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Let $ x,y,a,n \in {Z}>1$, Does the diophantine equation $ x^2-4a^n=y^2$ have any solutions? or How can I solve this?

2 Answers2

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Sure, let $x=a^n+1$ and $y=a^n-1$. There are many other solutions if $n$ is large and/or $a$ has many factors.

One can get all solutions in integers $x,y$ positive, negative, or $0$ by letting $x+y=2u$, $x-y=2v$ where $uv=a^n$, and solving for $x$ and $y$.

André Nicolas
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  • Brilliant, Andre. It works like a glove. I just tried to substitute $1$ with $ t \in {Z}\ neq 1$ It did not work out. How did you get to this? Thanks. –  Sep 30 '13 at 16:09
  • There is a not very complicated theory of what numbers are representable as a difference of two squares, and in how many ways. It is done in every introduction to number theory. In addition, the identity $(a+b)^2-4ab=(a-b)^2$ crops up all ovr. – André Nicolas Sep 30 '13 at 16:12
  • Excellent. Thanks for your feedback. –  Sep 30 '13 at 16:19
  • I like this approach as much as the basic one that you first mentioned. I noted that $(a^n+b^n)^2-4(ab)^n=(a^n-b^n)^2$. My question was a special case where $ b^n=1$. You have provided me with great insights into this problem. Thanks. –  Sep 30 '13 at 16:34
  • You are welcome. The factorizations of $a^n$ are more general, since in the above comment you are only using "special" factorizations $a^n=b^nc^n$. – André Nicolas Sep 30 '13 at 17:09
  • By letting $x+y=2u, x-y=2v, uv=a^n$, wouldn't we lose generality by implicitly considering the only cases where $ x,y$ have the same parity? –  Sep 30 '13 at 20:45
  • By writing $x+y$ and $x-y$ as in your comment, we are only considering cases where $x+y$ and $x-y$ have the same parity. But they always do. If we allowed $x+y$ even and $x-y$ odd (say) we would be making $x$ and $y$ non-integers. In general, the equation $x^2-y^2=c$ has solutions in integers when $c$ is odd, and also when $c$ is divisible by $4$. There are no solutions when $c$ is divisible by $2$ and not by $4$, precisely because of the parity issue. – André Nicolas Sep 30 '13 at 20:51
  • Thank you very much for your expert feedback. May I ask you, which theorem is that? Any advice on a college or pre-graduate algebra book that I can purchase to better myself. –  Oct 01 '13 at 13:12
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Note that if $n=2$ you can rewrite as $x^2=y^2+(2a)^2$ - which essentially means you are looking for a pythagorean triple since if positive integers satisfy $x^2=y^2+z^2$ at least one of $y, z$ must be even.

Mark Bennet
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  • Great, Mark. Thanks for your input. I did come up with this result as well. I was wondering if that was this only possible solution. –  Sep 30 '13 at 16:12
  • @Andy - it would have helped if you had mentioned this in the question - you asked for "any" solution. You could have put $n\gt 2$ for example. – Mark Bennet Sep 30 '13 at 16:21
  • You are absolutely right. I should have. Sincere apology. –  Sep 30 '13 at 16:25