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I have tried an elementary method to solve the equation $x^{2}-7^{y}=2$ and have given up this ideal at last. Then I try to solve it in $Q(\sqrt{2})$. Since $Q(\sqrt{2})$ has infinity many units $\pm(\pm 1+\sqrt{2})^{n}$, $n=0,1,2,\cdot\cdot\cdot$, the proof would become more difficult. Note that $(x,y)=(\pm 3,1)$ are integer solutions.

Update: An elementary number theory method on the equation which has the only integer solutions $(x,y)=(\pm 3,1)$ can be found in the direction of Pell equation and I have finished it many days before. This method has been mentioned in Mikael Jensen's comments below but there still needs much more other skills.

Now, my question is:

Does there exist a method of algebraic number theory to solve the equation?

Travis Wang
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  • y == ((8 - 3 Sqrt[7])^C[1] - (8 + 3 Sqrt[7])^C[1])/(2 Sqrt[7]), Element[C[1], Integers], C[1] >= 0}, {x == ((8 - 3 Sqrt[7])^C[1] + (8 + 3 Sqrt[7])^C[1])/2, y == -((8 - 3 Sqrt[7])^C[1] - (8 + 3 Sqrt[7])^C[1])/(2 Sqrt[7]), Element[C[1], Integers], C[1] >= 0}, {x == ((8 - 3 Sqrt[7])^C[1] + (8 + 3 Sqrt[7])^C[1])/2, y == ((8 - 3 Sqrt[7])^C[1] - (8 + 3 Sqrt[7])^C[1])/(2 Sqrt[7]), Element[C[1], Integers], C[1] >= 0}} – Mikael Jensen Sep 29 '13 at 12:26
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    One possible approach is the use of Pell's equation $u^2-7v^2=2$ where $v=7^n$. y must be odd, since otherwise $(x+7^(y/2))(x-7^(y/2))=2$. It then follows that $(u^2-1)^2-7(uv)^2=1$ and you must find solution $v=7^n$ I didn't find any other than your own (3,1). The rubbish above is Wolfram Alpha's solution for y (which I can't delete). Another approach is to consider that x must 3 or 4 (mod 7) and use brute force. I tried that in vain also. – Mikael Jensen Sep 29 '13 at 12:34
  • Related to http://en.wikipedia.org/wiki/Ramanujan%E2%80%93Nagell_equation See also http://en.wikipedia.org/wiki/Ramanujan%E2%80%93Nagell_equation#Equations_of_Ramanujan.E2.80.93Nagell_type as there is already a finiteness result. You will need to read some of the references listed there. – Will Jagy Sep 29 '13 at 18:31
  • more references at http://mathworld.wolfram.com/RamanujansSquareEquation.html In particular, i have Mordell's famous book, I will take a look at page 205. – Will Jagy Sep 29 '13 at 18:36
  • Hasse's paper on the original R-N equation can be downloaded as a free pdf at http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.nmj/1118801617 After looking at that and Mordell, I would agree that your difficulty is that you have $\mathbb Q (\sqrt 2)$ instead of $\mathbb Q (\sqrt {-2}).$ Still, difficult does not always mean impossible. – Will Jagy Sep 29 '13 at 19:29
  • @Mikael Jensen: Yes, I've done it just as what you had said. But one of the problem is that the diophantine equation $x^{2}-7y^{2}=2$ has infinity many solutions. – Travis Wang Sep 30 '13 at 08:57
  • @Will Jagy: Thank you for the paper you have mentioned. I'm a beginner in the field of algebraic number theory but I'll try my best to learn it. – Travis Wang Sep 30 '13 at 09:06
  • Wang, I put x^2 - 2 = 7^n into Google. It did better that I expected. I did find some other articles (usually by famous people) finding all solutions for specific $x^2 + a = b^y,$ where $a$ was generally positive. You will need to read some of the originals (one author was Cohn) to figure out whether $a$ positive is critical for the method or just a convenience. – Will Jagy Sep 30 '13 at 16:47

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Just some comments. There are primitive solutions to $$ x^2 - 2 z^2 = 7^y $$ for every positive $y.$ Indeed, if $z_n \equiv 5 x_n \pmod 7$ and $$ x_n^2 - 2 z_n^2 = 7^n, $$ we may take $$ x_{n+1} = 3 x_n + 2 z_n, \; \; z_{n+1} = x_n + 3 z_n $$ to get a primitive solution to $$ x_{n+1}^2 - 2 z_{n+1}^2 = 7^{n+1}, \; \; \; z_{n+1} \equiv 5 x_{n+1} \pmod 7. $$

Meanwhile, given any such $(x,z)$ pair, we get the same $7^n$ with either $$ (3 x + 4 z, 2 x + 3 z) $$ or $$ (3 x - 4 z, -2 x + 3 z). $$ After taking absolute values, you can get some pretty small values of $z$ this way: $$ 9^2 - 2 \cdot 4^2 = 49, $$ $$ 19^2 - 2 \cdot 3^2 = 343, $$ $$ 51^2 - 2 \cdot 10^2 = 2401, $$ and so on.

I will need to think about whether all solutions of $x^2 - 2 z^2 = 7^y$ arise this way. If so, it is a start.

Will Jagy
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