TITLE: Why can an integer written $2$ times in a row never be a perfect square?
QUESTION [5 upvotes]: It seems to be true for all natural numbers below $1,000,000$.
I am really stuck any kind of help will be appreciated!

REPLY [5 votes]: This is not true, though the conjecture is very plausible, as the smallest counterexample is 22 digits long (so, an 11-digit number repeated). These numbers are the content of OEIS A092118, and the first such numbers are:
$$1322314049613223140496, 2066115702520661157025, 2975206611629752066116, 4049586776940495867769, 5289256198452892561984, 6694214876166942148761, 8264462810082644628100, 183673469387755102041183673469387755102041.$$
The observation that the number given by concatenating the $k$-digit number $n$ with itself can be factored as $n \cdot (10^k + 1)$ leads to the observation that $10^k + 1$ must have a repeated prime factor (otherwise, in order for the concatenation to be a square, $n$ would have to admit all the same prime factors as $10^k + 1$, but this is impossible, as $n < 10^k + 1$).
Now, per my comment on ncmathsadist's answer, $10^n + 1$ cannot be divisible by $3$ (and hence not $3^2$) and for analogous reasons cannot be divisible by $5^2$. On the other hand, $10^k + 1$ can be divisible by $7^2$, but this only happens when $k \equiv 21 \bmod 42$. The smallest case, $k = 21$, leads to the $42$-digit example above. Similarly, $10^k + 1$ can be divisible by $11^2$, and this leads to the $22$-digit examples above.
This thread from 2004 seems to contain more good discussion of this problem.