Euler's Theorem, can anyone lend a hand?

Question

ive got this question that im struggling with,
Prove that 5 divides n^5 -n
using Euler's theorem if 5 does not divide n

Answer 1

Initial thoughts...
1) If 5 doesn't divide n, then that means that 5 and n are coprime.
2) φ(5) = 4 (This is Euler's totient function... φ(5) = 4, because there are 4 positive integers less than 5 that are coprime to 5 -> 1, 2, 3, and 4)

So, here's Euler's theorem:
a^φ(b) ≡ 1 (mod b)

So: n^φ(5) ≡ 1 (mod 5); or n^4 ≡ 1 (mod 5)

Now: n^5 -n = n (n^4 - 1) [Just factor out the "n"]

We know that 5 does not divide n (because that's given), so for 5 to divide this expression, it must divide "n^4 - 1"

Remember, we know that: n^4 ≡ 1 (mod 5)
In other words, when you divide n^4 by 5, you get a remainder of 1.
So, if you're dividing "n^4 - 1" by 5, it must divide perfectly.

Answer 2

Case 1: n^5 - n = n(n^4 - 1), therefore if 5 divides n, it divides n^5 - n. That was easy.

Case 2: if p is prime and p does not divide n, then it divides n^p - n. This is Fermat's Little Theorem, which is an earlier special case of Euler's Theorem.

Euler's Theorem deals with composite numbers as well as primes, so you do not need it here, just its special case for primes.

Answer 3

Sorry, I'm not familiar with Euler's theorem (which one? mathworld.wolfram.com describes several -- perhaps it's the totient theorem?), but I can prove that 5 divides n^5 - n.
If 5 does not divide n, then
n = 5p ± k, where k = 1 or 2.

Now n^5 - n
= n(n^2+1)(n+1)(n-1)

If k = 1, then either n + 1 or n - 1 is equal to 5p,
and so 5 divides n^5 - n.

If k = 2, then
n^2 + 1
= 25p^2 ± 10pk + k^2 + 1
= 5(5p^2 ± 2pk) + 4 + 1
= 5(5p^2 ± 2pk +1)

Q.E.D.

Answer 4

n^5-n = n(n^4-1) = n(n^2+1)(n^2-1) =

= n(n+1)(n-1)(n^2+1)

Now, if any one of the factors above is divisible by 5, then the entire product is also divisible by 5.

Since (n-1), n and (n+1) are consecutive integers, then none of them will divide by 5 only when:

n(mod 5) = 2 or
n(mod 5) = 3

Let's check (n^2 + 1) for each case

a) n=5k + 2, so
(5k+2)^2 + 1 = 25k^2 + 20k +4 + 1 = 5(5k^2 + 4k +1)

b) n=5k + 3, so
(5k+3)^2 +1 = 25k^2 + 30k + 9 +1 = 5(5k^2 + 6k + 2)

Both cases divide by 5. QED

Answer 5

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