Three Tri-Automorphic Numbers for each Number of Digits

Theorem
Let $d \in \Z_{>0}$ be a (strictly) positive integer.

Then there exist exactly $3$ tri-automorphic numbers with exactly $d$ digits.

These tri-automorphic numbers all end in $2$, $5$ or $7$.

Proof
Let $n$ be a tri-automorphic number with $d$ digits.

Let $n = 10 a + b$.

Then:
 * $3 n^2 = 300a^2 + 60 a b + 3 b^2$

As $n$ is tri-automorphic, we have:


 * $(1): \quad 300 a^2 + 60 a b + 3 b^2 = 1000 z + 100 y + 10 a + b$

and:
 * $(2): \quad 3 b^2 - b = 10 x$

where $x$ is an integer.

This condition is only satisfied by $b = 2$, $b = 5$, or $b = 7$

Substituting $b = 2$ in equation $(1)$:
 * $a = 9$

Substituting $b = 5$ in equation $(1)$:
 * $a = 7$

Substituting $b = 7$ in equation $(1)$:
 * $a = 6$