The problem at the title is one of my elementary number theory exercises.
I attempted to find a specific form of $n$ which satisfies the condition. I tried $n = p$ (prime) and $n = p^k$ case, but those two do not yield the answer. I also tried $n = pq$ case, but I was not able to decide that this works or not, since the expression gets too complicated.
Please help me if you know the form of $n$ which makes $\phi(n)^2 + n^2$ a perfect square. Thank you for any helps, answers, and hints.
Let $n=3^k\cdot 5$ for $k\ge 1$. Then we have $\phi(n)=8\cdot 3^{k-1}$, and hence $$ n^2+\phi(n)^2=3^{2k}\cdot 5^2+8^2\cdot 3^{2k-2}=(3^{k-1})^2(9\cdot 5^2+8^2)=(17\cdot 3^{k-1})^2. $$ This holds for all $k\ge 1$, so that we have infinitely many $n$ with $n^2+\phi(n)^2=c^2$.