2

Let $p$ be prime, $a,b \in \mathbb{Z}$. Prove that: $$(a+b)^p \equiv a^p + b^p\pmod p$$ How to deal with it.

user180834
  • 1,463
  • 1
    http://math.stackexchange.com/questions/1118819/proof-if-p-is-prime-and-0kp-then-p-divides-p-choose-k – Bman72 Jan 31 '15 at 13:15
  • 1
    This question is not a duplicate of @Ale's reference, even though one solution is to use that reference. So this question probably should not be closed as a duplicate. The OP should show some of his work, however. – Rory Daulton Jan 31 '15 at 14:13
  • @RoryDaulton I apologize if my comment has been seen as a report of a duplicate. It was not my intention :). I pointed that link out because it could have helped the OP in solving his problem. – Bman72 Jan 31 '15 at 17:46
  • 1
    @Ale: My comment was not directed at you but at whoever had already voted to close this question. I thought that perhaps he considered the question to be a duplicate based on your comment. I wanted people to close this question for the right reason: note that I also voted to close this question as off-topic, since the questioner has not shown that he has done any work of his own. – Rory Daulton Jan 31 '15 at 17:55

4 Answers4

6

Hint:-

$$(a+b)^p\equiv a+b \pmod p$$ $$a^p\equiv a \pmod p$$$$b^p\equiv b \pmod p$$

2

Hint: Show that ${p\choose k}$ is divisible by $p$ if $k\neq 0 \ or \ p$ so that ${p\choose k}a^kb^{p-k}=0$

mesel
  • 15,125
2

$\textbf{Hint:}$ First use Binomial theorem:

$$(a+b)^p=\sum_{k=0}^{p}{p \choose k}a^{k}b^{p-k}$$

Next prove that for all $k$ that $0<k<p$:

$$p\left|{p \choose k} \right.$$

You can do it using Multiplicative formula for binomial coefficient.

agha
  • 10,174
2

By binomial theorem $\displaystyle \left({a + b}\right)^p = \sum_{k \mathop = 0}^p \binom p k a^k b^{p-k}$

Also $\displaystyle \sum_{k \mathop = 0}^p \binom p k a^k b^{p-k} = a^p + \sum_{k \mathop = 1}^{p-1} \binom p k a^k b^{p-k} + b^p$

so $\displaystyle \forall k: 0 < k < p: \binom p k\equiv0\pmod p$ by (my answer)

therefore $\displaystyle \binom p k a^k b^{p-k}\equiv0\pmod p$ and then $\displaystyle \sum_{k \mathop = 1}^{p-1} \binom p k a^k b^{p-k}\equiv0\pmod p$

so $\displaystyle \sum_{k \mathop = 0}^p \binom p k a^k b^{p-k}\equiv a^p+b^p\pmod p$

Community
  • 1
user153330
  • 1,822
  • 18
  • 41