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Home » You Won’t Believe What Happens in Watch Dogs Legion –陌未为止！ - Anadea

You Won’t Believe What Happens in Watch Dogs Legion –陌未为止！ - Anadea

📅 May 17, 2026 👤 scraface

May 17, 2026

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📌 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.

📌 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 \(n - m = y\).

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