Elemantary number theory

Question

Let's $v$ the smallest positive integer > 1 for which $2 ^ {v} - 1$ divisible by $p$. Prove that $2 ^ {m} - 1$ ($m > v$) divisible by $p$ then and only then $v$ divide $m$?

There is a hint that enough to divide $m$ into $v$ with the remainder, but it gives me nothing.

See my answer here https://math.stackexchange.com/a/3117286/543867 — little o, Mar 23 '19 at 11:10
You are most welcome @Евгений Кондратенко. — little o, Mar 23 '19 at 11:15
@Dbchatto67 can you mark it as duplicate or I should close it? — Evgeny, Mar 23 '19 at 11:21
It is better to close this question to avoid future downvotes. — little o, Mar 23 '19 at 11:24
Please use descriptive titles. “Elementary number theory” (misspelled because somebody already used the insipid title elsewhere) does not help find the question later nor does it help tell readers what this is about. — Arturo Magidin, Mar 25 '19 at 03:33

score 0 · Answer 1 · answered Mar 25 '19 at 03:31

0

$m>v$

If $m=k.v$, then we have:

$2^m-1=[(2^v)^k-1]=(2^v-1)[(2^v)^{k-1}+(2^v)^{k-2}+ . . . +1]$

$p|2^v-1$ ⇒ $p|2^m-1$

answered Mar 25 '19 at 03:31

sirous

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Elemantary number theory

1 Answers1