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Home » \frac3628800288 = 12600 - Dyverse

\frac3628800288 = 12600 - Dyverse

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 But only pairs where \(a\) and \(b\) have the same sign and same parity contribute valid solutions.

📌 For \(x\) and \(y\) to be integers, \(a\) and \(b\) must both be even (since if both were odd, \(ab\) would be odd; if one odd and one even, \(a + b\) is odd). So both \(a\) and \(b\) must be even.

📌 Let \(a = 2m\), \(b = 2n\), then \(4mn = 2024 \Rightarrow mn = 506\).

📌 Now \(506 = 2 \cdot 11 \cdot 23\), so it has \((1+1)(1+1)(1+1) = 8\) positive divisors, giving 8 positive and 8 negative pairs \((m, n)\).

📌 Thus, 16 valid pairs \((m, n)\), each giving a unique \((x, y)\):

📌 x = \frac{2m + 2n}{2} = m + n, \quad y = \frac{2n - 2m}{2} = n - m

📌 Each such pair yields an integer solution. Therefore, there are 16 lattice points.

📌 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.

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