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📰 = 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:
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