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}. $   

If $x^2$ ends with $44$ then $x$ is even. Let $y=2x$. We are trying to solve
$$(2y)^2\equiv 44\pmod{100}$$
and this equation is equivalent to
$$y^2\equiv 11\pmod{25}$$
Since $6^2\equiv 11\pmod{25}$ this equation can be written as
$$(y-6)(y+6)\equiv 0\pmod{25}$$
It is not possible that both $y-6$ and $y+6$ are multiples of $5$, so the solutions are $y\equiv\pm6\pmod{25}$. The first positive values for $y$ are
$$6,19,31,44,56,69,81,94,106,\ldots$$
Multiply these numbers by $2$ to get the solutions for $x$.
