Solve Exercise with Modulo: Is this Correct?

[mathmari]

Hey!

I have the following exercise:

If $p$ is prime, $p \nmid a$, $p \nmid b$, prove that $$a^p \equiv b^p \pmod p \Rightarrow a^p \equiv b^p \pmod {p^2}$$

My idea is the following:
$$ a^p \equiv b^p \pmod p $$
$$ a^p \equiv a \pmod p $$
$$ b^p \equiv b \pmod p $$

$$ a \equiv b \pmod p \Rightarrow a=b+kp, k \in \mathbb{Z}$$

$$ a^p=(b+kp)^p=\sum_{m=0}^{p} \binom{p}{m} (kp)^mb^{p-m}=\binom{p}{0}b^p+\binom{p}{1}(kp)b^{p-1}+ \dots + \binom{p}{p}(kp)^p= \\ b^p+kp^2b^{p-1}+ \frac{1}{2}(p-1)p^3b^{p-2}k^2+\dots + k^pp^p$$

$$ p^2 \mid kp^2b^{p-1}+ \frac{1}{2}(p-1)p^3b^{p-2}k^2+\dots + k^pp^p
\Rightarrow p^2 \mid a^p-b^p $$

Is this correct?? (Wondering)

 

[Deveno]

That looks correct.

 

[mathmari]

That looks correct.

Do we not have to prove that $$kb^{p-1}+ \frac{1}{2}(p-1)pb^{p-2}k^2+\dots + k^pp^{p-2} \in \mathbb{Z} $$?? (Wondering)

 

[I like Serena]

Do we not have to prove that $$kb^{p-1}+ \frac{1}{2}(p-1)pb^{p-2}k^2+\dots + k^pp^{p-2} \in \mathbb{Z} $$?? (Wondering)

Yep. We do.
Can you? (Wondering)

 

[I like Serena]

Yep. We do.
Can you? (Wondering)

Consider that each of the binomial coefficients is of the form:
$$\binom p k = \frac{p \cdot (p-1) \cdot ... \cdot (p-k+1)}{1\cdot 2 \cdot ... \cdot k}$$
It is given that this is an integer.
So each of those factors in the denominator come back in the numerator somehow.
Can any of them contain a factor that divides $p$? (Wondering)

 

[Deveno]

One can prove that:

$\displaystyle \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$

which furnishes an easy inductive proof that for all $0 \leq k \leq n$ that $\displaystyle \binom{n}{k}$ is an integer, given:

$\displaystyle \binom{n}{0} = \binom{n}{n} = 1$, for all $n \in \Bbb Z^+$.

Then the only fact we need in the expansion of $(b + kp)^p$ is that $\displaystyle p|\binom{p}{1}$

since $p^2$ occurs in all the other terms of the expansion, no matter what the coefficient is.