Let $m^n-1$ be prime. What can $m$ be?

Question

Let $m^n-1$ be prime. What can $m$ be if $m$ and $n$ are not $1$?

How can I find $m$?

http://mathworld.wolfram.com/MersenneNumber.html – lab bhattacharjee May 01 '13 at 04:14 — lab bhattacharjee, May 01 '13 at 04:14

score 17 · Answer 1 · answered May 01 '13 at 01:14

17

Hint: $m^n-1 = (m-1)(m^{n-1}+m^{n-2}+\cdots+1)$.

answered May 01 '13 at 01:14

Abel

7,312

Still don't get it... – user75283 May 01 '13 at 01:17
m = 2 and m-1=1 .. – Halil Duru May 01 '13 at 01:20
@user75283 If $m>2$ then the RHS is a composite number. – Pedro May 01 '13 at 01:23
sorry for the dumb question, but what is RHS ? – Integral May 01 '13 at 01:24
1

"Right hand side" (of the equation). – Pedro May 01 '13 at 01:24

score 2 · Answer 2 · answered May 01 '13 at 01:31

2

If $m = 1$, the expression gives 0, not a prime. Ruled out.

If $m \ge 2$, we have the factorization $m^n - 1 = (m - 1) (m^{n - 1} + m^{n - 2} + \ldots + 1)$. This can only be a prime if $n = 1$ (in which case the second factor is 1) or if $m = 2$ (when the first factor is 1).

So, either $n = 1$ and $m - 1$ is a prime or $m = 2$ and $n \ge 2$ (not all $n$ work, mind you).

answered May 01 '13 at 01:31

vonbrand

27,812

2^4 - 1 = 15 and 7^2-1=13 – user75283 May 01 '13 at 01:45
So m could be 7? – user75283 May 01 '13 at 01:45
@user75283, $7^2 - 1 = (7 - 1) \cdot (7 + 1) = 48$. And no, $m$ is a prime + 1, i.e., $m = 1010$ works fine. – vonbrand May 01 '13 at 01:48
3+1=4, 4^2=16 , 16-1=15, 15 isn't prime – user75283 May 01 '13 at 01:58
It is one or the other pair of cases. You can have $n = 1$, in which case $m = p + 1$ for $p$ prime works. Or for $n > 1$ it has to be $m = 2$. – vonbrand May 01 '13 at 02:00

Let $m^n-1$ be prime. What can $m$ be?

2 Answers2