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Home » \[ r^2 = 36 \] - Baxtercollege

\[ r^2 = 36 \] - Baxtercollege

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 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 \(mn = 506\).

📌 But since 506 is not a perfect square and has only 8 positive divisors, the number of integer solutions \((m, n)\) with \(mn = 506\) is exactly 16: for each divisor \(m\), \(n = 506/m\), and there are 16 such ordered pairs (including negatives).

📌 But among these, we do **not** require \(m, n\) even — in fact, many are odd. But we already established that \(a = 2m\), \(b = 2n\), so both are even regardless.

📌 Wait — correction: the substitution \(a = 2m\), \(b = 2n\) ensures both are even, but for that, \(m, n\) must be integers such that \(2m \cdot 2n = 2024 \Rightarrow mn = 506\). So every integer solution to \(mn = 506\) gives a valid pair \((a,b) = (2m, 2n)\), and hence a valid \((x,y)\).

📌 There are exactly 16 such integer pairs \((m,n)\) (since 506 has 8 positive and 8 negative divisors), so 16 solutions.

📌 But each such \((m,n)\) gives:

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