TITLE: Is there an integer (clearly not 1) such that it has both integer square root and integer cube root?
QUESTION [6 upvotes]: By common sense, it looks impossible, and I believe that there is not such integer (except for 1). But it is hard (personally for now) to come up with an approach to the proof.
Formally speaking, this is the question:
"Is there an $a\in$$Z$ and $b\in$$Z$ such that there $\exists$$n\in$$Z$ where $a^2=n$ and $b^3=n$."

REPLY [14 votes]: Yes, $n = m^6$ for any integer $m > 1$ will do.
The smallest example is $64 = 2^6 = 4^3 = 8^2$.
In fact, there are no other possibilities (other than $0$ and $1$, which are their own sixth powers).  An easy way to see this is to note that the prime factorization of any square must include each prime an even number of times, while that of any cube must include each prime a multiple of three times.
Thus, the factorization of any number that is both a square and a cube must include each prime a number of times divisible by both two and three; all such numbers are multiples of six, and so the factorized number must be a sixth power.