Question

Prove that for each n in N, 5 divides n^5-n

Answer 1

by induction, so if n=0 then 0/5?

Answer 2

yes

Answer 3

so p(k) = k^5-k and p(k+1) = (k+1)^5-(k+1)

Answer 4

I'm not sure how to manipulate it

Answer 5

familiar with binomial theorem ?

Answer 6

(k+1)^5 = k^5 + 5M + 1

Answer 7

you can see that if you expand (k+1)^5,
all the middle terms will be multiples of 5

Answer 8

how this is helping ?

Answer 9

(k+1)^5-(k+1) = k^5 + 5M + 1 - (k+1)
= k^5 - k + 5M

Answer 10

oh induction

Answer 11

from induction assumption k^5-k is divisible by 5, so the proof is over.

Answer 12

so basically where it's +5M it shows that it is also divisible by 5?

Answer 13

yes, but i think it would make more sense if you expand it out

Answer 14

do you know how to expand (k+1)^5 ?

Answer 15

remember the (x+1)^n frormula or binomial theorem ?

Answer 16

uh like, k^5+5k^4+10k^3+10k^2+5k-k

Answer 17

exactly, notice that all the middle terms have a factor 5 in common

Answer 18

so they have to be divisible by 5! Perfect thanks!

Answer 19

(k+1)^5 = k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1

Answer 20

you got it!
but there are many other nice ways to prove this, are you supposed to do this using induction only is it ?

Answer 21

yes

Answer 22

okay then your work looks good, ust make sure you justify every step

Answer 23

Thank you :)

Answer 24

you're welcome!