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I believe that this may require the use of Fermat's Little Theorem. I rewrote it as $(a+1)^p - a^p \equiv 1 \pmod{p}$ because the right-hand side looks similar to Fermat's Little Theorem, but I was unable to figure out how I can get the left-hand side to become $a^{(p-1)}$.

FoiledIt24
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  • Welcome to Math Stack Exchange. You could expand $(a+1)^p$ using the binomial theorem – J. W. Tanner Apr 07 '19 at 15:41
  • Which version of little Fermat do you already know? – Bill Dubuque Apr 07 '19 at 15:46
  • I'm using a^(p-1) ≡ 1 (mod p). – FoiledIt24 Apr 07 '19 at 15:47
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    You left out the (necessary) hypothesis that $,p,$ is prime, and $,a\not\equiv 0,$ in little Fermat. – Bill Dubuque Apr 07 '19 at 15:56
  • Fixed it, thanks for noticing that. – FoiledIt24 Apr 07 '19 at 15:56
  • Btw, if you seek only proofs using little Fermat then you should explicitly emphasize that (else you may get swamped by other methods). This is surely a dupe but I couldn't find a good target with a quick search. – Bill Dubuque Apr 07 '19 at 16:03
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    Cf. this question – J. W. Tanner Apr 07 '19 at 16:11
  • @rtybase Not a good dupe target since it includes only 1 method - likely not the intended method since here the OP already knows little Fermat so there is no need to reprove it (see my hint below, or fleablood's elaboration). – Bill Dubuque Apr 07 '19 at 16:58
  • @J.W.Tanner See the prior comment. – Bill Dubuque Apr 07 '19 at 16:58
  • @BillDubuque I am not surprised, you disagree quite often lately. OP should have searched before asking. This (quite popular, including in literature) question was asked before, I just don't remember all the links .. – rtybase Apr 07 '19 at 17:09
  • ... also, 3 of the answers below (including the accepted one) use the same technique from the dupe I suggested. This reduces the cardinality of original methods for the answers to this question. Also, OP mentioned "I believe that this may require the use of ..." which doesn't sound too confident. Thus, my reason for the closing vote as a dupe. – rtybase Apr 07 '19 at 17:24
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    @rtybase I have no idea what "disagree quite often" refers too. But I do strive to find good dupe targets because I take pedagogical matters quite seriously. – Bill Dubuque Apr 07 '19 at 17:27

5 Answers5

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Hint: Expanding $(a+1)^p$ you get $\sum_{k=0}^{k=p}\binom{p}{k}a^k$. Now first prove that $p\mid\binom{p}{k}$, when $0<k\le p-1$ and $p$ is prime. After that you will left with $a^p+1\pmod{p}$.

tarit goswami
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  • But $p\not \mid {p \choose k}$. Example ${4\choose 2}= 6$. (And $(a + 1)^4 = a^4 + 4a^3 + 6a^2 + 4a + 1$.) – fleablood Apr 07 '19 at 15:54
  • $p$ is prime here. Otherwise, the equality OP trying to prove is not true. Thanks for mentioning it. – tarit goswami Apr 07 '19 at 15:54
  • I was able to show that p ∣ (p k)=(p(p-1)...(p-k-1))/k!, but then what is a^p+1 (mod p) congruent to? – FoiledIt24 Apr 07 '19 at 16:01
  • See that, $p$ does not divide $\binom{p}{0}a^0=1$ and $\binom{p}{p}a^p=a^p$. Except these two, all are divisible by $p$. So the remainder left after dividing by $p$ is $1+a^p$. If you want to apply Fermat Little thm.(given that $p$ does not divide $a$) you will have $a^p+1\equiv a+1\pmod{p}$, and when $a\equiv 0\pmod{p}$ then obviously we have $a^p+1\equiv 1\equiv a+1\pmod{p}$ . Got it? – tarit goswami Apr 07 '19 at 16:04
  • I think I get it now, thanks! – FoiledIt24 Apr 07 '19 at 16:06
  • @MattSchiff Since you already know little Fermat, it is highly unlikely that this is the intended proof since it essentially amounts to completely reprovng little Fermat. Rather, the result follows more simply (immediately!) by the method I hinted (now given in full detail in fleabloods answer). – Bill Dubuque Apr 07 '19 at 16:36
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$$(a+1)^{p}=_{0}^{p}\textrm{C}(a)^{p}+_{1}^{p}\textrm{C}(a)^{p-1}.......+1$$ in each combination you will get "prime p" in multiplication,but you are working under mod(p) so all the terms became zero except $$a^{p}+1$$ and this is your answer

gaurav saini
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Hint 1:

$$(a+1)^p=\sum_{i=0}^{p}a^i\binom{p}{i}$$

Hint 2:

What is $\binom{p}{i}~\text{mod}~p$?

cansomeonehelpmeout
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Hint $ $ Use little Fermat to show $\,\color{#c00}n^p \equiv \color{#c00}n\pmod{\!p}\, $ for all $\,n\,$ (separate into cases $\,n\equiv 0\,$ or not), then use that to show $\,(a+1)^p$ and $\,a^p+1\,$ are both $\,\equiv a+1\pmod{\!p}\ $ [by specializing $\,\color{#c00}n \equiv a+1\,$ and $\,\color{#c00}n\equiv a,\,$ above, while using the Congruence Power Rule].

Bill Dubuque
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Note: If $p$ is prime and $p|a$ then $a\equiv 0\pmod p$ and $a^p \equiv 0 \equiv a \pmod p$.

And if $p \not \mid a$ then $a^{p-1}\equiv 1 \pmod p$ by FLT.

So $a^p = a^{p-1}a \equiv a \pmod p$.

So for any integer $a^p \equiv a \pmod p$ whether or not $p\mid a$.

.....

So $(a+1)^p \equiv a+1 \pmod p$.

And $a^p + 1\equiv a+1 \pmod p$.

That's it.

fleablood
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