>>10938302sigma(n)/n = (p+2)/p for some n, is equivalent to
p*sigma(n) = n*(p+2) for some n.
This imposes the restriction that 2n = 0 (mod p) which means p would have to divide n.
Obviously n isn't p.
sigma(n)/n = sum[1/d; d|n] = 1 + 1/n + 1/p + sum[1/d;d|n and d isn't 1,p,n] = 1+ 2/p
1/n + sum[1/d;d|n and d isn't 1,p,n] = 1/p implies p is the smallest prime dividing n.
You could try writing n = m*p^k (k>0, p coprime to m)
and write sigma(n) = sigma(m)*(1-p^(k+1))/(1-p)
