Repunit Squares

Note that only numbers which are $\equiv 1 \pmod4$ or $\equiv 0 \pmod4$ can be perfect squares.

Now, for repunits which are greater than 1, $11\ldots11 \equiv 11\ldots00 + 11 \equiv 11 \equiv 3 \pmod4$ . Therefore no repunit greater than 1 is a perfect square. Since 1 is a square, The number of repunits which are perfect squares are $\boxed{1}$

To prove that $R_{n}$ can't be perfect squares except $n=1$ .

Given that $x^{2} = R_{n} = \sum\limits_{k=0}^n 10^{k}$ .

We can say that $x^{2} = 1 + 10(R_{n-1}) \equiv 1 \pmod{10}$ for $n > 1$ .

So we get $x \equiv \pm 1 \pmod{10}$ . (Since $x$ must have a unit of 1 or 9.)

Therefore, $x$ can be written as $10m \pm 1$ for some $m$ .

$x=10m \pm 1$ .

$x^{2} = (10m \pm 1)^{2} = 1+10+100+...10^{n}$

$100m^{2} \pm 20m + 1 = 1+10+100+...+10^{n}$

$10m^{2} \pm 2m = 1+10+100+...+10^{n-1}$ (cancel 1 and divide by 10)

$2(5m^{2} \pm m )= 1 + 2(5+50+...+5\times 10^{n-2})$

Since LHS is even but RHS is odd, contradiction!

Therefore, there exists only $\boxed{1}$ number. (which is 1)

only 1 can be square