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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.

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  • $\begingroup$ Is this an arbotrary $\phi$ with porperties (1),(2),(3), or is it the well-known Euler-phi (which is at least not ruled out)? $\endgroup$
    – Hagen von Eitzen
    Feb 11, 2015 at 13:05
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    $\begingroup$ @HagenvonEitzen Yes it is the Euler-phi $\endgroup$
    – Nadine
    Feb 11, 2015 at 13:09
  • $\begingroup$ Ah, I was just wondering because of the formualtion "Waht I am given"; after all (2) and (3) are well-known standard properties of $\phi$ $\endgroup$
    – Hagen von Eitzen
    Feb 11, 2015 at 13:10
  • $\begingroup$ What's your suggestion for $n$ ? Note that $2593$ is prime, and odd. $\endgroup$
    – Sary
    Feb 11, 2015 at 13:11
  • $\begingroup$ @Sary In my question I noted that I tried to take n as 5187 because it gave 2592. $\endgroup$
    – Nadine
    Feb 11, 2015 at 13:11

1 Answer 1

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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 $$\begin{align}\phi(2p+0)&=\phi(p)=2592\\ \phi(2p+1)&=\phi(5187)=2592\\ \phi(2p+2)&=\phi\left(4\cdot \tfrac{p+1}2\right)=2\phi\left(\tfrac{p+1}2\right)=p-1=2592\end{align}$$

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  • $\begingroup$ So what does $n$ equal to? $\endgroup$
    – Nadine
    Feb 11, 2015 at 13:17
  • $\begingroup$ 5187. Am I right? $\endgroup$
    – Nadine
    Feb 11, 2015 at 13:17
  • $\begingroup$ @MissNadine What do you suggest if you want to make $n,n+1,n+2$ behave like $2p,2p+1,2p+2$? $\endgroup$
    – Hagen von Eitzen
    Feb 11, 2015 at 13:17
  • $\begingroup$ Okay so since $p=2592+1$ then $2p=5186$, so we should make $n=5186$? $\endgroup$
    – Nadine
    Feb 11, 2015 at 13:19
  • $\begingroup$ So 5186,5187,5188? $\endgroup$
    – Nadine
    Feb 11, 2015 at 13:20

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