Three Tri-Automorphic Numbers for each Number of Digits

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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 $b = 7$


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In particular: Guess: Try proving for $n = 10 a + b$ and then by induction.
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Substituting $b = 2$ in equation $(1)$:

$a = 9$

Substituting $b = 5$ in equation $(1)$:

$a = 7$

Substituting $b = 7$ in equation $(1)$:

$a = 6$

$\blacksquare$


Examples

Tri-Automorphic Numbers with $4$ Digits

The $3$ tri-automorphic numbers with $4$ digits are:

\(\text {(6667)}: \quad\) \(\ds 3 \times 6667^2\) \(=\) \(\ds 133 \, 34 \mathbf {6 \, 667}\)
\(\text {(6875)}: \quad\) \(\ds 3 \times 6875^2\) \(=\) \(\ds 141 \, 79 \mathbf {6 \, 875}\)
\(\text {(9792)}: \quad\) \(\ds 3 \times 9792^2\) \(=\) \(\ds 287 \, 64 \mathbf {9 \, 792}\)


Tri-Automorphic Numbers with $10$ Digits

The $3$ tri-automorphic numbers with $10$ digits are:

\(\text {(6 666 666 667)}: \quad\) \(\ds 3 \times 6 \, 666 \, 666 \, 667^2\) \(=\) \(\ds 133 \, 333 \, 333 \, 34 \mathbf {6 \, 666 \, 666 \, 667}\)
\(\text {(7 262 369 792)}: \quad\) \(\ds 3 \times 7 \, 262 \, 369 \, 792^2\) \(=\) \(\ds 158 \, 226 \, 044 \, 98 \mathbf {7 \, 262 \, 369 \, 792}\)
\(\text {(9 404 296 875)}: \quad\) \(\ds 3 \times 9 \, 404 \, 296 \, 875^2\) \(=\) \(\ds 265 \, 322 \, 399 \, 13 \mathbf {9 \, 404 \, 296 \, 875}\)


Sources

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