An application of Fermat's little theorem

Question

Given that $p$ be a prime and $a$ and $b$ are integers not divisible by $p$, prove that $$(a+b)^{p} \equiv( a^p + b ^p ) \pmod{p}$$

My approach : By Fermat's little theorem, $$a^p \equiv a \pmod{p}$$ and $$b^p \equiv b \pmod{p}$$ Then $$a^p + b^p \equiv (a + b) \pmod{p}$$ Either $p \mid (a+b)$ or $p \nmid (a+b) $. In the former, $$(a+b)^p \equiv 0 \equiv (a+b) \pmod{p}$$ In the latter, $$(a+b)^p \equiv (a+b) \pmod{p}$$ follows from F.L.T.

1. Is this approach correct ?

2. How to prove the result starting with the binomial theorem: $$(a+b)^p = a^p + \sum_{k=1}^{p-1} a^{k} b^{p-k} {p \choose k} + b^p \equiv (a+b) + \bigg\{ \sum_{k=1}^{p-1} a^{k} b^{p-k} {p \choose k} \bigg\} \pmod{p}$$ How do we show that the term within the curly braces is $0 \pmod{p}$?

Any help is much appreciated.

Answer 1

Your approach is right. To solve it using binomial theorem, note that for any $0<n<p$ we have: $${p \choose n} =\frac{p!}{n! (p-n)! }$$ Since $n, p-n<p$, the numerator has a factor of $p$ while the denominator doesn't. This means that the whole expression in the curly brackets in your question is divisible by $p$.

Answer 2

Yes, your approach is right, but you do not need to consider the case $a+b$ divisible by $p$: the equation $a^p\equiv a\pmod p$ holds for all integer $a$.

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To show that $$ \sum_{k=1}^{p-1}a^kb^{p-k}\binom pk=0, $$ it suffices to show that $\binom pk=0,\,\forall k=1,\ldots,p-1$.

Write $\binom pk=\frac{p!}{k!(p-k)!}=\frac{p\cdot(p-1)\cdots(p-k+1)}{k!}$.

Now $p$ divides the numerator but does not divide the denominator of the fraction, so $p\mid\binom pk$.

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Hope this helps.