Math

There exist several positive integers x such that
$\dfrac{1}{x^2 + 2x}$ is a terminating decimal.
What is the second smallest such integer?

Terminating decimal, if $2,5 | x^2 + 2x$

$\begin{array}{|rcll|} \hline x^2 + 2x &=& 0 \pmod{2} \quad | \quad +1 \\ x^2+2x+1 &=& 1 \pmod{2} \\ (x+1)^2 &=& 1 \pmod{2} \\ \sqrt{ (x+1)^2 } &=& \sqrt{ 1 } \pmod{2} \\ x+1 &=& \pm 1 \pmod{2} \\ \hline x+1 &=& +1 \pmod{2} \quad | \quad -1 \\ x &=& 0 \pmod{2} \qquad \text{All even numbers: } x=0,~2,~4,~6,~\dots \\ \hline x+1 &=& -1 \pmod{2} \quad | \quad -1 \\ x &=& -2 \pmod{2} \\ x &=& -2+2 \pmod{2} \\ x &=& 0 \pmod{2} \qquad \text{All even numbers: } x=0,~2,~4,~6,~\dots \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline x^2 + 2x &=& 0 \pmod{5} \quad | \quad +1 \\ x^2+2x+1 &=& 1 \pmod{5} \\ (x+1)^2 &=& 1 \pmod{5} \\ \sqrt{ (x+1)^2 } &=& \sqrt{ 1 } \pmod{5} \\ x+1 &=& \pm 1 \pmod{5} \\ \hline x+1 &=& +1 \pmod{5} \quad | \quad -1 \\ x &=& 0 \pmod{5} \qquad x= 0,~5,~10,~15,~\dots \\ \hline x+1 &=& -1 \pmod{5} \quad | \quad -1 \\ x &=& -2 \pmod{5} \\ x &=& -2+5 \pmod{5} \\ x &=& 3 \pmod{5} \qquad x=3+5n,~ n \in \mathbb{Z} \\ \hline \end{array}$

positive integers $x= 2,~ 3,~ 4,~ 5,~ 6,~ 8,~ \dots$

The second smallest such integer is 3