# Thread: proof of a congruence modulo prime powers

1. ## proof of a congruence modulo prime powers

let the symbol = denote congruence (a=b mod n means a is congruent to b mod n)

let p be a prime number, m,k be two positive integers, p does not divide a

Prove that if p>2 or m>1 and $a=1 mod p^m$ is true, a=1 mod p^(m+1) is false, then:

a^(p^k)=1 mod p^(m+k) is true (I have already proved this)

a^(p^k)=1 mod p^(m+k+1) is false (this is the one I need help on)

thank you! and sorry about the sloppy notation, I don't know how to do the imaging stuff

2. $a=1+rp^m\ \Rightarrow\ a^{p^k}=\left(1+rp^m\right)^{p^k}\equiv1+rp^{m+k}\ pmod{p^{m+k+1}}$

If $a^{p^k}\equiv1\pmod{p^{m+k+1}}$ we would have

___ $rp^{n+k}\equiv0\pmod {p^{m+k+1}}$

$\Rightarrow\ p\mid r$

$\Rightarrow\ a=1+rp^m\equiv1\pmod{p^{m+1}}$ (contradiction)

$\therefore\ a^{p^k}\not\equiv1\pmod{p^{m+k+1}}$

also, why does this fail when p=2 and m=1?

4. Originally Posted by minivan15
In the binomial expansion of $\left(1+rp^m\right)^{p^k}$ the first two terms are 1 and $\left(p^k\right)\left(rp^m\right)=rp^{m+k}$ and the last term is $r^{p^k}p^{mp^k};$ all the other terms contain $p$ to powers greater than $m+k.$ If $m>1$ or $p\ge3$ then $mp^k\ge m+k+1$ as well. Hence $\left(1+rp^m\right)^{p^k}\equiv1+rp^{m+k}\pmod{rp^ {m+k+1}}$ if $m>1$ or $p\ge3.$
However, when $p=2$ and $m=1$ it is possible for $mp^k e.g. $2<1+1+1.$ Hence the congruence $\left(1+rp^m\right)^{p^k}\equiv1+rp^{m+k}\pmod{rp^ {m+k+1}}$ breaks down if $p=2$ and $m=1.$