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What are the integer solutions to the diophantine equation

$$n^2 + 1 = 2 \times 5 ^m? $$

We have $(n,m) = (3,1), (7, 2) $ as solutions. Are there any more?

This seems like it would be a well known diophantine equation, but I can't seem to find any information about it.

Calvin Lin
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  • Idea for $m\geq 2$, the last two digits of $2\times 5^m-1$ are $249$. You want to prove that you find perfect squares $n>2$ with these last two digits. For $n>7$, the numbers can have the last three digits $a3$ or $b7$. For $\overline{\dots a3}^2=\overline{\dots a^200}+60a+9$ so $a$ has to be $4$ or $9$ which do not work. For $\overline{\dots b7}^2=\overline{\dots b^200}+140b+49$ so $b$ has to be $0$ which does not work or $5$ which works. I do not know it it helps. – Iulia Feb 16 '15 at 01:56
  • @Iulia Unfortunately it does not seem likely to be helpful. Hensel's lemma tells us that there will be a solution to the condition, but the issue is that the solution is too large, and hence violates the next condition that we need to check. – Calvin Lin Feb 16 '15 at 02:41
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    @CalvinLin: It is called a Ramanujan-Nagell-type equation $$x^2+C = AB^m$$ where $A,B,C$ are constants. By a result of Seigel, there should only be a finite number of solutions. (It would have been nice if his proof included an upper bound for $m$.) – Tito Piezas III Feb 16 '15 at 03:03
  • Having just asked a similar question, I found that there are indeed no further solutions, with reference to the paper "On the number of solutions of the generalized Ramanujan-Nagell equation" by Bugeaud and Shorey, and thus note such here. – Sridhar Ramesh Jun 04 '17 at 03:18
  • The only solutions $(m,n)$ are $\quad(0,\pm1)\quad(1,\pm3)\quad(2,\pm7)$, You can verify this in a spreadsheet using $n = \pm\sqrt{2*5^m - 1}$ – poetasis Oct 08 '20 at 00:59
  • @poetasis How did you check the infinitely many possible values of $(m, n)$ with a spreadsheet? – Calvin Lin Oct 08 '20 at 16:35
  • @Calvin Lin Your point is well taken. The solutions offered were from a spreadsheet with a limit of $15$ digits. I have no proof that there are no more solutions. I only know that all solutions other than zero have $m$ ending in $3$ or $7$. – poetasis Oct 08 '20 at 17:06
  • You might try this paper – poetasis Oct 08 '20 at 17:16

1 Answers1

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By factoring the given equation over the Gaussian integers, we are looking for solutions of: $$a_m=\operatorname{Re}\left((1\pm i)(2+i)^m\right)=\pm 1.\tag{1}$$ It is straighforward to check that $(1)$ is fullfilled by $m=0,m=1$ and $m=2$.

In order that $(1)$ holds, $\frac{\pi}{4}+m\arctan\frac{1}{2}\pmod{2\pi}$ must be close to $0,\pm\frac{\pi}{2}$ or $\pi$.

That happens for $m=15,22,29,36,\ldots$ and many others $m$ such that: $$\left\{m\cdot\frac{1}{\pi}\arctan\frac{1}{2}\right\}\text{ is close to }\frac{1}{4}\text{ or } 0.\tag{2}$$ where $\{\cdot\}$ stands for the fractional part. So we have that other solutions of $(1)$ may occur for $m=q$ or $m=q/4$, where $\frac{p}{q}$ is a convergent of the continued fraction of $\alpha=\frac{1}{\pi}\arctan\frac{1}{2}$ and $4\mid q$ if needed.

The first of such good candidates is $m=83$: in such a case the absolute value of the ratio between the real part and the imaginary part of $(1+i)(2+i)^m$ is something like $0.00175$, but the real part is still a $27$-digits number.

It is reasonable to conjecture that no other solutions exist apart from the trivial ones; however, that looks hard to prove (at least to me) since the continued fraction of $\alpha$ does not have a good structure (like for the golden ratio, the constant $e$ or some values of Bessel's functions).

An alternative approach (but not so much) may be to prove by arithmetical arguments that the only solutions of the Pell equations $$ n^2-2N^2=1,\qquad n^2-10 N^2=1\tag{3}$$ with $N=5^m$ are the trivial ones. That problem depends on the convergents of $\sqrt{2}$ and $\sqrt{10}$ that are way nicer. It is probably not to difficult to derive an explicit formula for the $5$-adic height of a lambda number $P_n$ and deduce that there are a finite number of $N$s of the form $5^m$ that may fulfill $(3)$.

Jack D'Aurizio
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