prove that $(1+pt)^{p^{r-1}} \equiv 1 \pmod {p^r}$

Question

I need to prove that $(1+pt)^{p^{r-1}} \equiv 1 \pmod {p^r}$

the original question is this:

and I'm following the second answer there. I'm trying to use the binomial theorem and having hard time..

any help will be appriciated

What happens when you expand $(1+pt)^{p^{r-1}}$ with the binomial theorem? Try writing $(1+x)^n$ then substitute $x=pt$ and $n=p^{r-1}$. — Thomas Andrews, Jun 20 '15 at 22:22
i tried both ways. Thomas when i try i'm stuck with the denominators. I understand each number in the sum has $p^r$ in it but it is multiplied by a fraction . and Geoff I couldn't make it by induction. thanks anyway — user2993422, Jun 20 '15 at 22:52
No, it is not multiplied by a fraction, $\binom{p^r}{i}$ is always an integer. @user2993422 — Thomas Andrews, Jun 20 '15 at 23:18
but i need to use the numerator to get p^r so im not sure it stays an int after that — user2993422, Jun 20 '15 at 23:25
The answer by Thomas Andrews indicates how to make the induction work. — Geoff Robinson, Jun 22 '15 at 19:24

score 2 · Accepted Answer · answered Jun 20 '15 at 23:20

2

Here's a good lemma to try to prove. If $s\geq 1$ then:

$$\left(1+kp^s\right)^p\equiv 1\pmod {p^{s+1}}$$

This will let you prove the result above by induction.

answered Jun 20 '15 at 23:20

Thomas Andrews

1 Answers1