Sum of square roots of integers Let $x, y$ be integers and consider the equation $$\sqrt{x}+\sqrt{y}= 8 \sqrt{31}.$$ 

It is claimed that this implies $\sqrt{x}= a\sqrt{31}$ and $\sqrt{y}=b \sqrt{31}$ for $a,b$ integers.

While this is intuitively obvious, can the claim be proved by elementary means?
Square both sides and rearrange to get $2\sqrt{xy} = 64 \times 31 - x - y$.

Let the greatest common factor between x and y be a. Since xy is a square, x/a and y/a must both be squares.

Let $x = ab^2$, $y = ac^2$.

This gives us:

$2abc = 64 \times 31 - a(b^2 + c^2)$.

Looking at things modulo a, $64 \times 31$ must be a multiple of a.

$b^2 + c^2 + 2bc = 64 \times 31/a$.

So $(b+c)^2 = 64 \times 31/a$.

Since 64 * 31/a must be a square, a must be a multiple of 31.

Moreover, since 64 * 31/a is a square, it means that 31/a must also be a square (as 64 is a square). So the factors of x and y other than 31 can be taken out of the sqrt sign.
Note that since $x, y$ both are integers, so either $\sqrt x, \sqrt y$ are both integers or at least one of them are irrational. 

But if both of them are integers, then their sum should be integers whereas here RHS is irrational. Hence at least one of them must be irrational. 

If $\sqrt x$ is irrtional and $\sqrt y$ is integer then their sum will be of the form $m+\sqrt n$ where $m\in \mathbb Z, \sqrt n\in \mathbb R-\mathbb Q$. Which will be contradiction again cause RHS is simply of the form $0+8\sqrt{31}.$

Thus it follows that both of them must be irrational of the form $\sqrt{n}$. If $\sqrt x=a\sqrt m, \sqrt y=b\sqrt n$ then we must have $a+b=8$ and $\sqrt m+\sqrt n=\sqrt{31}$. Which immediately shows $m=n=31$.

thus $x=a\sqrt{31}, y=b\sqrt{31}$ with $a+b=8$