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
So, the last digit of 786^{5} is 786, right?
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.
Hello, AndroidMachine!
I'll get you started . . .
$\displaystyle n$ 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.