You are given an array of integers "array []" and an input "x". you have to find if there are any two numbers in your array such that the sum of their square equals to x2 . in other words, you have to find out if there is any pair of i, j such that array[i]2 + array[j]2 = x 2 . if there are such pairs, you should print all such (i, j). otherwise print "there are no such pairs". example: array []: 6, -4, 6, 3, 9, 0, -1, -9, 2, 6 x: 5 you have 1 pair: (-4) 2 + (3) 2 = 5 2 so, your output would be corresponding array indices: (1, 3) note: the array size is 10 and you have to get the elements of the array from

Given an integer array a, size n, and an input integer value x..  Need to find all pairs of members a[i], a[j] such that (a[i],a[j],x) form a Pythagoras triplet, with x>a[i], a[j].

We need a double loop, the outer loop for a[i], and the inner loop finds a[j].
Here's a pseudocode.

int i, j, x, i2, x2, count=0; 
input x;
x2=x*x;   // economizing on cycles
if x<0 {x=-x};
for i=1 to n-1 {
  if x>=i {      // skip processing the impossible
    i2=i*i;      // economizing on cycles     
    for j=2 to n {
      if x>=j {     // skip processing the impossible
         if i2+j*j==x2 {
            count++;
            print(i,j);
         }
      }
    }
  }
}
if count==0 { print("there are no matched pairs");

It will have a similar complexity as a bubble sort, n^2/2.