Question

show that for all natural numbers, n

1. n^5 - n is divisible by 5.

any help please?

Answer 1

but that's not true...
for example, if n=2, a natural number, then n^5=32.
but 32 is NOT divisible by 5.

can you check if the problem is written correctly?

Answer 2

oops!!. mistake!!

EDIT: n^5 - n is divisible by 5

Answer 3

ok... maybe it'll help if we factor first:

\(\large n^5-n=n(n^4-1)=n(n^2+1)(n^2-1)=n(n^2+1)(n+1)(n-1) \)

Answer 4

i'm just thinking aloud but:
notice that n, (n+1), (n-1) are three consecutive natural numbers...

Answer 5

but if n = 4 then this will be come

4, 5, 3 which are not consecutive....

Answer 6

well, not in that order that you wrote. but 3, 4, 5 are consecutive

Answer 7

and when n=4, since one of the factors is a 5, that proves that when n=4, n^4-n is divisible by 5.

Answer 8

so if n = 10 i would get
9, 10, 11

i haven't shown n^5 - n is divisible by 5 then :(

Answer 9

yes... the proof has to be more formal than this but if n=10, 10 is divisible by 5 so 10^5-10 is

Answer 10

*also

Answer 11

we know this. but HOW do i show it? how do i make it "formal"?

Answer 12

i'm sorry... i'm hitting dead ends with whatever i'm trying...

Answer 13

is this for an analysis class?

Answer 14

here's a proof by induction of the expression you want:
http://www.physicsforums.com/showthread.php?t=245925

Answer 15

thank you very much...