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I'm interested in the following diophantine eqaution: $(5^n-1)/4=y^2$.

It turns out that this is a special case of the Nagell-Ljunggren equation, where $x=5$ and $q=2$

It has been shown that for x=5 this has no solutions but I'm looking for an elementary solution of this special case.

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    $\begingroup$ First, you say that there are no solutions, but $(n,y)=(1,\pm1)$ are solutions. Second, did you try factoring $5^n=(2y)^2+1$ as $ (2+i)^n(2-i)^n = (2y+i)(2y-i)$? This might lead to an elementary proof (if you consider using the fact that $\mathbb{Z}[i]$ is a PID elementary). $\endgroup$
    – Joe Silverman
    Commented Dec 29, 2013 at 15:53
  • $\begingroup$ I was referring to non-trivial solutions. Besides $(n,y)=(1,±1)$ are always solutions of the Nagell-Ljunggren diophantine. $\endgroup$
    – user44801
    Commented Dec 29, 2013 at 16:07
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    $\begingroup$ When writing something mathematical, it's always best to be precise, since the reader may not be able to guess what you mean. So if you mean "no non-trivial solutions", then that's what you should write. (I hope that you will take this as the constructive criticism that it's meant to be. I'm paraphrasing advice that I give to my PhD students; that they should always proofread what they've written as if they are someone seeing the material for the first time. So every statement needs to be precise.) $\endgroup$
    – Joe Silverman
    Commented Dec 29, 2013 at 16:56
  • $\begingroup$ Ok, I appreciate your advice sir. $\endgroup$
    – user44801
    Commented Dec 29, 2013 at 21:36

1 Answer 1

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This kind of problem usually requires a little algebraic number theory. Joe Silverman sketches one possible approach in the comments. Here is another. Let's rewrite as $$ (2y)^2-5^n=-1. $$ If $n$ is even then the left-hand side is a difference of two squares, which quickly gives a contradiction. So write $n=2m+1$. Then $$ (2y+5^m \sqrt{5})(2y-5^m \sqrt{5})=-1. $$ Thus $2y+5^m \sqrt{5}$ is a unit in $\mathbb{Z}[(1+\sqrt{5})/2]$. A fundamental unit is $\epsilon=(-1+\sqrt{5})/2$. It follows that $$ 2y+5^m \sqrt{5}=\pm \epsilon^{t}. $$ Conjugating $$ 2y-5^m \sqrt{5}=\pm \mu^t, \qquad \mu=(-1+\sqrt{5})/2. $$ Taking differences and dividing by $\pm \sqrt{5}$ we have $$ \pm 2 \cdot 5^m= \frac{\epsilon^t-\mu^t}{\sqrt{5}}. $$ The right-hand side is the $t$-th Fibonacci number. Thus the equation becomes $$ F_t=\pm 2 \cdot 5^m. $$ Let's rule out the case $m \ge 1$, which leaves you with $F_t= \pm 2$. If $m \ge 1$ then $F_t \equiv 0 \pmod{10}$. Now write out the Fibonacci sequence modulo $10$ and convince yourself that this forces $t \equiv 0 \pmod{15}$. But then $F_{15} \mid F_t$. However, $F_{15}=2 \cdot 5 \cdot 61$, which does not divide $2 \cdot 5^m$, giving a contradiction.

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