Answer:
It's true
Step-by-step explanation:
Operating the square we have:
[tex](n+1)^{2} -1= (n^{2} +2n+1)-1\\ n^{2} +2n+1-1= n^{2} +2n[/tex]
Here we can factorize n:
[tex]n^{2} +2n=n*(n+2)[/tex]
The last line means our number ([tex](n+1)^{2} -1[/tex]) is divisible by n.
The clause of n>2 is true, but even its true for [tex]n\geq 0[/tex] because when n=1 the theorem means that the expression is divisible by 1, but this it's true for every integer, and for n=0 the expression will be 0 and 0 it´s divisible by 0 (Following the definition a|b if a=nb, with a, b and n integers).
