121 is Square Number in All Bases greater than 2
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Theorem
Let $b \in \Z$ be an integer such that $b \ge 3$.
Let $n$ be a positive integer which can be expressed in base $b$ as $121_b$.
Then $n$ is a square number.
Proof
Consider $11_b$.
By the Basis Representation Theorem:
- $11_b = b + 1$
Thus:
\(\ds {11_b}^2\) | \(=\) | \(\ds \paren {b + 1}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b^2 + 2 b + 1\) | Square of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds 121_b\) |
Thus:
- $121_b = {11_b}^2$
and so is a square number.
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $121$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $121$