Maths Olympiad Prep

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Number theory Difficulty 5.1 AIME, harder Prove it China

Find the integer solutions of the function x22xy+126y2=2009x^2 - 2xy + 126y^2 = 2009.

Solution

Suppose that the integers x,yx, y satisfy
x22xy+126y22009=0. x^2 - 2xy + 126y^2 - 2009 = 0.
Looking at this as a quadratic function of xx,
Δ=4y24×(126y22009)=500(42y2)+36 \Delta = 4y^2 - 4 \times (126y^2 - 2009) = 500(4^2 - y^2) + 36
should be a square number.
If y2>42y^2 > 4^2, then Δ<0\Delta < 0. So y2<42y^2 < 4^2, when y2{0,12,22,32}y^2 \in \{0, 1^2, 2^2, 3^2\}, Δ{8036,7536,6036,3536}\Delta \in \{8036, 7536, 6036, 3536\} is not a square number.
For y2=42y^2 = 4^2, Δ=500(42y2)+36=62\Delta = 500(4^2 - y^2) + 36 = 6^2. According to y=4y = 4, one can get x=1x = 1 or 77; according to y=4y = -4, one can get x=1x = -1 or 7-7.
All the integer solutions are (x,y)=(1,4),(7,4),(1,4),(7,4)(x, y) = (1, 4), (7, 4), (-1, -4), (-7, -4).

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