How many copied numbers? When I was doing my math team training, I encountered a difficult question. The question is 

If there is exist a natural number $n$ such that a number equal $n$ written twice, then $n$ is a copied number. $123123$ is a copied number because $123123$ is $123$ written twice. How many copied number that are perfect squares are smaller than $10^{100}$?

I didn't know how to do, so I tried all the copied number less than $10^4$ and fond none of them are perfect squares.
Thanks for any help!
 A: Edited to add: Those comments have now been moved to chat. Err...thanks, Aloizio...
Just to clear things up for anybody baffled by all those long numbers in the comments:
The idea is that if $10^d+1$ has a repeated prime factor, then we can use that to express $(10^d+1)n$ as a perfect square. And if $n$ is in the right range, with precisely $d$ digits, this will give us a copied number.
For example, $10^{11}+1= 11^2\times 23\times 4093\times 8779$. So if $n$ is equal to $826446281=23\times 4093\times 8779$ multiplied by a perfect square, then $(10^{11}+1)n$ is a perfect square. So now all we need to do is find out which perfect squares, when multiplied by $826446281$, give an $11$-digit number. And it turns out that any square from $16$ to $100$ does the job.
Note that if $10^d+1$ does not have a repeated prime factor, this approach doesn't work, because then for $(10^d+1)n$ to be a perfect square would require $n\ge 10^d+1$, which has $d+1$ digits. And $-$ roughly speaking $-$ $(10^d+1)n$ would no longer be a copied number.
