show $p\mid 2^n-n$ for infinitely many $n$

Question

Show that $p\mid 2^n-n$ for infinitely many $n$. $p$ is a prime and $n$ is an integer.

I tried using Fermat's little theorem and got $2^p-p\equiv2\pmod p$ and $2^{p-1}-(p-1)\equiv2\pmod p$. So I can't even find one $n$ that satisfy the condition.

Any helps and hints appreciated. (even just one $n$ that works for every $p$)

@TheSimpliFire The problem says it's a prime. If $p=2$ then the problem is trivial. So we can assume it's an odd prime. — abc..., Feb 17 '19 at 09:55
@TheSimpliFire I meant for every $p$, there exists some $n$, because I thought prove for a particular $p$ won't be very helpful in this problem... I just edited my post. — abc..., Feb 17 '19 at 09:58

score 10 · Accepted Answer · answered Feb 17 '19 at 10:03

10

Put $$n = (p-1)(kp-1)$$

where $k$ is arbitrary positive integer.

Then $$ 2^n-n = (2^{p-1})^{kp-1}-(p-1)(kp-1)\equiv 1-1 =0 \pmod p$$

answered Feb 17 '19 at 10:03

nonuser

91,557

Nice! How did you come up with this? What's the intuition here? – abc... Feb 17 '19 at 10:07
1

First thought was, how to get out $2^n$. This is easy by FLT, put $n= (p-1)\times$ something. Then we have $1-(p-1)\times$ something must be divisible by $p$. When would that be? – nonuser Feb 17 '19 at 10:09

show $p\mid 2^n-n$ for infinitely many $n$

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