2023/033) Show that if 5 does not divide n then 5$|n^4-1$

We have $n^4-1 = (n^2+1)(n^2-1)$ difference of 2 squares

 $= (n^2+ 1)(n+1)(n-1)$ difference of 2 squares

$= (n^2-4 + 5) (n+1)(n-1) as$n^2+ 1=n^2-4 + 5$ 

$= (n^2-4)(n+1)(n-1)+ 5(n+1)(n-1)$

$= (n-2)(n+2)(n+1)(n-1)+ 5(n+1)(n-1)$

$\frac{(n-2)(n-1)n(n+1)(n+2)}{n} + 5(n+1))(n-1)$

  $5(n+1))(n-1)$ is divisible by 5

$(n-2)(n-1)n(n+1)(n+2)$ being product of 5 consecutive numbers is divisible by 5

As n is not dvisible by 5 so $\frac{(n-2)(n-1)n(n+1)(n+2)}{n}$ is divisible by 5

Hence the sum is divisible by 5 and so the given expression

Note: this can be done faster using mod 5 arithmetic but I have chosen a different way.

Aditionaly this is direct from fermats little theorem