PRE$-$condition: n $>=0$

$1.$ long square$($int n$)\{$

$2.$

if $($n $==0)$ return $0$;

$3.$

return square$($n$-1)+2*$ n$-1$;

$4.\}$

POST$-$condition: Return the quantity n $*$ n$.$

In other words, assuming N $=$ n$,$ the square algorithm should return N$2.$

Let P$($N$)=$ square$($N$)$ returns N$2,$ for any integers N $>=0.$

Use P$($N$)$ as it$$s defined to prove, using strong mathematical induction, that square

correctly satisfies the POST$-$condition