Find a number $n \neq 2017$ such that $\phi(n) = \phi(2017)$

Find a number $$n \neq 2017$$ such that $$\phi(n) = \phi(2017)$$, as above. I know the formula for a general $$\phi$$ function, but I cannot see how this is helpful here. Any help would be appreciated!

• You could work out $\phi(2017)$ for a start? May 24 '20 at 4:47
• @AnginaSeng Yes, I think I can figure this out, but I'm quite stuck after here.. May 24 '20 at 4:48
• if $p$ is odd then $\phi(2p)=\phi(2)\phi(p)=\phi(p)$
– Kat
May 24 '20 at 4:52

Hint: $$2017$$ is a prime number.
Solution: Fortunately, the totient function $$\phi$$ has the nice property that $$\phi(p) = \phi(2p)$$, where $$p$$ is prime. This is because $$2p$$ is relatively prime to all odd numbers less than it except $$p$$. There are $$p$$ even numbers less than $$2p$$, and discounting the fact that $$\gcd(p, 2p) \neq 1$$, we see that $$\phi(2p) = p - 1 = \phi(p)$$ You could have alternatively seen this from the multiplicative property of $$\phi$$, namely that $$\phi(mn) = \phi(m)\phi(n)$$. Hence, $$\phi(4034) = \phi(2017)$$, because $$\phi(2) = 1$$.
• Thank you, I think I see that. So $\phi(2017) = 2016$. How does this fact help answering this question? May 24 '20 at 4:47
• I saw that $2017$ is prime but didn't think of $4034.$ I found $\phi(29\cdot 73)=(28)(72)=(2^5)(7)(9)=2016.$ May 24 '20 at 10:41