I was skimming again through Dummit & Foote's Abstract Algebra and I came across this exercise:
Prove that for any given positive integer $N$ there exist only finitely many integers $n$ with $\varphi(n)=N$, where $\varphi$ denotes Euler $\varphi$-function. Conclude in particular that $\varphi(n)$ tends to infinity as $n$ tends to infinity.
I don't doubt that $\varphi(n) \to \infty$ as $n \to \infty$. However, if I recall Erdös proved that if there is an $x_0$ such that $\varphi(x_0)=N$, then there must be infinitely many integers such that $\varphi(x)=N$. See this article for a mention of this (it's in the very first $2$ lines).
Am I reading something wrong here? Has has Dummit and Foote actually asked the impossible? Or does the MathWorld site contain an error?
The result you refer to does not say what you think it does. What it says is that if there is an $m_0$ such that $\varphi(x)=m_0$ has exactly $k$ solutions, then there exist infinitely many $m$ such that $\varphi(x)=m$ has exactly $k$ solutions.
For example, let $k=3$. There is an $m_0$, namely $2$, such that $\varphi(x)=m_0$ has exactly $k$ solutions. These are $3$, $4$, and $6$. We can conclude that there are infinitely many $m$ such that $\varphi(x)=m$ has exactly $3$ solutions.