Power to the 5

**AndroidMachine** · June 7th 2012, 04:20 AM

Hi everyone

I am having with this question:

n is an integer.

Show that n⁵ - n is divisible by 30 for all integers n.

If it helps, in the questions before this one, we proved that the last digit of n⁵is n.

Thanks

**emakarov** · June 7th 2012, 06:21 AM

Originally Posted by AndroidMachine

If it helps, in the questions before this one, we proved that the last digit of n⁵is n.

So, the last digit of 786⁵ is 786, right?

One way is to factor n⁵ - n into 4 factors and then consider 5 cases depending on the remainder of n when divided by 5. In each case, find the factors among the 4 ones that are divisible by 2, 3 and 5.

Another way is to use mathematical induction.

**Soroban** · June 7th 2012, 06:27 AM

Hello, AndroidMachine!

I'll get you started . . .

$n$ is an integer.
Show that $N \:=\:n^5- n$ is divisible by 30 for all integers $n.$

We must show that $N$ is divisible by 2, 3 and 5.

Factor: . $n^5 - n \;=\;n(n^4-1)$
. . . . . . . . . . . . $=\;n(n^2-1)(n^2+1)$
. . . . . . . . . . . . $=\;n(n-1)(n+1)(n^2+1)$

$\text{We have: }\:N \:=\:\underbrace{(n-1)\cdot n\cdot(n+1)}_{\text{3 consecutive integers}}\cdot(n^2+1)$

Facts: .The product of two consecutive integers is divisible by 2.
n . . . . The product of three consecutive integers is divisible by 3.

Now you must show that one of the four factors will be divisible by 5.

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