Repunit cannot be Square
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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 $\ds \sum_{k \mathop = 0}^{r - 1} 10^k$ 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.
$\blacksquare$
Sources
- 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): $1,111,111,111,111,111,111$