# For what values of n is $2^{2n} -1$ divisible by $4n+1$

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.

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