For every integer $ n\ge 0$, show that
$$ n^5 -n $$ is divisible by 5.
This problem is a tutorial about induction. Notice that the statement is true for n=0. Suppose it is true for n. Let us show it for n+1. Notice that
$$
(n+1)^5-(n+1)=n^5+5 n^4+10 n^3+10 n^2+4 n=n^5+5 n^4+10 n^3+\ 10 n^2+5 n-n= \ (n^5-n)+5 n^4+10 n^3+10 n^2+5 n

$$
It is clear that the second side is divisible by 5.

Question

For every integer $ n\ge 0$, show that 
$$ n^5 -n $$ is divisible by 5. 
This problem is a tutorial about induction. Notice that the statement is true for n=0. Suppose it is true for n. Let us show it for n+1. Notice that
$$
(n+1)^5-(n+1)=n^5+5 n^4+10 n^3+10 n^2+4 n=n^5+5 n^4+10 n^3+\ 10 n^2+5 n-n= \ (n^5-n)+5 n^4+10 n^3+10 n^2+5 n

$$
It is clear that the second side is divisible by 5.

anonymous · Answer

Missing a vital part of a proof by induction: a conclusion.

anonymous · Answer

sometimes I just find induction so magical

ganeshie8 · Answer

nice :) division algorithm also gives a neat proof :

n^5-n = n(n^4-1) = n(n^2-1)(n^2+1) = (n-1)n(n+1)(n^2+1)
clearly it is divisible by 5 for n = 5k, 5k+1, 5k+4
for$n = 5k \pm 2$,   5 | (n2+1) ending the proof

anonymous · Answer

That is a nice proof @ganeshie8