Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $n \in \mathbb N$. Suppose that $p$ is an odd prime number that divides $n^2+1$. Show that if $p=4k+3$ for some integer $k \ge 0$, then $n^{p−1}\equiv−1 \pmod p$.

The question says to use "if $p=4k+3$ then $\frac {(p-1)} 2$ is odd".

So we know $n^2 \equiv -1 \pmod p$

Do I manipulate that into $n^{4k+2} \equiv -1 \pmod p$ ? I tried that and ended up with $n^{4k+2} \equiv -1^k \pmod p$.

How do I use "if $p=4k+3$ then $\frac {p-1} 2$ is odd" ?

share|cite|improve this question
up vote 2 down vote accepted

Modulo $p$ we get $n^{p-1} \equiv (n^2)^{(p-1)/2} \equiv -1^{(p-1)/2} \equiv -1$, which gives the implication you are looking for. By the way this implication is vacuously true, as the odd prime divisors of $n^2 + 1$ are all of the form $4k + 1$.

share|cite|improve this answer
    
Wow that's so simple, thank you. – Bob Nov 5 '13 at 5:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.