Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $n \equiv 3 \pmod 4$

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

1 Answer 1

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 $n = 15$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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