Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.6 AIME Find the answer

If $x y=5$ and $x^{2}+y^{2}=21$ , compute $x^{4}+y^{4}$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

We have $441=\left(x^{2}+y^{2}\right)^{2}=x^{4}+y^{4}+2(x y)^{2}=x^{4}+y^{4}+50$ , yielding $x^{4}+y^{4}=391$ .

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