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Edit: Dietrich Burde kindly gave me a hint but I, in my inexperience, am not able to understand the logic. I've tried but failed. So, right now my problem is understanding him. So, if you can explain the logic to me, I would be grateful.

For which positive integer $n$, is $\frac{2^n-1}{3}$ a factor of $4m^2+1$ for some integer $m$? So, I've gotten that $n=2^q,\ q\in\mathbb{Z}_+$ is one solution, and I'm pretty certain that it is the only solution, but I don't know how to prove this (correct or wrong).

If you're interested on how I got to this point, well here's briefly what I've done: $n$ must be an even integer, as $3\mid 2^n-1$ as $2^{2k}-1=4^k-1\equiv 1^k-1\equiv 0\ (mod\ 3)$

$n=2$ is a quite trivial solution.

$\frac{2^n-1}{3}$ must be a factor of $4m^2+1$, which gives us this (Diophantine) equation: $$4m^2+1=\frac{d(2^n-1)}{3}$$ For $d=1$, we get $4m^2=\frac{2^n-4}{3}$ and because $n$ must be even, we get: $$4m^2=\frac{4^k-4}{3}$$ $$m^2=\frac{4^{k-1}-1}{3}$$ $$3m^2=(2^{k-1})^2-1^2$$ $$3m^2=(2^{k-1}-1)(2^{k-1}+1)$$ From this you can quite easily get $k=1, m=\pm 1$ as a solution so $n=4$.

But what about for other values of $d$?

Well, we have $n=2,4=2^1, 2^2$. Maybe it works for all $n=2^q$. I proved this with induction using the fact that $2^{(2^{l+1})}=(2^2)^{2^l}$.

But is there other solutions? I don't know. I feel like there is no other ones because I haven't been able to find anything for the integers in between.

110112345
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2 Answers2

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Hint: Every divisor $d$ of $4m^2+1$ satisfies $d\equiv 1\bmod 4$. And $\frac{2^n-1}{3}$ is an integer if and only if $n$ is even. Suppose that $n=2k$ is not a power of $2$. Then $\frac{2^n-1}{3}$ has a prime divisor $p$ with $p\equiv 3\bmod 4$, a contradiction.

Dietrich Burde
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  • Would this in itself answer the original question? This seems much more elegant than mine? – 110112345 Jan 08 '21 at 17:34
  • Also can't the divisors be also $3 mod 4$ if there are two of them as $3^2\equiv 1 mod 4$? – 110112345 Jan 08 '21 at 18:03
  • After thinking for a while, it feels like I'm missing something. a) I don't see why every divisor (and thus prime factor) of $4m^2+1$ is d mod 4 (I know it is but I can't prove it)

    b) I don't see the logic that $(n\neq 2^q\wedge p\mid \frac{2^n-1}{3})\Rightarrow p\equiv 3\ \bmod\ 4$. Or even that when it is a power of two that the prime factor would be 1 mod 4.

     – 110112345 Jan 08 '21 at 18:46
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    For a) see this duplicate with $n=2m$. Try to find the others for b) and c) on this site, too. – Dietrich Burde Jan 08 '21 at 20:01
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$$\frac{4m^2+1}{\frac{2^n-1}{3}}=\frac{3 (4 m^2 + 1)}{(2^n - 1)}$$

This answer does not address the theoretical solutions to this problem but, $-1500\le n \le 1500$ reveals:

$$n=\in\{\pm1,\pm2\}\implies m\in\{\mathbb{Z}-0\}$$ $$ n=\pm 4\implies m\in\{\pm1,\pm4,\pm6,\pm9,\pm11,\pm14,\pm16,\pm19,\cdots\}\\ \text{(OEIS A047209)}$$

$$n=\pm 8 \implies |m|\in\{19,36,49,66,104,121,134,151,\cdots\}$$ $$n=\pm 16 \implies |m|\in\{3606,5646,7461,9501,\cdots\}$$

For $n\in\{32,64,128,\cdots\}$ no solutions were found using $15$-digit precision.

poetasis
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  • Well in the range of $n$ there are still solutions like I noted. $n=2,4,8,16,32...$ – 110112345 Jan 08 '21 at 19:19
  • Good point, I had quit looking at $n=7$ thinking it was a lost cause. The list will never be complete but, perhaps I can reveal a pattern. – poetasis Jan 08 '21 at 19:34
  • Yeah, there is a pattern ($n=2^q$) and I think I proved it. The problem I had was to prove that those are the only solutions. – 110112345 Jan 09 '21 at 12:59
  • Whoever downvoted me: I indicated up front that I was not trying to provide "the" answer. I was providing empirical data to support any "theory" provided by others. I thought this was useful. If anyone disagrees, please let me know why it was not useful. Thanks – poetasis Jan 09 '21 at 23:30