I need to find a positive integer $n$ such that $ϕ(n) = ϕ(n + 1) = ϕ(n + 2)$ where $ϕ(n)$ denotes Euler's totient function.
What I am given:
(1) You may take $ϕ(n) = 2592$.
(2) $ϕ(2n) = ϕ(n)$ provided that $n$ is odd.
(3) $ϕ(p) = p − 1$ for p a prime.
What I did:
I thought that I can take in a value for $ϕ(n)$, so I tried to take $ϕ(n) = 2592$. I found that $ϕ (5187) = 2·6·12·18 = 2592$ So can someone verify that I meet all the requirements that I am give, and that my answer is correct.
Taking the hint (1), we observe that $p=2592+1$ is prime and also $\frac{p+1}2=1297$ is prime. Hence together with your computation $$\begin{align}\phi(2p+0)&=\phi(p)=2592\\ \phi(2p+1)&=\phi(5187)=2592\\ \phi(2p+2)&=\phi\left(4\cdot \tfrac{p+1}2\right)=2\phi\left(\tfrac{p+1}2\right)=p-1=2592\end{align}$$