But from the substitution, $m n = 506$, and 506 is even, but not divisible by higher powers — all its factor pairs consist of one even and one odd? No:

February 22, 2026

$But from the substitution, $m n = 506$, and 506 is even, but not divisible by higher powers — all its factor pairs consist of one even and one odd? No:$

But note: each solution $(x, y)$ is counted once for each factor pair. Since $ (a,b) $ and $ (b,a) $ yield potentially different $(x,y)$, but in our setup $a = x - y$, $b = x + y$, so order matters in assignment. However, fixing $a, b$ as both even and same parity already covers all valid factorizations.
Wait: we set $ab = 2024$, and we require both $a$ and $b$ even.
Total factor pairs (positive and negative) where both are even: Since 2024 is divisible by 4, and all factorizations into two evens are valid.
= 2 × 253, and 253 is odd. So in the factor pairs of 506, one factor (the first in a pair) is even, the other odd — unless both are even. But 506 is not divisible by 4?
Wait: 506 ÷ 2 = 253, so 506 ≡ 2 mod 4. So it has exactly one factor of 2. Therefore, in any factor pair $(m, n)$, one is even, one is odd — so $m$ and $n$ have alternating parity. But we need both $m$ and $n$ even for $a = 2m$, $b = 2n$ both even, but $mn = 506$, which has only one factor of 2, so in any factorization, one is odd, one is even.
Therefore, $a = 2m$ is even, $b = 2n$ is even, but $m$ and $n$ have opposite parity — so $a + b = 2(m + n)$, $b - a = 2(n - m)$ — wait, no:

But from the substitution, \(m n = 506\), and 506 is even, but not divisible by higher powers — all its factor pairs consist of one even and one odd? No:

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