Hi everyone

I am having with this question:

n is an integer.

Show that n^{5}- 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 n^{5 }is n.

Thanks

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- Jun 7th 2012, 04:20 AMAndroidMachinePower to the 5
Hi everyone

I am having with this question:

n is an integer.

Show that n^{5}- 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 n^{5 }is n.

Thanks - Jun 7th 2012, 06:21 AMemakarovRe: Power to the 5
So, the last digit of 786

^{5}is 786, right? (Smile)

One way is to factor n^{5}- 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. - Jun 7th 2012, 06:27 AMSorobanRe: Power to the 5
Hello, AndroidMachine!

I'll get you started . . .

Quote:

$\displaystyle n$ is an integer.

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

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

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

. . . . . . . . . . . . $\displaystyle =\;n(n^2-1)(n^2+1)$

. . . . . . . . . . . . $\displaystyle =\;n(n-1)(n+1)(n^2+1)$

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

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.