1. ## Prove problem

Hi,
Could someone please explain how to do this:

Prove that n^5 - n is divisible by 10 for all positive integer n.
[Hint: a number is divisible by 10 if and only if it is divisible by 5 and ....]

thanks!

2. Originally Posted by quah13579
Hi,
Could someone please explain how to do this:

Prove that n^5 - n is divisible by 10 for all positive integer n.
[Hint: a number is divisible by 10 if and only if it is divisible by 5 and ....]

thanks!
(your hint is incorrect with the double implication, while it is true that if a number is divisible by 10 then it is divisble by 5, it is not true to say that if a number is divisible by 5 then it is divisible by 10 for ex:25, though I'm guessing a 2 shoulda been after that "and")

Proof by induction

Base case n=1, $1^5-1=1-1=0$ and 0 is divisible by 10 since 0=10*0
(an integer a is divisible by an integer b if there exists some c such that a=bc)

Now assume $n^5-n$ is divisible by 10... which means $n^5-n=10k$ for some $k\in\mathbb{Z}$

Consider $(n+1)^5-(n+1)=n^5+5n^4+10n^3+10n^2+5n+1-n-1=n^5-n+5n^4+10n^3+10n^2+5n$ (we just rearranged here and cancelled the 1's)

By our assumption we can substitute in for the first part of the expression so

$=10k+5n^4+10n^3+10n^2+5n=10(k+n^3+n^2)+5(n^4+n)$

Now show $n^4+n$ is divisible by 2 for all positive integers, then the 5*2 will give us a 10 and the entire expression will be divisible by 10

Can you do this induction?

3. Originally Posted by quah13579
Hi,
Could someone please explain how to do this:

Prove that n^5 - n is divisible by 10 for all positive integer n.
[Hint: a number is divisible by 10 if and only if it is divisible by 5 and ....]

thanks!
$n^5-n$ is divisible by 5 by Fermat’s little theorem. It is also even (as $n^5$ and $n$ are either both even or both odd). Hence $n^5-n$ is divisible by 10.