<s>[INST] prove that there are infinite positive numbers $\overline{a_{1}a_{2}a_{3}\cdots a_{n}a_{1}a_{2}a_{3}\cdots a_{n}}$ is square number 
Show that there are infinitely many positive integer  number $A=\overline{a_{1}a_{2}a_{3}\cdots a_{n}}$,and $0\le a_{i}\le 9$,such that
  $\overline{a_{1}a_{2}a_{3}\cdots a_{n}a_{1}a_{2}a_{3}\cdots a_{n}}$ is a square number.

Here, $\overline{a_{1}a_{2}a_{3}\cdots a_{n}}=10^{n-1}a_1+10^{n-2}a_2+ \dots 10a_{n-1}+a_n$. 
How to prove ? Thank you
 [/INST] Here is an observation: $10^{11} + 1$ is divisible by $11^2$. We can use this to get infinitely many numbers of the required form.
Pick and arbitrary natural number $k$. Note that $x = 10^{11(2k+1)}+1$ is divisible by $10^{11}+1$, which in turn means that $x$ is divisible by $11^2$.
For convenience, let's denote $n = 11(2k+1)$. We see that $x=10^n+1$ has $n+1$ digits and looks like $x = \overline{10\ldots01}$, with $n-1$ zeroes in the middle.
Now, look at number $z = x/11^2$. Note that $z$ is an integer, and that $z$ is clearly less that $10^n$. Therefore, $z$ has at most $n$ digits:
$$
z = \overline{a_1 a_2 \ldots a_n}.
$$
Note: several leading digits here can happen to equal to $0$, but that seems to be ok.
Now think of what happens when we multiply $x$ by $z$. On the one hand, $xz = x^2 / 11^2 = (x/11)^2$, so $xz$ is clearly a square. On the other hand, $xz = z(10^n+1) = 10^n z + z$. In decimal notation, $10^n z$ looks like $z$ with $n$ zeroes appended on the right. Therefore,
$$
xz = 10^n z + z = \overline{a_1 \ldots a_n a_1 \ldots a_n}.
$$
There it is, we have built an arbitrarily large number $xz$ which is a square and whose decimal representation has the desired property.
EXAMPLE: To make things easier, look at the minimal example of this construction, i.e. what happens when we pick $k=0$ (some might say it's not natural, but whatever). Then we have $x = 100000000001$, $x/11 = 9090909091$ and $z = x/11^2 = 826446281$. And our big number:
$$
xz = 82644628100826446281 = 9090909091^2
$$
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