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Let $k\ge 1, m\ge 1.$ Show that if $x\equiv 1 \pmod {m^k}, $then $x^m \equiv 1\pmod{m^{k+1}}$.

First I noticed that the assumption would imply $x^m \equiv 1 \pmod{m^k}$, but that doesn't seem to work out. Then I tried:

$x\equiv 1 \pmod {m^k}$ implies that $x=m^k n+1$ for some $n$. If I raise it to the $m$-th power, I would get: $x^m=(m^k n+1)^m=\sum_{j=0}^m C(m,j)m^{kj}n^j$.

Now if the proposition is true, then $m^{k+1}$ must divide $1+\sum_{j=0}^m C(m,j)m^{kj}n^j$. Each term in the binomial expression are divisible by $m^{k+1}$ except for when $j=0$ and $j=1$. Take them out, we get $1+C(m,0)+C(m,1)m^kn=2+m^{k+1}n$. Everything look perfect except for that $2$...

3x89g2
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  • "now if the proposition is true, then $m^{k+1}$ must divide " should be followed by $1 - \Sigma$ not $1 + \Sigma$. – Callus - Reinstate Monica Apr 22 '15 at 04:23
  • Aw man...so then I would simply get $-m^{k+1} n$, which is exactly what I want, correct? – 3x89g2 Apr 22 '15 at 04:26
  • Yes, that is correct. Except, I just noticed another small error. I don't think your statement about divisibility of the $C(m,j)$ is quite correct. For example, $C(m,m)=1$ – Callus - Reinstate Monica Apr 22 '15 at 04:43
  • But then $m^{mk}n^{m}$ is divisible by $m^{k+1}$, right? – 3x89g2 Apr 22 '15 at 05:15
  • Okay, maybe I misunderstood what you meant by binomial expression. It is true that $m^{k+1}$ divides every term $C(m,j)m^{kj}n^j$ except when $j=0$. – Callus - Reinstate Monica Apr 22 '15 at 05:19
  • And also when $j=1$ I guess? Sorry I'm too sleepy right now, but I'm trying to say that we pull out those terms that are NOT divisible by $m^{k+1}$ and combine them with the $-1$ and we should get what we want. – 3x89g2 Apr 22 '15 at 05:20

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Hint. If $$x=1+nm^k$$ then by the binomial theorem $$x^m=(1+nm^k)^m=1+\binom m1nm^k+\binom m2(nm^k)^2+\cdots\ .$$ Do a bit of simplification and see what you notice about every term (except the first) on the RHS.

In your solution, near the end, you have a $+$ sign instead of a $-$.

David
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