1. ## Dividable by 5

I got to proof that
n^5 - n
is dividable by 5 for n >= 0, n in Natural numbers..
And that by Induction..
The "skelet" was succesfull, now I got to proof that
P(i) => P(i+1)
..
Any ideas?

2. Originally Posted by kirschplunder
I got to proof that
is dividable by 5 for n >= 0, n in Natural numbers..
And that by Induction..
Say that $\displaystyle K^5-K$ is divisible by five.
Then expand this $\displaystyle (K+1)^5-(K+1)$.
After collecting terms, what do you see?

3. This is the famous Fermat' small Theorem for the case n=5 in Number Theory.

4. Originally Posted by Plato
Say that $\displaystyle K^5-K$ is divisible by five.
Then expand this $\displaystyle (K+1)^5-(K+1)$.
After collecting terms, what do you see?
I did that, then I made
K^5 + 1^5 - K - 1 = K^5 - K
But I didn't think that was really the right way?

5. Originally Posted by kirschplunder
I did that, then I made
K^5 + 1^5 - K - 1 = K^5 - K
But I didn't think that was really the right way?
To do these you must understand basic algebra.
$\displaystyle (K+1)^5=K^5+5K^4+10K^3+10K^2+5K+1$
The basic operations are standing in your way.

6. Originally Posted by Plato
To do these you must understand basic algebra.
$\displaystyle (K+1)^5=K^5+5K^4+10K^3+10K^2+5K+1$
The basic operations are standing in your way.
I'm a bit dumb xD Sowwy.

So.

$\displaystyle K^5+5K^4+10K^3+10K^2+5K+1 - (K-1) - (K^5 - K)$
is
$\displaystyle 5K^4+10K^3+10K^2+5K$I got this by substracting P(K) from it, but I don't really get why.

7. $\displaystyle (K+1)^5-(K+1)=(K^5-K)+(5K^4+10K^3+10K^2+5K)$
The sum of two multiples of five is a multiple of five.

8. ## Thank you alot

Thank you, really
I was thinking way too difficult, gotta get rid of that

9. Suppose K leaves a remainder of r when divided by 5,
Then $\displaystyle K^5$ has the same remainder as $\displaystyle r^5$.
So what we need to do is Just check whether the conclusion hold when r take the value from {0,1,2,3,4}.