What is the solution to a Putnam Mathematical Competition problem from 1995?

[Ackbach]

Here is this week's POTW:

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Evaluate $\displaystyle\sqrt[8]{2207-\dfrac{1}{2207-\dfrac{1}{2207-\cdots}}}$. Express your answer in the form $\dfrac{a+b\sqrt{c}}{d}$, where $a,b,c,d$ are integers.

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[Ackbach]

Re: Problem Of The Week # 218 - May 31, 2016

This was Problem B-4 in the 1995 William Lowell Putnam Mathematical Competition.

Congratulations to kiwi and kaliprasad for their correct answers. kiwi's solution follows:

Spoiler

$$\sqrt[8]{2207-\frac{1}{2207-\frac{1}{2207-\cdots}}}$$Let \(k=2207-\frac{1}{2207-\frac{1}{2207-\cdots}}=2207-\frac{1}{k}\)

So \(k^2-2207k+1=0\)

So \(k=\frac{2207 \pm \sqrt{2207^2-4}}{2}=\frac{2207 \pm \sqrt{987^2.5}}{2}=\frac{2207 \pm 987\sqrt{5}}{2}\)

Now assume:
\(\sqrt[2]{k}=\sqrt[2]{\frac{2207 \pm 987\sqrt{5}}{2}}=\frac{a+b\sqrt{5}}{2} \)

So
\(2(2207 \pm 987\sqrt{5})= ({a+b\sqrt{5})^2= (a^2+2ab\sqrt{5}+5b^2})\)

So \(a^2+5b^2=2 \times 2207\) and \(2ab=\pm 2 \times 987\)

Now \(987=3\times7\times47\) so a and b are each 3,7,47,21,141 or 329. By trial and error a=47 and b=21. Giving:
\(\sqrt[2]{k}=\frac{47+21\sqrt{5}}{2} \)

Now resetting a and b we write:
\(\sqrt[4]{k}=\sqrt[2]{\frac{47+21\sqrt{5}}{2}}=\frac{a+b\sqrt{5}}{2} \)

So
\(2(47+21\sqrt{5})=({a+b\sqrt{5})^2=(a^2+2ab\sqrt{5}+5b^2})\)

So \(a^2+5b^2= 2 \times 47\) and \(2ab=\pm 2 \times 21\)

Which has solutions a=7, b=3 so:

\(\sqrt[4]{k}=\frac{7+3\sqrt{5}}{2} \)

Once again resetting a and b we write:
\(\sqrt[8]{k}=\sqrt[2]{\frac{7+3\sqrt{5}}{2}}=\frac{a+b\sqrt{5}}{2} \)

So
\(2(7+3\sqrt{5})= ({a+b\sqrt{5})^2= (a^2+2ab\sqrt{5}+5b^2})\)

So \(a^2+5b^2=2 \times 7\) and \(2ab=\pm 2 \times 3\)

Which has solutions a=3, b=1 so:

\(\sqrt[8]{k}=\frac{3+\sqrt{5}}{2} \)

or

\(\sqrt[8]{2207-\frac{1}{2207-\frac{1}{2207-\cdots}}}=\frac{3+\sqrt{5}}{2} \)