Math Help - The sum of postive divisors Function

1. The sum of postive divisors Function

a) Let p and p+2 be twin primes. Prove that sigma(P+2)=sigma(p) + 2
b)Prove or disprove the converse of part a above.

For Part a i just can seem to equate the 2.

I get that sigma(P+2) is [(P+2)^2 - 1]/[(P+2)-1] and rearranging i get
P^2 + 4p +4 / P+1

and then i get that sigma(P) +2 is [P^2-1/P-1] + 2 and rearranging i get
2p^2 +4p -6 / 2(P-1)

Maybe i am approaching it the wrong way?

2. The only two divisors of a prime is 1 and itself and so, the sum of its divisors is p + 1: $\sigma (p) = p + 1$

Now apply this to both sides:
$\sigma (p+2) = (p+2) + 1$
$\sigma (p) + 2 = (p + 1) + 2$