Pairs of integeres for which the arithmetic mean exceeds the geometric mean exactly by $2$

Question

Suppose $0<x<y<2015$ are integers. How many pairs of $x$ and $y$ are there for which the arithmetic mean exceeds the geometric mean exactly by $2$?

Progress

Obtained the equation $(x+y)=2[(xy)^{1/2}+2]$; how to solve it?

Answer 1

$$\frac{x+y}{2} \ge \sqrt{xy} + 2$$ $$x+y-2\sqrt{xy} \ge 4$$ $$(\sqrt{x}-\sqrt{y})^2 \ge 4$$ $\sqrt{x}-\sqrt{y} \ge 2 $ or $\sqrt{x}-\sqrt{y} \le -2$

Answer 2

Let me now write the rest of the answer.For x < y,we have y=(x^1/2+2)^2

Since y < 2015,we see x can take any value from 1 to 42×42 i.e. all perfect squares between 1 and 2015.

So the desired pairs are(1,9), (4,16),......,(42×42,1936).Hence the total number of such pairs is 42.