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Belorusija 2012

Belarus 2012 number theory

Problem

Find all pairs of integers and satisfying the equality .
Solution
Answer: .

Multiplying the given equality by , we obtain Since are all possible factorizations of , it suffices to consider the following cases. \begin{array}{l@{\quad}l@{\qquad}l} 1) \left\{ \begin{array}{l} 2m - 2n + 1 = 1, \\ 2m + 2n + 3 = 39 \end{array} \right. & \text{which gives } \left\{ \begin{array}{l} 4m + 4 = 40, \\ 4n + 2 = 38 \end{array} \right. & \text{so } \left\{ \begin{array}{l} m = 9, \\ n = 9. \end{array} \right. \\ 2) \left\{ \begin{array}{l} 2m - 2n + 1 = 3, \\ 2m + 2n + 3 = 13 \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{l} 4m + 4 = 16, \\ 4n + 2 = 10 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m = 3, \\ n = 2. \end{array} \right. \\ 3) \left\{ \begin{array}{l} 2m - 2n + 1 = 39, \\ 2m + 2n + 3 = 1 \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{l} 4m + 4 = 40, \\ 4n + 2 = -38 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m = 9, \\ n = -10. \end{array} \right. \\ 4) \left\{ \begin{array}{l} 2m - 2n + 1 = 13, \\ 2m + 2n + 3 = 3 \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{l} 4m + 4 = 16, \\ 4n + 2 = -10 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m = 3, \\ n = -3. \end{array} \right. \\ 5) \left\{ \begin{array}{l} 2m - 2n + 1 = -1, \\ 2m + 2n + 3 = -39 \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{l} 4m + 4 = -40, \\ 4n + 2 = -38 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m = -11, \\ n = -10. \end{array} \right. \\ 6) \left\{ \begin{array}{l} 2m - 2n + 1 = -3, \\ 2m + 2n + 3 = -13 \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{l} 4m + 4 = -16, \\ 4n + 2 = -10 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m = -5, \\ n = -3. \end{array} \right. \\ 7) \left\{ \begin{array}{l} 2m - 2n + 1 = -39, \\ 2m + 2n + 3 = -1 \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{l} 4m + 4 = -40, \\ 4n + 2 = 38 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m = -11, \\ n = 9. \end{array} \right. \\ 8) \left\{ \begin{array}{l} 2m - 2n + 1 = -13, \\ 2m + 2n + 3 = -3 \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{l} 4m + 4 = -16, \\ 4n + 2 = 10 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m = -5, \\ n = 2. \end{array} \right. \end{array}
Final answer
[(-10, -11), (-10, 9), (-3, -5), (-3, 3), (2, -5), (2, 3), (9, -11), (9, 9)]

Techniques

Techniques: modulo, size analysis, order analysis, inequalitiesFactorization techniques