If we have that $\bigg(\dfrac{m}{5}\bigg)^2\equiv0,1,4 \text{ mod } 5$, and we're given that $5|m^2$, then why does it follow that $m^2\equiv0,25,100 \text{ mod } 125$?
The implication was used in the user "Ivan Lohs" answer to this question. Proving that a number with digits 1...9 in some order, ending in 5, is not a perfect square.
I assume you are referring to the rational number $\frac m5$.
If we write $m=5^aM$ with $5\,\nmid M$ then $m^2=5^{2a}M^2$ .
We remark that either $M^2=5k+1$ or $M^2=5k+4$ since the only non-zero squares $\pmod 5$ are $1$ and $4$.
Now $a>1\implies m^2\equiv 0 \pmod {125}$
If $a=1$ then either $m^2=125k+25$ or $m^2=125k+100$ and we are done.