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.
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.
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.
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
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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 $n=6,14.$
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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$
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.