let n = 1
then f(1) = 1^1 - 1 + 2 = 2 so it is true for n = 1
for the next number after n ( n+1) we have f(n+1) =
(n+1)^2 - (n+1) + 2
= n^2 + 2n + 1 - n - 1 + 2
= n^2 + n + 2
= n(n+1) + 2
Now n(n+1) must be divisible by 2 because either n is odd and n+1 is even OR n is even and n+1 is odd and odd & even always = an even number.
So the function is divisible by 2 for n+1 We have shown that its true for n = 1 Therefore it must be true for n = 1,2,3,4 ...
True for all positive integers
