An exercise asked to give an infinite set of values so that $\varphi(n)$ is a square. One simple answer was to take $n=2^{2k+1}.$ Here $\varphi(n)$ is Euler's totient function.
My question is whether every even square is the value of $\varphi(n)$ for some $n.$ [I found it so for several small even squres.]
There are infinitely many counterexamples.
We'll build counterexamples of the form $A_p=(2p)^2$ where $p$ is an odd prime. We note that the factors of $A_p$ are $\{1,2,4,p,p^2,2p,2p^2,4p,4p^2\}$.
Suppose that $A_p=\varphi(n)$ for some $n$. Clearly $n$ can't be a power of $2$ so there must be some odd prime $q$ dividing $n$. It is easy to eliminate the cases in which the only odd primes dividing $n$ are $3,5$ so assume that $q>5$. In that case $q-1\,|\,\varphi(n)$ so $q-1$ must be a divisor of $A_p$. To get a counterexample we just need $p$ such that no odd prime $q$ is one greater than any divisor of $A_p$. Of course $p+1,p^2+1$ are even. To ensure that $2p+1,2p^2+1$ are composite take $p\equiv 1 \pmod 3$. To ensure that $4p+1,4p^2+1$ are composite take $p\equiv 1 \pmod 5$. To ensure all of the above, take $p\equiv 1 \pmod {15}$. By Dirichlet, there are infinitely many such primes and we are done.