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I was just wondering if there are any integer solutions to the Diophantine equation:

$x^2 - (n^2 - 2)y^2 = -1 \ \ $ for $n > 2$

I don't think there are any but can't prove why.

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    Sorry! It is meant to be $-1$ not $1$!2012-05-25

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Note that if $d$ is divisible by a prime $p$ of the form $4k+3$, then the equation $x^2-dy^2\equiv -1$ cannot have a solution, for $x^2\equiv -1\pmod{p}$ does not have a solution.

If $n>2$ is odd, then $n^2-2\equiv -1\pmod{4}$, so $n^2-2$ is divisible by a prime of the form $4k+3$.

If $n$ is divisible by $4$, then again $n^2-2$ is divisible by a prime of the form $4k+3$. But this leaves the possibility $n\equiv 2\pmod{4}$, where $n^2-2$ need not have a prime divisor of the form $4k+3$.

Remark: Will Jagy has settled the problem in general, by observing that the continued fraction of $\sqrt{n^2-2}$ has period $4$. (If $\sqrt{d}$ has continued fraction with even period, then the equation $x^2-dy^2=-1$ has no integer solutions.)

There is an approach that does not use properties of continued fractions, but instead uses basic properties of Pell equations. Note that $x=n^2-1$, $y=n$ is a solution of the Pell equation $x^2-(n^2-2)y^2=1$. If there were solutions of $x^2-(n^2-2)y^2=-1$, there would be a fundamental solution $(a_0,b_0)$, and $(n^2-1,n)$ would be an "even power" of $(a_0,b_0)$, in the sense that $n^2-1+n\sqrt{n^2-2}=(a_0+b_0\sqrt{n^2-2})^{2k}$ for some positive integer $k$. This is not possible, for if $(a_0+b_0\sqrt{n^2-2})^{2k}=a+b\sqrt{n^2-2}$, then $a \ge n^2-1$, and we cannot have equality.

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    @AndréNicolas : Sir my bounty on this [question](http://math.stac$k$exchange.com/questions/15831$2$/proving-a-statement-regarding-a-diophantine-equation) is going to get expired within 3 days. Do you have any suggestions sir ?2012-06-26
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EDIT: in simple terms, this follows from the fact that the continued fraction for $\sqrt{n^2 -2}$ has period $4,$ with coefficients $ [n-1;1,n-2,1,2n-2] $

There are no solutions for $ x^2 - (n^2-2)y^2 = -1 $ with $n>2.$The small values of $ x^2 - (n^2-2)y^2,$ given by continued fractions or by this, the method of neighboring reduced forms, are $ 1, \, 2, \; 3 - 2 n. $

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$      jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./Pell Input n for Pell  7  0  form   1 4 -3   delta  -1 1  form   -3 2 2   delta  1 2  form   2 2 -3   delta  -1 3  form   -3 4 1   delta  4 4  form   1 4 -3   disc   28 Automorph, written on right of Gram matrix:   2  9 3  14    Pell automorph  8  21 3  8  Pell unit  8^2 - 7 * 3^2 = 1   ========================================= jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./Pell Input n for Pell  14  0  form   1 6 -5   delta  -1 1  form   -5 4 2   delta  2 2  form   2 4 -5   delta  -1 3  form   -5 6 1   delta  6 4  form   1 6 -5   disc   56 Automorph, written on right of Gram matrix:   3  20 4  27    Pell automorph  15  56 4  15  Pell unit  15^2 - 14 * 4^2 = 1   ========================================= jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./Pell Input n for Pell  23  0  form   1 8 -7   delta  -1 1  form   -7 6 2   delta  3 2  form   2 6 -7   delta  -1 3  form   -7 8 1   delta  8 4  form   1 8 -7   disc   92 Automorph, written on right of Gram matrix:   4  35 5  44    Pell automorph  24  115 5  24  Pell unit  24^2 - 23 * 5^2 = 1   ========================================= jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus  

Since Andre is asking about n \equiv 2 \pmod 4$ when $n^2 -2$ may happen to have no prime divisors $q \equiv 3 \pmod 4,$ I have also run $n=6,14.

========================================= jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus      jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./Pell Input n for Pell  34  0  form   1 10 -9   delta  -1 1  form   -9 8 2   delta  4 2  form   2 8 -9   delta  -1 3  form   -9 10 1   delta  10 4  form   1 10 -9   disc   136 Automorph, written on right of Gram matrix:   5  54 6  65    Pell automorph  35  204 6  35  Pell unit  35^2 - 34 * 6^2 = 1   ========================================= jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./Pell Input n for Pell  194  0  form   1 26 -25   delta  -1 1  form   -25 24 2   delta  12 2  form   2 24 -25   delta  -1 3  form   -25 26 1   delta  26 4  form   1 26 -25   disc   776 Automorph, written on right of Gram matrix:   13  350 14  377    Pell automorph  195  2716 14  195  Pell unit  195^2 - 194 * 14^2 = 1   ========================================= jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$  
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    Hmmm. I give a pretty good introduction, including how to do the algorithm in detail, at http://math.stackexchange.com/questions/90406/how-to-detect-when-continued-fractions-period-terminates/90565#90565 and http://mathoverflow.net/questions/22811/upper-bound-of-period-length-of-continued-fraction-representation-of-very-composi/23014#230142012-05-26
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This can be a form of Pell's equation if $\,n^2-2\,$ is not a square, and it always has non-trivial solutions with $\,y>0\,$ by a theorem of Lagrange.

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    Just as a matter of interest $n^2 -2$ is never a square when $n$ is an integer, since for positive integers x >y, $x^2 - y^2$ is expressible as a sum of distinct odd numbers.2012-05-26