Prove that if n is a product of twin primes, then Φ(n)σ(n) = (n+1)(n-3).?

Question

Φ(n) is Euler's function denoting the number of divisors relatively prime to n that are less than n.

σ(n) denotes the sum of all the divisors of n.

Answer 1

If n is a product of twin primes, then n = (2k-1)(2k+1) and p = (2k-1) is prime and q = (2k+1) is prime. So

σ(n) = 1 + p + q + n = 1 + 4k + 4k^2 - 1 = 4k(k+1)

Now, I think you have the definition of the Euler totient wrong [1]. Using the proper definition, recall that the totient function is multiplicative: if two numbers a and b are coprime, then φ(ab) = φ(a)φ(b). This means that

φ(n) = φ(p)φ(q)

But since p and q are primes, φ(p) = p-1 = 2k-2 and φ(q) = q-1 = 2k, so

φ(n) = 2k(2k-2) = 4k(k-1)

This implies that

φ(n)σ(n) = (4k)^2(k-1)(k+1)

Now (n+1)(n-3) = (4k^2)(4k^2 - 4) = (4k)^2(k^2-1)

So you see that the claimed equality is in fact true.