Why can an integer written $2$ times in a row never be a perfect square? It seems to be true for all natural numbers below $1,000,000$.
I am really stuck any kind of help will be appreciated!
 A: This is not true in general.
Here is a counterexample: 4049586776940495867769 is the square of 63636363637
Why did I have to come up with such a huge counterexample? The reason is that the "conjecture" indeed works for small numbers. Here is why : Consider $\overline{NN}$ where $N$ is a $d$-digit number so that $\overline{NN}$ has $2d$ digits. $\overline{NN} = N \times (10^d+1)$. If $10^d+1$ is squarefree, $\overline{NN}$ has to be at least $(10^d+1)^2$ which has more than $2d$ digits, a contradiction. Thus we need $10^d+1$ to be the multiple of the square of some prime for $\overline{NN}$ to have the possibility of being a perfect square. $d=11$ is the smallest value for which this works.
A: That happens just because $10^n+1$ is squarefree for small $n$s.
However, $10^{11}+1$ is not squarefree, so we can just take its square-free part, $23\cdot 4093\cdot 8779=826446281$, multiply it by some square till it reaches eleven digits,
$$ 16\cdot 23\cdot 4093\cdot 8779 = 13223140496 $$
then append this number to itself, getting a square:
$$ 1322314049613223140496 = 2^{4}\cdot 11^{2}\cdot 23^2\cdot 4093^2\cdot 8779^2.$$
