Penholodigital Square Equation

Theorem
The following equations, which include each digit from $1$ to $9$ inclusive, are the only ones of their kind:

Proof
The square of a $2$-digit integer cannot have more than $4$ digits:
 * $99^2 = 9801$

The square of a $4$-digit integer has at least $7$ digits:
 * $1000^2 = 1 \, 000 \, 000$

Hence we only need to inspect $3$-digit integers, with a corresponding $6$-digit square.

A lower bound is given by $\ceiling {\sqrt {123456}} = 352$, where $123456$ is the smallest $6$-digit integer without repeating digits or $0$.

An upper bound is given by $\floor {\sqrt {876543}} = 936$, where $876543$ is the largest $6$-digit integer without repeating digits or $9$.

It is seen that it is not necessary to investigate $6$-digit squares beginning with $9$, because its square root would also begin with $9$.

It is noted that integers ending in $1$, $5$ or $6$ have squares ending in those same digits.

Such numbers can be eliminated from our search, as they will duplicate the appearance of those digits.

We also cannot have $0$ as a digit.

Moreover, observe that an integer $\bmod 9$ and its square $\bmod 9$ have the following pattern:
 * $\begin{array}{|c|c|c|c|c|c|c|c|c|c|}

\hline n \bmod 9 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline n^2 \bmod 9 & 0 & 1 & 4 & 0 & 7 & 7 & 0 & 4 & 1 \\ \hline \end{array}$

By Congruence of Sum of Digits to Base Less 1, the sum of digits of $n$ and $n^2 \pmod 9$ are congruent to $n$ and $n^2 \pmod 9$ respectively.

We must require their sum to be congruent to $0 \bmod 9$ since the sum of all their digits is:
 * $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 \equiv 0 \pmod 9$

Hence $n \equiv 0$ or $8 \pmod 9$.