# Finding factor with primes [closed]

If $p$ is a prime other then $2$, express the general prime factor of $2^p-1$ in terms of p and some other integer.

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Let prime $q\mid (2^p-1)\implies ord_q2\mid p$

If $ord_q2=1,2^1\equiv 1\pmod q\implies q\mid (2-1)$ which is impossible.

SO, $ord_q2=p$ and $ord_q2\mid\phi(q)\implies p\mid\phi(q)\implies p\mid(q-1)$

For prime $q>2,q-1$ is even as $q$ must be odd.

So, $2\mid (q-1)\implies lcm(2,p)\mid (q-1)$

But $lcm(2,p)=2p$ as $p$ is odd.

So, $2pk=q-1\implies q=2pk+1$ for some natural number $k$

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Hey, I've just started studying number theory. When you refer to $\phi(q)$ do you mean the Euler $\phi$-function? which counts the number of integers smaller then q which have a gcd of 1 with q? Also what does $ord_q2|p$ actually represent? –  user48133 Nov 4 '12 at 11:58
@user48133, yes its Euler $\phi$ function. $ord_qa$ multiplicative/modulo order(mathworld.wolfram.com/MultiplicativeOrder.html) –  lab bhattacharjee Nov 4 '12 at 12:14