Program to find number of solutions for given equations with four parameters in Python

Suppose we have four numbers a, b, c and d and We have to find number of pairs (x, y) can be found such that that follows the equation: x^2 + y^2 = (x*a) + (y*b) where x in range [1, c] and y in range [1, d]

So, if the input is like a = 2 b = 3 c = 2 d = 4, then the output will be 1 because one pair is (1, 1).

To solve this, we will follow these steps −

• ans := 0
• for x in range 1 to c, do
  • l := x*(x-a)
  • det2 := b*b - 4*l
  • if det2 is same as 0 and b is even and 1 <= floor of (b/2) <= d, then
    • ans := ans + 1
    • go for next iteration
  • if det2 > 0, then
    • det := integer part of the square root of det2
    • if det^2 is same as det2 and (b+det) is even, then
      • if 1 <= floor of (b+det)/2 <= d, then
        • ans := ans + 1
      • if 1 <= floor of (b-det)/2 <= d, then
        • ans := ans + 1
• return ans

Example

Let us see the following implementation to get better understanding −

def solve(a, b, c, d):
   ans = 0
   for x in range(1,c+1):
      l = x*(x-a)

      det2 = b*b - 4*l
      if det2 == 0 and b%2 == 0 and 1 <= b//2 <= d:
         ans += 1
         continue
      if det2 > 0:
         det = int(round(det2**0.5))
         if det*det == det2 and (b+det) % 2 == 0:
            if 1 <= (b+det)//2 <= d:
               ans += 1
            if 1 <= (b-det)//2 <= d:
               ans += 1
   return ans

a = 2
b = 3
c = 2
d = 4
print(solve(a, b, c, d))

Input

2, 3, 2, 4

Output

1