# 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

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 $109 \, 376$:

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