If $p$ and $2p + 1$ are odd primes and $n = 4p$, prove that $φ(n + 2) = φ(n) + 2$.

Question

Prove: if $p$ and $2p + 1$ are odd primes and $n = 4p$, show that $φ(n + 2) = φ(n) + 2$.

I'm stuck on this simple question about Euler's theorem. Any help would be welcome.

Thanks in advance!

Answer 1

$\varphi(n+2)=\varphi(2*(2p+1))=\varphi(2)*\varphi(2p+1)=1*(2p+1-1)=2p$

$\varphi(n)=\varphi(4*p)=\varphi(4)*\varphi(p)==2*(p-1)=2p-2$

$\varphi(n+2)=2p=(2p-2)+2=\varphi(n)+2$

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