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Home » Black Labs & Pit Bulls Mixed? You Won’t Believe What This Powerful Hybrid Looks Like! - jntua results

Black Labs & Pit Bulls Mixed? You Won’t Believe What This Powerful Hybrid Looks Like! - jntua results

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 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.

📌 But each pair \((m,n)\) with \(mn = 506\) gives a unique \((x,y)\)?

📌 Suppose \((m,n)\) and \((m',n')\) give same \((x,y)\): then \(m + n = m' + n'\) and \(n - m = y\).

📌 From second: \(n = m + y\), substitute: \(m(m + y) = 506\), and \(m(n') = m'(m + y)\). But complicated.

📌 Instead, note that \(x = m + n\), \(y = n - m\), so \(x^2 - y^2 = (m+n)^2 - (n - m)^2 = 4mn = 4 \cdot 506 = 2024\), so every such pair satisfies the equation.

📌 And different \((m,n)\) may give same \((x,y)\)?

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