Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.8 AIME Find the answer

Let $f(x)=x^{2}+x^{4}+x^{6}+x^{8}+\cdots$ , for all real $x$ such that the sum converges. For how many real numbers $x$ does $f(x)=x$ ?

Your answer

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

Solution

Clearly $x=0$ works. Otherwise, we want $x=x^{2} /\left(1-x^{2}\right)$ , or $x^{2}+x-1=0$ . Discard the negative root (since the sum doesn't converge there), but $(-1+\sqrt{5}) / 2$ works, for a total of 2 values.

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