Puzzle 1121 Phi(p)/Phi(p-2)=2

Sebastián Martín Ruiz sent the following nice puzzle:

I have obtained the following results for the Fermat Numbers, p=2^(2^n)+1, being Phi the Euler Totient Function:

n= 0 Phi[p]/Phi[p-2]= 2

n= 1 Phi[p]/Phi[p-2]= 2

n= 2 Phi[p]/Phi[p-2]= 2

n= 3 Phi[p]/Phi[p-2]= 2

n= 4 Phi[p]/Phi[p-2]= 2

n= 5 Phi[p]/Phi[p-2]= 261735/131072 =1.99688...

n= 6

Phi[p]/Phi[p-2]= 18764930006193/9367860543488 = 2.00312...

n= 7

Phi[p]/Phi[p-2]= 9615364552842273224968401793233/4800163769175502470780823797760 = 2.00313...

It is observed that in the case where p is prime, that is, from n=0 to 4

Phi(p)/Phi(p-2)=2

and if p is composite, that is, from n=5 to 7 Phi(p)/Phi(p-2) is not an exact number close to 2.

Q) Let p a Fermat Number. Prove p is a Fermat prime if and only if Phi(p)/Phi(p-2)=2, or find a counterexample.