Square numbers in squares

$\overline{a} = v^2$

$\overline{b6} = x^2$

$\overline{cde} = y^2$

$\overline{f0g5} = z^2$

Since $5|z^2$ , $5|z$ , but $z$ is not a multiply of $10$ . So $z$ has $10$ possible values: $5, 15, 25, \dots, 95$ . From these the $55^2$ and $95^2$ could be $\overline{f0g5}$ .

If $\overline{f0g5}=3025$ , then $x$ has only one possible value: $x=4$ and $\overline{b6}=16$ . A square number's possible endings: $0, 1, 4, 5, 6, 9$ , so $a$ and $e$ are $4$ and $9$ in a way. If $a=4$ , then $\overline{cde}$ would be $789$ or $879$ , but non of these is a square number. So $a=9$ and $\overline{cde}$ only could be $784$ . So if $\overline{f0g5}=3025$ , then there are one possible placing. If $\overline=9025$ , then $\overline{b6}$ only can be $36$ , because if it would be $16$ , then from the $3, 4, 7, 8$ digits we would have to make two square numbers, but square numbers' endings are from the $0, 1, 4, 5, 6, 9$ digits. So because neither $784$ nor $874$ is a square number, $a=1$ , and because $874$ is not a square number, $\overline{cde}=784$ .

So there are two possible numbering.