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The 1995 Putnam B4 asks to evaluate some infinite expression and I got this value after some computations:$$\left({{\frac{2207+(2207^2-4)^{1/2}}2}}\right)^{1/8}$$ However I am supposed to write it in form:$$\frac{a+b\sqrt{c}}{d}$$. What do I do?

(The answer is: $\frac{3+\sqrt{5}}{2}$ which is the same as mine if you put in a calculator.)

danielleontiev
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mtheorylord
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1 Answers1

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Let $\,x \gt 1\,$ be the expression in question, then:

$$\require{cancel} 2x^8=2207 + \sqrt{2207^2-4} \\ (2x^8-2207)^2=2207^2-4 \\ 4x^{16}-4\cdot 2207 x^8+\cancel{2207^2}=\cancel{2207^2}-4 \\ x^{16}-2207 x^8 + 1 = 0 \\ x^8 + \frac{1}{x^8}=2207 \\ \left(x^4+\frac{1}{x^4}\right)^2 = 2209 \\ x^4+\frac{1}{x^4} = 47 \\ \left(x^2+\frac{1}{x^2}\right)^2 = 49 \\ x^2+\frac{1}{x^2} = 7 \\ \left(x+\frac{1}{x}\right)^2 = 9 \\ x+\frac{1}{x} = 3 \\ x^2-3x+1=0 $$

dxiv
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    Nice trick $x^{2n} + x^{-2n} = (x^n + x^{-n})^2-2$. –  Aug 03 '17 at 00:07
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    @j4nbur53 Thanks. Initial clue here was to derive the equation $x^{16}-2207 x^8+1 = 0$ and recognize that the LHS is a palindromic polynomial of even degree, which then invites substitutions in powers of $x+1/x$. – dxiv Aug 03 '17 at 00:28