Repunit cannot be Square

Theorem

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

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

Thus for $r \ge 2$:

$\ds m$

$=$

$\ds 11 + 100 s$

for some $s \in \Z$

$\ds $

$=$

$\ds \paren {2 \times 4} + 3 + 4 \times \paren {25 s}$

$\ds $

$=$

$\ds 3 + 4 t$

for some $t \in \Z$

Hence:

$m \equiv 3 \pmod 4$

and so cannot be square.

$\blacksquare$

1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.1$ The Division Algorithm: Problems $2.1$: $7$
1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $1,111,111,111,111,111,111$
1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $1,111,111,111,111,111,111$