I got 100/100 point.

My Solution: we need calculate how many pair (a, b) with: a^2 - b^2 <= n

=> (a - b)(a + b) <= n

Let: a = b + k (with b > 0; k> 0; and k is even, means: k % 2 == 0)

=> k(2b + k) <= n

=> b <= (n/k - k)/2

The result is sum of all b.

We need 1 FOR-loop: from 1 to sqrt(n):

long caculate(n, k) {
	return (long) ((double) n / k - k) / 2;
}
// Main:
sum = 0;
for (int k = 2; k < sqrt(n); k += 2) {
   b = calculate(n, k);
   if (b > 0) 
   	sum += b;   
}
return sum;

n,k,b.. is Long, not Integer!

GOOD LUCK!!