Can a number consisting only of numbers $8$ and $6$ be a square of an integer?

Question

Can a number consisting only of numbers $8$ and $6$ be a square of an integer?

I'm confused. Which approach can be used in solving this?

Answer 1

The last three digits must be divisible by 8, which means the possibilities are $888$ and $688$, as they are the only 3 digit permutations of $6,8$ that are divisible by $8$. However, a number ending in $8$ can't be a square, so this concludes the proof.

Answer 2

Can a number ending in $8$ be a square? If a number ends in $6$ and has an even tens digit, how many times does $2$ divide into it?