Let $n$ be a natural number such that $\phi(n^4+1)=8n$ where $\phi(n)$ is the number of numbers less than $n$ and coprime to $n$. Find all such $n$?
I know that the only solution is $2$(through hit and trial) but I don't know why. I also know that the function $\phi(n^4+1)-8n$ is also strictly increasing but I don't know why. Can anybody explain it to me?
Note: there may well be a straight forward algebraic approach to this question, but, elegant or not, looking at crude estimates suffices to get the job done.
Use the conclusions of this question to see that $$\varphi (n^4+1)≥\frac {\sqrt {n^4+1}}{\sqrt 2}≥\frac {n^2}{\sqrt 2}$$
Now, it is easy to see that $$n≥12\implies \frac {n^2}{\sqrt 2}>8n$$
so we only need to check $n\in \{ 2, \cdots, 11\}$ which is easily done.
