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:

\(\displaystyle {11_b}^2\) \(=\) \(\displaystyle \paren {b + 1}^2\)
\(\displaystyle \) \(=\) \(\displaystyle b^2 + 2 b + 1\) Square of Sum
\(\displaystyle \) \(=\) \(\displaystyle 121_b\)

Thus:

$121_b = {11_b}^2$

and so is a square number.

$\blacksquare$


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