# Count of (x,y) pairs that satisfy the equation x^2+y^2 = n^2

Given (n) , what is the Count of (x,y) pairs that satisfy the equation x^2+y^2 = n^2. Is there any way I can re-write this code just without using nested loop?

int counter=0;
int x=0;

for (int i =(int) n-1; i>=0 ; i--){

if(i == x)
break;

for (int j = 1; j< (int)n ; j++){

if (Math.pow(i,2)+Math.pow(j,2) == Math.pow(n,2)){
counter++;
x = j;
}
}
}
System.out.print(counter);
• This question looks like off-topic. Any way, it fits Stack Overflow probably more. – Apass.Jack Mar 20 at 3:43

Your approach is to try every possible $$x$$ and $$y$$ and see if $$x^2+y^2=n^2$$. However, $$n$$ is fixed and, for any $$x$$, either $$n^2-x^2$$ is a perfect square or it isn't. You can calculate what $$y$$ is, instead of looping and trying every possible value.
Actual code is off-topic but it's confusing that you're trying to solve a problem about $$x$$ and $$y$$, but you've represented those quantities in variables called i and j. And even worse that you have a variable called x that represents something else entirely. (In fact, I'm not sure it's even correct. You seem to be using x to avoid reporting, e.g., $$x=3$$, $$y=4$$ and $$x=4$$, $$y=3$$ as separate solutions for $$n=5$$. But they are separate solutions and the question, at least as you've summarized it, doesn't say that they should be counted as the same.)