ has exactly 1005 solutions. Since $x$ and $y$ are uniquely determined by $\alpha$ and $\beta$, it follows that the solutions to the initial equation are also exactly 1005.

(Di Trani)