Maths Olympiad Prep

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Algebra Difficulty 5.7 AIME, harder Find the answer

10 Find all integers x,yx, y such that x2+xy+y2=1x^{2} + xy + y^{2} = 1.

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Solution

10. Multiply both sides by 4, then complete the square, to get
(2x+y)2+3y2=4(2 x+y)^{2}+3 y^{2}=4

Thus, 4−3y24-3 y^{2} is a perfect square, which requires y2=0y^{2}=0 or 1, corresponding to (2x+y)2=4,1(2 x+y)^{2}=4,1. Solving these respectively yields
(x,y)=(±1,0),(0,±1),(1,−1),(−1,1)(x, y)=( \pm 1,0),(0, \pm 1),(1,-1),(-1,1)

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.