How many $N$ in $[100,400]$ such that $N^N$ is perfect square?

Question

How many $N$ in $[100,400]$ such that $N^N$ is perfect square? A perfect square is a number that can be represented as $N=a^2$ for some $a$.

How would you approach this? I'm struggling to see a clear path to answering this?

Answer 1

Between $100$ to $400$ there are $151$ even numbers (including $100$). These numbers are perfectly divisible by $2$. Therefore $N^N$ can be written as $N^{\frac{N}2\cdot 2}$. Square root of this $N^{N/2}$. Therefore if $N$ is even $N^N$ is a perfect square.

Now lets take the case of remaining $149$ odd numbers $(101, 103, \ldots, 399)$. $\sqrt{N^N} = N^{N/2} = (N^{0.5N})$. If $N$ is a odd number and perfect square, then $N^N$ is perfect square. Between $101$ and $399$ we have $5$ odd perfect squares ($11^2 = 121, 13^2 = 169, 15^2 = 225, 17^2 = 289, 19^2 = 361$).

Therefore in total we have $156$ numbers ($151$ even + $5$ odd) between $[100 400]$ such $N^N$ is perfect square