Can you prove that n^5 - n is divisible by 5 for all n in ZZ by the principle of mathematical induction?

1 Answer
Feb 1, 2017

See proof below

Explanation:

Let the statement be P(n)=n^5-n

P(1)=0, this is divisible by 5, the statement is true for n=1

P(k)=k^5-k=5m, where m in ZZ

aassume that the statement is true for n=k

Then,

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

=k^5+5k^4+10k^3+10k^2+5k+cancel1-k-cancel1

=(k^2-k)+5(k^4+2k^3+2k^2+k)

=5m+5(k^4+2k^3+2k^2+k)

=5(m+(k^4+2k^3+2k^2+k))

Therefore, P(k+1) is divisible by 5, the statement is true

We proved that the statement is true for all n in ZZ