Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.5 AIME Prove it Soviet Union

Problem:
If rationals $x$ , $y$ satisfy $x^{5} + y^{5} = 2x^{2}y^{2}$ show that $1 - x y$ is the square of a rational.

Solution

Solution:
Put $y = k x$ , then $x^{5}(1 + k^{5}) = 2k^{2}x^{4}$ , so $x = \dfrac{2k^{2}}{1 + k^{5}}$ , $y = \dfrac{2k^{3}}{1 + k^{5}}$ and $1 - x y = \dfrac{(1 - k^{5})^{2}}{(1 + k^{5})^{2}}$ . $x$ and $y$ are rational, so $\dfrac{1 - k^{5}}{1 + k^{5}}$ is rational.

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