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I encountered this question (posed by Fermat) in a letter from Fermat to Carcavi and was wondering what would be the best elementary way to solve it.

Solve in positive integers$$(2x^2-1)^2=2y^2 - 1$$

Any help will be appreciated.
Thanks in advance.

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  • $\begingroup$ Is that letter somewhere on the Internet? I would like to see it. Thank you! $\endgroup$ – Janko Bracic Feb 26 '15 at 14:25
  • $\begingroup$ Thanks! I see now, the guy to whom Fermat sent the letter was Carcavi. For Caviar I didn't hear in this context :) $\endgroup$ – Janko Bracic Feb 26 '15 at 14:47
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The solution of the Pell equation: $$ X^2-2y^2 = -1 \tag{1}$$ are given by $X=1,7,41,239,1393,8119,\ldots $ (OEIS A002315), hence we are looking for integers of the form $2x^2-1$ among terms of the sequence given by: $$ X_1 = 1,\qquad X_2 = 7,\qquad X_{n+2}=6 X_{n+1}-X_n \tag{2}$$ or: $$\begin{eqnarray*} X_n = \frac{(1+\sqrt{2})^{2n-1}+(1-\sqrt{2})^{2n-1}}{2}&=&\sum_{j=0}^{2n-2}(-1)^j(1+\sqrt{2})^{2n-2-j}(1-\sqrt{2})^{j}\\&=&\sum_{k=0}^{n}\binom{2n-1}{2k-1}2^{n-k}. \tag{3}\end{eqnarray*}$$ It is easy to check that $n=1,2$ give two solutions; not obvious that $n=1,2$ give the only solutions; probably the property $ X_n^2-X_{n-1}X_{n+1}=8$ can be exploited in some tricky way.

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For $y$ a prime, the simplest solution I've ever seen is this answer in another thread. I have no doubt that (very elementary) method could be adapted to solve the fully-general theorem. Another solution, given by Mahoney, I replicated here.

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