Do there exists a number with $r$ repeats which forms a perfect square? Take a number $x=\overline{a_1a_2a_3\cdots a_n}$. Its repeated form is like $\overline{\underbrace{a_1a_2a_3\cdots a_n}_{\text{x}}\underbrace{a_1a_2a_3\cdots a_n}_{\text{x}}}$
And $\exists$ $s$ such that the repeated form is a perfect square, that is, $\overline{\underbrace{a_1a_2a_3\cdots a_n}_{\text{x}}\underbrace{a_1a_2a_3\cdots a_n}_{\text{x}}}=s^2$. In fact, I have done it and that's a pretty common problem. 
Still, here's my solution:

Solution for $r=2$ (just a super gist):
  Let the number be $n$ with $m$ digits.. Its repeated form is $n(10^m+1)$ and we have to show that there exists $k$ such that $(10^m+1)n=k^2$. Since $n$ has $m$ digits and $10^m+1$ has $m+1$ digits, $n<10^n+1$. If GCD$(10^m+1,n)=1$, we will easily get a contradiction as $10^m+1$ has to be individually a square and that's not possible. Similarly, we will encounter a few more cases (I won't prove them as they are just easy and simple parity checking) and finally, we can claim that $10^m+1=a^2b$ and $n=bc^2$ where $a,b,c > 1$. We can check by Wolfram Alpha or simply make a program (C++, Python perhaps) and ensure that in case of $2^{11}+1$ we get such $a,b$ where $a=11, b=826446281$. Solve it to get $4\leq c\leq10$. Let's show that it holds for infinitely many. Just take any $\text{odd }x$ and it will hold for $10^{mx}+1$ as $10^m+1\mid 10^{mx}+1$ since $m,x$ are $\text{odd numbers}$. The value of $c$ will change. That's it!! 

But what if the repeat is done $r$ times, that is, the repeated number is $\overline{\underbrace{a_1a_2a_3\cdots a_n}_{\text{1st set}}\underbrace{a_1a_2a_3\cdots a_n}_{\text{2nd set}}\cdots \underbrace{a_1a_2a_3\cdots a_n}_{r\text{-th set}}}$
In that case, do there exists a perfect square?

Let $r\ge2$ be the number of repetitions. We look for $n$-digit numbers $a$ such that
$$
\frac{10^{rn}-1}{10^n-1}\,a=\square.
$$
Let $s$ be the square free part of $(10^{rn}-1)/(10^n-1)$, that is
$$
\frac{10^{rn}-1}{10^n-1}=s\,m^2
$$
and $s$ has no square divisors (other than $1$.) Then we must have
$$
a=s\,k^2\quad\text{for some integer }k.
$$
In particular, $s$ must have at most $n$ digits.
Doing a search when $r=2$, I found solutions with $11$, $21$, $33$, $39$, $55$ and $63$ digits. For instance
$$
\frac{10^{2\times63}-1}{10^{63}-1}\times20408163265306122448979591836734693877551020408163265306122449=\\ (142857142857142857142857142857142857142857142857142857142857143)^2.
$$
Edit: that number has $62$ digits. To get $63$ digits multiply it by $9$, $16$, $25$ and $36$.
I have found no solutions for $3\le r\le15$.
