Find the second smallest integer such that its square's last two digits are $ 44 $ Given that the last two digits of $ 12^2 = 144 $ are $ 44, $ find the next integer that have this property.
My approach is two solve the equation $ n^2 \equiv 44  \pmod{100}, $ but I do not know how to proceed to solve that equation.
I try a different path by letting $ n = 10x + y $ for some integers $ x, y, $ where $ 0 \le y \le9. $ Then $ n^2 \equiv 44 \; \pmod{100} $ can be reduced to $ 20xy + y^2 \equiv 44  \pmod{100}. $ At this point I let $ x $ run from $ 0, 1, 2, \dots $ and find the integer $ y \in \mathbb{Z}_{100} $ such that $ y^2 + 20xy - 44 = 0. $
My question is is there an alternative way to tackle this problem without having to try each $ x $ and $ y? $ Maybe try to solve the initial congruence equation $ n^2 \equiv 44  \pmod{100}. $   

$n^2 \equiv 44 \pmod {100}$
By Chinese remainder theorem, $n^2 \equiv 0 \pmod 4$ and $n^2 \equiv 19 \pmod {25}$ is an solution for the formula above.
$$n^2 \equiv 0 \pmod 4 \tag{1}$$
$$n^2 \equiv 19 \pmod {25} \tag{2}$$
By (1), $n = 2k$. Substitude into (2), 
$$k^2 \equiv 19 \times 4^{-1} \equiv 19 \times 19 \equiv 11 \pmod {25} \tag{3}$$
By (3), $k \equiv 6 \pmod {25}$ or $k \equiv 19 \pmod {25}$,
when $k = 25w +6$, $n = 50w + 12$.
when $k = 25w +19$, $n = 50w + 38$.
Addendum:  The procedure to get the inverse 4 under module 25 is same to get quadratic residue for 11 under module 25.
Here is the manifestation for the latter one.
Since $n^2 \equiv 11 \equiv 1 \pmod 5$, by numerating, $n \equiv 1,4 \pmod 5$.
When $n = 5k+1$, $n \equiv 1, 6, 11, 16, 21 \pmod {25}$
Check one after another and eliminate, gets $n \equiv 6 \pmod{25}$.
When $n = 5k+4$, same procedure plays. 
So $k \equiv 6,19 \pmod{25}$ such that $k^2 \equiv 11 \pmod{25}$
