# How to find $(a,n)$ such that : $5^a+1 \equiv 0 \pmod {3\cdot 2^n-1}$ and $3\cdot 2^{n-1}-1 \equiv 0 \pmod a$?

Is it possible to find such integer pair $(a,n)$ that :

$\begin{cases} 5^a+1 \equiv 0 \pmod {3\cdot 2^n-1} \\ 3\cdot 2^{n-1}-1 \equiv 0 \pmod a\\ \end{cases}$

where $n \equiv 3 \pmod 4$

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I still wonder every now and then why it was that you wanted to know this. –  Tara B May 13 '12 at 13:00

Yes, for example $n = 3$, $a = 11$. Why did you want to know?
EDIT: I guess that only answers the 'Is it possible?' part. As for how to find $(a,n)$, well actually I used my computer, but $(11,3)$ is the smallest pair that could possibly work, so it would have made sense to check it first. So far my computer has only found one other pair: $(191,7)$. My code probably isn't very good, though.
UPDATE: Also $(3071,11)$. Note that $191 = 3*2^6 - 1$ and $3071 = 3*2^{10} - 1$, but it is not the case that the pair $(3\cdot 2^{n-1} - 1, n)$ works for every $n\equiv 3 \mod 4$. This fails already for $n = 15$.