12
$\begingroup$

The problem comes from a problem I encountered when I wrote the article

Find all positive integer $m$ such $$2^{m}+1\mid5^m-1$$ it seem there no solution. I think it might be necessary to use quadratic reciprocity knowledge to solve this problem. If $m$ is odd then $2^m+1$ is divisible by 3 but $5^m-1$ is not. so $m$ be even, take $m=2n$, then $$4^n+1\mid25^n-1.$$ if $n$ is odd,then $4^n+1$ is divisible by $5$, but $25^n-1$ is not so $n$ is even,take $n=2p$, we have $$16^p+1\mid625^p-1.$$

$\endgroup$
24
$\begingroup$

Here is a proof.

Theorem. $2^m+1$ never divides $5^m-1$.

Assume that there is some $m$ such that $2^m+1$ divides $5^m-1$. We already know that $m$ must be divisible by $4$. Let $m = 2^n a$ with an odd integer $a$ and $n \ge 2$. The $n$th Fermat number $$F_n = 2^{2^n} + 1$$ is congruent to $2$ mod $5$ (this uses $n \ge 2$), so it has a prime divisor $p$ such that $$p \equiv \pm 2 \pmod 5.$$ We know that $p-1 = 2^{n+1}k$ for some integer $k$. Since $\left(\frac{5}{p}\right) = \left(\frac{p}{5}\right) = -1$, we have that $$5^{2^n k} = 5^{(p-1)/2} \equiv -1 \pmod p,$$ so $$5^{mk} = (5^{2^n k})^a \equiv -1 \pmod p$$ as well. In particular, $$5^m \not\equiv 1 \pmod p.$$ On the other hand, $$2^m = (2^{2^n})^a \equiv (-1)^a = -1 \pmod p.$$ Thus $p$ divides $2^m+1$, but does not divide $5^m-1$, a contradiction.

$\endgroup$
  • 1
    $\begingroup$ @Joe Thanks for improving the layout! $\endgroup$ – Michael Stoll Dec 14 '18 at 16:58
13
$\begingroup$

There might well be a very elementary argument for this, but in the spirit of taking a hammer to a fly, one can prove that the number of $m$ such that $$ 2^m+1 \mid 5^m-1 $$ is finite by invoking a theorem of Bugeaud, Corvaja and Zannier [Math. Z. 2003] which implies, in this context that, given $\epsilon > 0$, $$ \gcd (4^m-1, 5^m-1) \leq e^{\epsilon m} $$ for suitably large $m$. Schmidt's Subspace Theorem is used here, so the result is ineffective.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.