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}. $   

We can use your idea: Let $n=10x+y$. If the last digit of $n^2$ is $4$ then the last digit of $n$ is either $2$ or $8$, so $y=2$ ot $y=8$.
First let's assume $y=2$. We then have $40x+4\equiv 44\mod100$, or equivalently $40x+4=44+100t$ for some $t$. For $t=0$ we get $x=1$, so $n=12$, and this is not what we want. For $t<0$ we get $x<0$ and these are not solutions. For $t=1$ we get $x=3.5$, not a solution and for $t=2$ we get $x=6$, so $n=62$ is a possible solution.
Now let's assume $y=8$, so, just as above, we need $160x+64=44+100t$ for some $t$, or equivalently $8x+1=5t$, for some $t,x$ intergers. The smallest positive integer solution is $x=3$, $t=5$, and then $n=38$ is a solution for our problem.
$n=38$ is therefore the second smallest positive integer with square ending in $44$.
