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Okay. I'll check.

]]>There are recurrences to get the continued fraction from a quadratic surd. You could raise the LHS up the 8th power and work with that one. It will produce a repeating cf and according to Legendre it will do it in 10 or less iterations.

]]>The only weak part is that you need to prove that is the cf of your irrational number.

That's the only part I need to know!

]]>This is not rigorous enough for a putnam problem but I did not know that when you posted it.

The cf for

You can replace the infinite portion of the cf with y - 2.

Now just solve for y.

The simplification was done by the continued fraction for us. The only weak part is that you need to prove that is the cf of your irrational number.

]]>No hurry.]]>

That is how I solved it. One way was with a PSLQ, the other was with a continued fraction. Please hold on, I am swamped. I will write up what I did.

Putnam questions are really hard, so do not be suprised if I am wrong.

]]>It's actually a putnam question. The original question was a continued fraction, and it needs the answer to be in that form.

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