congruence proof modulo a prime power

[dts5044]

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 and 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

 

[stapel]

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

To learn how to use LaTeX, use any of the on-topic links in the "Forum Help" pull-down menu at the very top of every forum page.

To learn how to use standard web-safe textual formatting, follow either of the formatting links you saw when you read the "Read Before Posting" thread.

Thank you!

Also, kindly please confirm or correct the following as being the exercise:

\(\displaystyle \mbox{Let }\, p\, \mbox{ be a prime number, let }\, m,\, k\,\mbox{ be two}\)

\(\displaystyle \mbox{positive integers, and suppose that }\, p\, \mbox{ does not divide }\, a\)

\(\displaystyle \mbox{Prove that, if }\, p\, >\, 2\, \mbox{ or }\, m\, >\, 1,\, \mbox{ and }\, a\, \equiv\, 1\, \mbox{mod }\, p^m\)

\(\displaystyle \mbox{is true but }\, a\, \equiv\, 1\, \mbox{mod }\, p^{m+1}\, \mbox{ is false, then:}\)

\(\displaystyle \mbox{a. }\, a^{p^k}\, \equiv\, 1\, \mbox{mod }\, p^{m+k}\, \mbox{ is true.}\)

\(\displaystyle \mbox{b. }\, a^{p^k}\, \equiv\, 1\, \mbox{mod }\, p^{m+k+1}\, \mbox{ is false.}\)

 

[dts5044]

yeah, that's it!