= 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?

February 22, 2026

= 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?

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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.
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:
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:
\(x = m + n\), \(y = n - m\) — both \(x, y\) are integers as long as \(m, n\) are integers. But \(m\) and \(n\) are integers with \(mn = 506\), so they can be any integer solution to \(mn = 506\).

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