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\( 36 + 3 = 39 \) - go-checkin.com

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 But since 506 ≠ 0, no solution has \(y = 0\) (would require \(m = n\), \(m^2 = 506\), not square), so all solutions come in pairs \((x,y)\) and \((x,-y)\), except if \(y = 0\), which doesn’t occur.

📌 But each \((m,n)\) is distinct, so 16 solutions.

📌 But let’s test a small one: \(m = 1\), \(n = 506\):

📌 \(a = 2\), \(b = 1012\), then \(x = (2+1012)/2 = 507\), \(y = (1012 - 2)/2 = 505\).

📌 Check: \(507^2 - 505^2 = (507-505)(507+505) = 2 \cdot 1012 = 2024\), correct.

📌 Another: \(m = -1\), \(n = -506\): \(x = -507/2\), not integer — wait! But \(m = -1\), \(n = -506\), \(mn = 506\), but \(a = 2m = -2\), \(b = 2n = -1012\), then \(x = (-2 -1012)/2 = -507\), \(y = (-1012 + 2)/2 = -505\), so \((-507, -505)\), also valid.

📌 But when we list all divisor pairs \((m,n)\) with \(mn = 506\), we include both positive and negative, and all 16 are valid and give distinct \((x,y)\)?

📌 No: for each factorization, we get a point. But since \(m\) and \(n\) are determined, and \(x = m+n\), \(y = n-m\), and since \(m\) runs through all 16 divisors (positive and negative), we get 16 values of \(x\), 16 of \(y\), but not necessarily distinct points.

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