is a number of the below form ever a perfect square if a number is of the form $10^x + 1$ can it ever be a perfect square
such numbers are 
$$11, 101, 1001, 10001$$
I have looked modulo 10 and numbers which square to leave a remainder of 1 are modulo 1 and 9
 A: Is it known whether there is a square (other than a power of $10$) whose base 10 representation has only $0$'s and $1$'s?  It was unknown to Mahler in 1989: see K. Mahler, The representation of squares to the base 3, Acta Arith. Vol. 53, Issue 1 (1989), p. 99-106. For base $7$ there is $20^2 = 1111_7$, for base $8$
there is $11677^2 = 1010111111_8$, 
and for base $12$ there is $521^2 = 11101_{12}$.
A: If $x$ is a non-zero integer, then $10^x + 1$ can't be a perfect square.
Tom Cooney's comment provides the necessary clue, to look at this numbers modulo $3$: verify that $10^x \equiv 1 \pmod 3$. Provided you mind the "wrap-around" effect, you can do arithmetic on congruences much the same way you do with $\mathbb{Z}$. Then its obvious that $10^x + 1 \equiv 2 \pmod 3$. If $n \equiv 1 \pmod 3$, then $n^2 \equiv 1 \pmod 3$ also, but if $n \equiv 2 \pmod 3$, then $n^2 \equiv 1 \pmod 3$ as well!
As a bonus, notice also that if $x$ is even, then $10^x$ is itself a square, $10^x = (10^{\frac{x}{2}})^2$ to be precise.
