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For which $n\in\mathbb{N}$ $$4\cdot 7^{n} - 3 = m^{2}$$ for some $m\in\mathbb{N}$? Here $\mathbb{N} = \mathbb{Z}\cap[0;+\infty)$.

First thing to notice is that the case with even $n$ is pretty obvious, since there are not many squares that differ by $3$. In fact, there is only one such pair: $1$ and $4$, which yields $(n,m) = (0,1)$. Now we are left to deal with the odd values of $n$. It turns out that this case is more interesting, as there are at least two solutions: $n=1$ and $n=3$.

I thought of proving that these are the only solutions in this case, but didn't succeed. It was not much use to apply modular arithmetic, since it is more helpful when proving that there are no solutions. It can give something like $m = 2k+1$, and maybe some algebraic manipulations can transform the equation to a more obvious one, but I don't see that right away. Also, it seems that $4\cdot 7\cdot 7^{n-1} - 3 = 28N^{2}-3 = m^{2}$ has infinitely many solutions in $(N,m)\in\mathbb{N}\times\mathbb{N}$, so one must use the fact that $N$ is a power of $7$ somehow.

I wouldn't like to see a full solution, but rather just hints or strategies.

chirico
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    What is the context of this problem, what kind of tools are allowed/expected? For example, this is easy with a computer (using elliptic curves), but that's of course not feasible in, say, a contest setting. – anankElpis Apr 14 '25 at 12:27
  • @anankElpis I was playing around with expressions like that, ended up with this problem.

    All kinds of approaches are welcomed, but I prefer those that are as elementary as possible. If you have a solution (even using a computer), you may post it, why not

     – chirico Apr 14 '25 at 12:35
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    Equivalent to $7^n=k^2+k+1$. That gives $k\not\equiv1\bmod3$. – mr_e_man Apr 14 '25 at 12:51
  • An idea, for $n=2k+1$ you can multiply by $7$ to get $x^2-7m^2=21$ a generalised Pell equation where $x=14\cdot 7^k$ and then use the theory to get solutions. – zwim Apr 14 '25 at 13:09
  • Hint : Use Elliptic Curves – Guruprasad Apr 14 '25 at 13:31
  • @zwim There's no need to multiply by $7$. Just write the equation as $m^2-7x^2=-3$ where $x=2 \times 7^k$ – jjagmath Apr 14 '25 at 13:31
  • @DietrichBurde Indeed, I wonder why this didn't pop up in the list of questions, when I was looking for something similar... – chirico Apr 14 '25 at 13:54
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    Note: The title of the linked question can be confusing, since it seems to say that there are no solutions. But in the body it is specified that $y>3$, so the result agrees with the examples in this question. – Milten Apr 14 '25 at 13:59

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