Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.9 AIME Find the answer

143. For which $n \in \boldsymbol{N}$ does the equality

$\sqrt[n]{17 \sqrt{5}+38}+\sqrt[n]{17 \sqrt{5}-38}=\sqrt{20} ?$

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

S o l u t i o n. Denoting the first term by $a$ , we will have $a+\frac{1}{a}=\sqrt{20}$ , from which $a=\sqrt{5}+2$ . Since $(\sqrt{5}+2)^{3}=17 \sqrt{5}+$ $+38=(\sqrt{5}+2)^{n}$ , then $n=3$ .

A n s w e r: $n=3$ .

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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.