http://www.artofproblemsolving.com/Forum/viewtopic.php?p=365111&sid=64dca94beff8264a3a14829dfbb3d032#p365111

choose $p\in\mathbb{Z^+}$ such that $v_{7}(p)=1,~3|p$ from L.T.E we'll get that $v_{7}(10^p+1)=v_{7}(1000+1)+v_{7}(\frac{p}{3})=2$ let $n=\frac{10^p+1}{7^2}\in\mathbb{Z^+}$ we can observe that $\frac{10^p+1}{10^2}>n>\frac{10^p+1}{10}$ $\Rightarrow 10^{p-2}+\frac{1}{10^2}>n>10^{p-1}$ since $n\neq 10^{p-2}$ we'll get that $n$ has $p-2$ digits let $m=n(10^{p+2}+10^2)=\overline{n00n00}\in\mathbb{Z^+} $ hence, $m$ is a double number since $m=\frac{10^p+1}{7^2}(10^2)(10^p+1)=(\frac{(10^p+1)(10)}{7})^2$ which means $m$ is a perfect square double number $\because $ we can choose infinte number of $p$ which satisfy the condition $\therefore$ there is infinitely many double numbers which are perfect square $~~\square$ lmk if there is any mistakes

A kinda known PROMYS admission problem, as well