It is clear that if $n$ is an even number then, $n$ to power $n$ is surely a perfect square.
Now consider the case that n is an odd number. Let us do prime power factorization of n.
$$\text{so }n = a_1^{p_1} \times a_2^{p_2} \times a_3^{p_3} \cdots \times a_k^{p_k}$$.
Where, $a_1,a_2,\cdots,a_k$ are primes and $p_1,p_2,\cdots,p_k$ are maximum power of the corresponding prime present in $n$.
Now,
$$n^n = a_1^{p_1 \times n} \times a_2^{p_2 \times n} \times a_3^{p_3 \times n} \times a_k^{p_k \times n} $$
Now, we just have to use the fact that a number is a perfect square, iff all the powers in its prime-power-factorization are even numbers.
This suggests that all of $(p_1 \times n), (p_2 \times n), (p_3 \times n),\cdots, (p_k \times n)$ must be even to make $n^n$ a perfect square.
However, $n$ is an odd number as per our initial assumption. So it is clear that all of $p_1, p_2, p_3, \cdots, p_k$ must be even. (Since odd$ \times $odd cannot be even).
But $p_1,p_2,p_3,\cdots,p_k$ are the powers of prime-power-factorization of $n$. Since these are all even, it suggests that $n$ must also be a perfect square.
So, $n^n$ is a perfect square iff $n$ is even or $n$ itself is a perfect square.
For $n$ from $1$ to $100$ inclusive, there are $50$ even numbers. And $1, 9, 25, 49, 81$ are the five odd numbers which are perfect squares.
Hence, the answer is $55$.
Look at the numbers 16, 17, 18: We have 16 < 16+x < 36, 16 < 17+y < 36, 16 < 18+z < 36. Therefore 16+x = 17+y = 18+z = 25, making x = 9, y = 8, z = 7. Therefore each of these numbers has only one possible neighbour and therefore must be the first or last of the sequence. Which is not possible, since we have three of them.
(Sorry, doesn't quite answer the question. )
Best Answer
No, it's not correct: If $a$ is an odd perfect square, then so is $a^a$. For example,
$$9^9 = (3^2)^9 = (3^9)^2$$