Dividable by 5

**kirschplunder** · November 26th 2009, 07:31 AM

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?

**Plato** · November 26th 2009, 08:02 AM

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 $K^5-K$ is divisible by five.
Then expand this $(K+1)^5-(K+1)$ .
After collecting terms, what do you see?

**Shanks** · November 26th 2009, 09:28 AM

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

**kirschplunder** · November 26th 2009, 10:01 AM

Originally Posted by Plato

Say that $K^5-K$ is divisible by five.
Then expand this $(K+1)^5-(K+1)$ .
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?

**Plato** · November 26th 2009, 10:24 AM

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.
$(K+1)^5=K^5+5K^4+10K^3+10K^2+5K+1$
The basic operations are standing in your way.

**kirschplunder** · November 26th 2009, 11:03 AM

Originally Posted by Plato

To do these you must understand basic algebra.
$(K+1)^5=K^5+5K^4+10K^3+10K^2+5K+1$
The basic operations are standing in your way.

I'm a bit dumb xD Sowwy.

So.

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

**Plato** · November 26th 2009, 11:40 AM

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

**kirschplunder** · November 26th 2009, 12:45 PM

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

**Shanks** · November 26th 2009, 07:11 PM

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

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Dividable by 5

Thank you alot

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