12. There are two two-digit numbers, their difference is 58, and the last two digits of their square numbers are the same. Then these two two-digit numbers are $\qquad$ .
12.79 and 21.

Let these two two-digit numbers be $m, n (m<n)$.
Since $n-m=58, n^{2}$ and $m^{2}$ have the same last two digits, then
$$
n^{2}-m^{2}=(n+m)(n-m)=58(n+m)
$$

is a multiple of 100.
Also, $(58,100)=2$, so, we have
$n+m=50$ or $n+m=100$ or $n+m=150$.
Thus, we get the system of equations
Solving them respectively yields $n=54, m=-4$ (not valid, discard); $n=79, m=21 ; n=104, m=46$ (not valid, discard).
In conclusion, these two two-digit numbers are 79 and 21.