Repunit cannot be Square

Theorem

A repunit (apart from the trivial $1$) cannot be a square.

Proof

Let $m$ be a repunit with $r$ digits such that $r > 1$.

By definition, $m$ is odd.

Thus from Square Modulo 4, if $m$ were square it would be of the form:

$m \equiv 1 \pmod 4$.

$m$ is of the form $\displaystyle \sum_{k \mathop = 0}^{r - 1} 10^k$ where $r$ is the number of digits.

Thus for $r \ge 2$:

 $\displaystyle m$ $=$ $\displaystyle 11 + 100 s$ for some $s \in \Z$ $\displaystyle$ $=$ $\displaystyle \paren {2 \times 4} + 3 + 4 \times \paren {25 s}$ $\displaystyle$ $=$ $\displaystyle 3 + 4 t$ for some $t \in \Z$

Hence:

$m \equiv 3 \pmod 4$

and so cannot be square.

$\blacksquare$