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, 2012

Number theory Difficulty 5.7 AIME, harder Prove it Belarus

Find all pairs (n,m)(n, m) of integers nn and mm satisfying the equality n2+m=m2+2n−9n^2 + m = m^2 + 2n - 9.

Solution

Answer: (−10,−11),(−10,9),(−3,−5),(−3,3),(2,−5),(2,3),(9,−11),(9,9)(-10, -11), (-10, 9), (-3, -5), (-3, 3), (2, -5), (2, 3), (9, -11), (9, 9).

Multiplying the given equality by 44, we obtain
4n2+4n=4m2+8m−36⇔4n2+4n+1=4m2+8m+4−39⇔ 4n^2 + 4n = 4m^2 + 8m - 36 \Leftrightarrow 4n^2 + 4n + 1 = 4m^2 + 8m + 4 - 39 \Leftrightarrow
(2n+1)2=(2m+2)2−39⇔(2m+2)2−(2n+1)2=39⇔(2m+2−2n−1)(2m+2+2n+1)=39⇔(2m−2n+1)(2m+2n+3)=39. (2n+1)^2 = (2m+2)^2 - 39 \Leftrightarrow (2m+2)^2 - (2n+1)^2 = 39 \Leftrightarrow (2m+2-2n-1)(2m+2+2n+1) = 39 \Leftrightarrow (2m-2n+1)(2m+2n+3) = 39.
Since 39=1⋅39=39⋅1=3⋅13=13⋅3=(−1)⋅(−39)=(−39)⋅(−1)=(−3)⋅(−13)=(−13)⋅(−3)39 = 1 \cdot 39 = 39 \cdot 1 = 3 \cdot 13 = 13 \cdot 3 = (-1) \cdot (-39) = (-39) \cdot (-1) = (-3) \cdot (-13) = (-13) \cdot (-3) are all possible factorizations of 3939, it suffices to consider the following cases.
1){2m−2n+1=1,2m+2n+3=39which gives {4m+4=40,4n+2=38so {m=9,n=9.2){2m−2n+1=3,2m+2n+3=13⇔{4m+4=16,4n+2=10⇔{m=3,n=2.3){2m−2n+1=39,2m+2n+3=1⇔{4m+4=40,4n+2=−38⇔{m=9,n=−10.4){2m−2n+1=13,2m+2n+3=3⇔{4m+4=16,4n+2=−10⇔{m=3,n=−3.5){2m−2n+1=−1,2m+2n+3=−39⇔{4m+4=−40,4n+2=−38⇔{m=−11,n=−10.6){2m−2n+1=−3,2m+2n+3=−13⇔{4m+4=−16,4n+2=−10⇔{m=−5,n=−3.7){2m−2n+1=−39,2m+2n+3=−1⇔{4m+4=−40,4n+2=38⇔{m=−11,n=9.8){2m−2n+1=−13,2m+2n+3=−3⇔{4m+4=−16,4n+2=10⇔{m=−5,n=2. \begin{array}{lll} 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}

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