A Problem on Euler’s totient function.

elementary-number-theorygcd-and-lcmprime numberstotient-function

Find all m,n such that φ(mn)=φ(n)

here φ is the euler's totient function.

My attempt:
So i split this up into two cases, note: { (m,n) = gcd(m,n) }

Case 1: (m,n)=1 , Then we'd get φ(m)φ(n)=φ(n) which would imply m=1 and n is any integer or m=2 and n is an odd integer. Simple enough,

Case 2: (m,n) is not 1. Well here, things get tricky. I could not figure out much so i checked a lot of examples and came up with φ(m)φ(n) = $φ(mn) / (m,n)$ if (m,n) is a prime. I am not even sure if this result is correct or not. And if (m,n) was composite, well then i could't understand what to even do.

Any help/hint would be appreciated, thanks!

Best Answer

There is no solution in case 2.

If there is a prime number $p$ that divides both $m$ and $n$, then we would have $\varphi(pn)\mid \varphi(mn)$, and also $\varphi(pn) = p\varphi(n)$.

This means that $\varphi(mn)$ cannot be equal to $\varphi(n)$ in that case.