Math Help - A proof using the phi-function

1. A proof using the phi-function

Given a positive integer k, show there are at most a finite number of integers n for which Φ(n)=k

2. Bounding the function

It is already known that the totient function is bounded as follows:

$\sqrt{n} < \phi(n) < n-\sqrt{n}$

So, given an arbitrary positive integer $k$, $\phi(n)>k$ for all $n>k^2$. So the highest number $n$ satisfying $\phi(n)=k$ must still be less than $k^2$. Since there is such an upper bound, the total number of integers n satisfying this equality is finite.