For what values of n will the expresion $2^{2n} -1$ be divisible by $4n+1$.

I have checked using a computer and the values of 2n I get are 8,20,36,44,48,56,68,96,116,120,128,140,156,168,170,176 $\cdots$etc [For $2n<200$] .

I don't seem to find a relation between these values.I am curious to see if we can find a relation by applying simple number theory.

  • 2
    $\begingroup$ OEIS doesn't find a plausible option for a sequence containing those four numbers, for what that's worth. $\endgroup$ – lulu Sep 27 at 10:51
  • 1
    $\begingroup$ Are your values correct? I get $\{4,10,18,22,24,28,34,48,\cdots\}$, which also fails to appear in OEIS. $\endgroup$ – lulu Sep 27 at 11:00
  • $\begingroup$ @lulu I got that too. $\endgroup$ – José Carlos Santos Sep 27 at 11:01
  • $\begingroup$ The values are of 2n. $\endgroup$ – The Demonix _ Hermit Sep 27 at 11:01
  • $\begingroup$ Why would you compute $2n$ instead of $n$? Not that it matters much in terms of identifying the sequence, but it seems very misleading. $\endgroup$ – lulu Sep 27 at 11:02

Writing $p:=4n+1$, your question gets restated as follows: $$ \text{For what values of $p$ is $\ 2^{\frac{p-1}2}\equiv 1\!\!\!\pmod p$?} $$ For $p$ prime, this means that $2$ is a quadratic residue mod $p$; that is, $p\equiv 1\pmod 8$. For $p$ composite, this implies $2^{p-1}\equiv 1\pmod p$; that is, $p$ is a base-$2$ Fermat pseudoprime. There are infinitely many Fermat's pseudoprimes, but they are rare. I doubt it is possible to characterize Fermat pseudoprimes $p$ with $2^{\frac{p-1}2}\equiv 1\pmod p$.


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