Hint

Keith50 wrote: A perfect square ends with the same two digits. How many possible values of this digit are there? IGNORANCE REDACTED

Let $k = 10a + b$, since we only care about the last two digits. We want to find $k^2$ modulo 100. $(10a + b)^2 = 100a^2 + 20ab + b^2$ which reduces to $20ab + b^2$ modulo 100. We want pair $(a,b)$ such that $20ab + b^2$ modulo 100 is two same digits. The units digit will be given by $b^2$ so we are reduced to $0, 1, 4, 5, 6, 9$ as possibilities. $0$ is trivially possible. To get $1$ we need $b = 1$ or $b = 9$, neither of which work. To get $4$ you need $b = 2$ or $b = 8$. Trying out $b = 2$, we get $40a + 4 \equiv 44 \mod (100)$. We see $a = 6$ works, indeed $62^2 = 3844$. To get $5$ we need $b = 5$ but this gives $100a + 25$ so we never get $55$. To get $6$ we need $b = 4$ or $b = 6$, neither of which work. Finally to get $9$ we need $b = 3$ giving $60a + 9$, again no solutions. So the digits can be $0$ or $4$.