Probability that $5$ divides $a^2+b^2$ is $\frac9{25}$ 
Two positive integers $a$ and $b$ are randomly selected with replacement, then prove that the probability of $(a^2 +b^2)/5$ being positive integer is $9/25$.

I found out the pattern of the last digits of $a^2$ , $b^2$  but then the sample space isn't finite so I dropped idea of thinking of finding the a's and b's such that their square's last digit is either 0 or 5. Please help.

$1^2 mod  5 \equiv 1$
$2^2 mod 5 \equiv 4$
$3^2 mod 5 \equiv 4$
$4^2 mod 5 \equiv 1$
$5^2 mod 5 \equiv 0$ 
We want $a^2+b^2 mod 5 \equiv 0$, 
That happens for (1,3)  (1,2)  (4,3)  (4,2)  (3,1)  (2,1)  (3,4)  (2,4)  and (5,5)
where (x,y) represents (a mod 5, b mod 5)
5*5 ways to express (x,y)
So the probability is 9/25
