51st Ukrainian National Mathematical Olympiad, 3rd Round

Since $a-b=2011$ and $2011$ is prime, then $(a,b)$ is $1$ or $2011$.

1 Case. If $(a,b)=1$ then $a=c^2$, $b=d^2$, and we have $a-b=c^2-d^2=(c-d)(c+d)=2011=1\cdot 2011$ or $$\{\begin{array}{l}c-d=1,\\ c+d=2011.\end{array}$$ Solving this system we get the answer.

2 Case. If $(a,b)=2011$, then $a=2011c$, $b=2011d$, where $(c,d)=1$. We have $ab=2011^{2}cd$, hence $c=e^2,d=f^2$. We have $a-b=2011(c-d)=2011$, or $c-d=(e-f)(e+f)=1$, which is impossible.