Smallest Penholodigital Square

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Theorem

The smallest penholodigital square number is:

$11 \, 826^2 = 139 \, 854 \, 276$


Proof

Let $n$ be the smallest positive integer whose square is penholodigital.

First it is noted that the smallest penholodigital number is $123 \, 456 \, 789$.

Hence any square penholodigital number must be at least as large as that.

Thus we can can say that:

$n \ge \ceiling {\sqrt {123 \, 456 \, 789} } = 11 \, 112$

where $\ceiling {\, \cdot \,}$ denotes the ceiling function.

It remains to be demonstrated that no positive integer between $11 \, 112$ and $11 \, 826$ has a penholodigital square.


This theorem requires a proof.
In particular: This is over $700$ numbers. The task can be filtered by, for example, disregarding all $n$ ending in $1$ and $9$ because their squares will both begin and end in $1$ at this low range of the $10000$s, and of course all $n$ ending in $0$ because their squares will end in $0$.
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Historical Note

$139 \, 854 \, 276$ was remarked upon by John Hill in his $1716$ work Arithmetick, Both in the Theory and Practice, where he stated:

This number $139854276$ is a very remarkable Number: First, It's a Square Number; Secondly it contains $9$ Places, and they are $9$ Digits, and I think there is not another that does.


Sources

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