Find all integers $n \ge 2$ such that $n$ divides $2^n-3$.
This is a variant of the classic problem find all $n$ such that $n$ divides $2^n \pm 1$, which can be done using orders. However I've tried the same methods on this variant without success.
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Yes, for $n^2\pm 1$ there are nice solutions, see here. It looks like that there is no $n\ge 2$ dividing $2^n-3$. – Dietrich Burde Mar 31 '16 at 14:39
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@DietrichBurde I don't think that the method of solving the case $2^n\pm1$ work here. – Mar 31 '16 at 17:07
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@DietrichBurde there are such $n$, but they are all big. See the answer I have posted. – Gerry Myerson Sep 11 '20 at 05:06
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A later question asked for the smallest such $n$, https://math.stackexchange.com/questions/3035344/what-is-the-smallest-integer-n-such-that-2n-equiv-3-mod-n – Gerry Myerson Sep 11 '20 at 05:27
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@GerryMyerson Very interesting (+1), and this after four years! – Dietrich Burde Sep 11 '20 at 09:36
2 Answers
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This case is harder than the case $2^n\pm1$ and I test this for all $n<7,000,000$ and in all cases $n \nmid 2^n-3$.
When $n$ is a prime number $n\nmid2^n-3$ by Fermat's little theorem. And I find that it's sufficient to check this for all square-free numbers. Because $$p|2^q-3\iff p|2^{q.p^k}-3$$
when $p$ is a prime number. Because $(2^q)^{p}-3\equiv 2^q-3 \pmod{p}$ .
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Integers $n\ge2$ such that $n$ divides $2^n-3$ are tabulated here. Only five are currently known. The smallest is $4700063497$ so it's no surprise that testing up to $7,000,000$ failed to find any. Finding all such $n$, as OP desires, is hopeless. Even deciding whether there are infinitely many is hopeless.
Gerry Myerson
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