To solve this problem, we need to find the number of integer pairs (x, y) such that their sum equals a given integer (a) and their greatest common divr (gcd) equals another given integer (b).

Approach

1. Key Insight: If the gcd of (x) and (y) is (b), then (x = b \times m) and (y = b \times n) where (gcd(m, n) = 1) (coprime). Substituting into (x + y = a) gives (b(m + n) = a), so (a) must be divisible by (b) (otherwise, there are no valid pairs).
2. Reduction: If (a) is divisible by (b), let (k = \frac{a}{b}). We need pairs ((m, n)) such that (m + n = k) and (gcd(m, n) = 1). Since (gcd(m, k - m) = gcd(m, k)), we need (gcd(m, k) = 1).
3. Euler's Totient Function: The number of integers (m) (coprime with (k)) is given by Euler's Totient Function (\phi(k)). Each valid (m) gives two ordered pairs ((b \times m, b \times (k - m))) and ((b \times (k - m), b \times m)), so the total number of pairs is (2 \times \phi(k)).

Solution Code

def compute_phi(n):
    result = n
    i = 2
    while i * i <= n:
        if n % i == 0:
            while n % i == 0:
                n = n // i
            result -= result // i
        i += 1
    if n > 1:
        result -= result // n
    return result

a, b = map(int, input().split())
if a % b != 0:
    print(0)
else:
    k = a // b
    phi = compute_phi(k)
    print(2 * phi)

Explanation

1. Compute Euler's Totient Function: The function compute_phi(n) calculates (\phi(n)) using the formula: (\phi(n) = n \times \prod_{p|n} (1 - \frac{1}{p})) where (p) are distinct prime factors of (n).
2. Check Divisibility: If (a) is not divisible by (b), output 0 (no valid pairs).
3. Calculate Result: If (a) is divisible by (b), compute (k = \frac{a}{b}), then the number of valid pairs is (2 \times \phi(k)).

This approach efficiently computes the result using number theory concepts, ensuring optimal performance even for large values of (a) and (b). The example input (a = 12) and (b = 3) gives (k = 4), (\phi(4) = 2), so the result is (2 \times 2 = 4), which matches the expected output.

Answer: The code will output the correct number of pairs as described. For the given example, the output is 4. So the final answer is (\boxed{4}).

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