# Left-Truncated Automorphic Number is Automorphic

## Theorem

Let $n$ be an automorphic number, expressed in some conventional number base.

Let any number of digits be removed from the left-hand end of $n$.

Then what remains is also an automorphic number.

## Proof

Let $n$ be an automorphic number of $d$ digits, expressed in base $b$.

By Definition of Automorphic Number, we have:

- $n^2 \equiv n \pmod {b^d}$

Let some digits be removed from the left-hand end of $n$, so that only $d'$ digits remain.

This only makes sense when $d' < d$.

Define this new number as $n'$.

Then we have:

- $n \equiv n' \pmod {b^{d'} }$

Thus we have:

\(\ds n^2\) | \(\equiv\) | \(\ds n\) | \(\ds \pmod {b^{d'} }\) | Congruence by Divisor of Modulus: from $n^2 \equiv n \pmod {b^d}$ and $b^{d'} \divides b^d$ | ||||||||||

\(\ds n^2\) | \(\equiv\) | \(\ds \paren {n'}^2\) | \(\ds \pmod {b^{d'} }\) | Congruence of Powers: from $n \equiv n' \pmod {b^{d'} }$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {n'}^2\) | \(\equiv\) | \(\ds n'\) | \(\ds \pmod {b^{d'} }\) |

Hence $n'$ is an automorphic number of $d'$ digits in base $b$.

$\blacksquare$

## Also see

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Similar proofs can give similar results for other similarly defined numbers, e.g. Trimorphic Numbers, Tri-Automorphic Numbers.

## Examples

### Left-Truncation of $1 \, 787 \, 109 \, 376$

We have that $1 \, 787 \, 109 \, 376$ is automorphic:

- $1 \, 787 \, 109 \, 376^2 = 3 \, 193 \, 759 \, 92 \, \mathbf {1 \, 787 \, 109 \, 376}$

Hence so is $109 \, 376$:

- $109 \, 376^2 = 11 \, 963 \, \mathbf {109 \, 376}$

## Sources

- 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $1,787,109,376$ - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $1,787,109,376$