6 & 0 fever

http://burningmath.blogspot.in/2013/09/how-to-check-if-number-is-perfect-square.html

Case 1: The number ends in $2n$ zeroes.

Then it is divisible by $10^{2n}$ , which implies it is divisible by $2^{2n}$ . However, since there is a $6$ to the left, it is then divisible maximally by $2^{2n+1}$ , which means it can never be a square.

Case 2: The number ends in $2n+1$ zeroes.

Then it is divisible by $10^{2n+1}$ , which implies it is divisible by $5^{2n+1}$ . However, since there is a $6$ to the left, it is then divisible maximally by $5^{2n+1}$ , which means it can never be a square.

Thus, $\boxed{0}$ solutions exist.