# Congruence Modulo Proof

1. Feb 27, 2014

### knowLittle

1. The problem statement, all variables and given/known data
Suppose n is an integer which is not a divisor of 5.
Prove that $n^{4} \equiv 1 mod5$

3. The attempt at a solution
I know that 16 mod 5 is equivalent to 1 mod5.

$16 = 2^{4}$

2 is not a divisor of 5. How do I prove this for the general case?

I know
$n^{4} -1 = 5k,$ for some k $\in Z$

2. Feb 27, 2014

### tiny-tim

hi knowLittle!

(you mean, not a multiple of 5 ! )
if you know that, then that's just another way of saying $n^{4} \equiv 1 mod5$

(otherwise, you know it for n = 2, and obviously for n = 1, so how many more cases do you need to prove?)

3. Feb 27, 2014

### knowLittle

Hi TINY-tim,

The problem states divisor. Also, there's a typo in the $\equiv$ symbol; in the paper it shows '='.

I understand that n=2 and n=1 and satisfies the statement. But, how do I prove this, I am confused as to how to prove this. I assume they want me to prove a general case?

4. Feb 27, 2014

### tiny-tim

hi knowLittle!
yes

i don't know how much you know about mod calculations

since you can prove it for n = 2 (because 16 - 1 = 15), does that help you with any other values of n ?

and since it's obviously true for n = 1, does that help you with any other values of n ?

and are there any other cases left?

(alternatively, do you know Fermat's little theorem?)

5. Feb 27, 2014

### knowLittle

We didn't explicitly covered Fermat's theorem, but I will take a look. I am sure that the solution requires elementary knowledge, for we have not covered much.

I am still lost :/

6. Feb 27, 2014

### tiny-tim

if 24 - 1 is divisble by 5, what can you say about (5n + 2)4 - 1 ?

7. Feb 27, 2014

### knowLittle

If 5n+2 is a multiple of 2, then $(5n+2)^{4} -1$ would be divisible by 5. Right?

8. Feb 27, 2014

### tiny-tim

right!

and $(5n+1)^{4} -1$ ?

and $(5n-1)^{4} -1$ ?

and $(5n-2)^{4} -1$ ?

what other cases do you need?