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From: Ross on 9 Apr 2010 00:24
Automorphic numbers are defined as those n such that n=n^2 mod 10^k
for some k
If leading zeroes are allowed, there are two of each length greater
than 1. For k=6, they are 890625 and 109376. The Wikipedia page says
that if you have one of length k, you can get one of length 2k by
n'=3n^2-2n^3 mod 10^(2k)
I have checked small values and it works, but there are no references
and the ones on MathWorld are inaccessible. Is there an easy proof of
this? I am prompted to look for other bases than 10 due to an Euler
Project problem (which I have solved) and if I understand where it
comes from it will help.
From: Link on 9 Apr 2010 04:10
On Apr 8, 9:24 pm, Ross <rmill...(a)pacbell.net> wrote:
> Automorphic numbers are defined as those n such that n=n^2 mod 10^k
> for some k
> If leading zeroes are allowed, there are two of each length greater
> than 1.  For k=6, they are 890625 and 109376.  The Wikipedia page says
> that if you have one of length k, you can get one of length 2k by
> n'=3n^2-2n^3 mod 10^(2k)
> I have checked small values and it works, but there are no references
> and the ones on MathWorld are inaccessible. Is there an easy proof of
> this?  I am prompted to look for other bases than 10 due to an Euler
> Project problem (which I have solved) and if I understand where it
> comes from it will help.

Coudn't you just check some larger values with a program like this?

nextInt(); if (d >= n) { if ((n * n) % d == n)
{ System.out.println("Automorphic Number"); } else
{ System.out.println("Not Automorphic Number");

My stab at a proof:

Proof.

From [10.1]
main result of [11]

two numbers in: e(ρ(2,S)
ratio=n−1

n is odd
q−ε1≡4(mod 8)

t(2,S)=3

write rn−2 as r
nk=2(n−1)k
2k=(3n−4)k
n−1=n/(3n−4)

mmm
From: Tim Little on 9 Apr 2010 21:19
On 2010-04-09, Ross <rmillika(a)pacbell.net> wrote:
> Automorphic numbers are defined as those n such that n=n^2 mod 10^k
> for some k
> If leading zeroes are allowed, there are two of each length greater
> than 1. For k=6, they are 890625 and 109376. The Wikipedia page says
> that if you have one of length k, you can get one of length 2k by
> n'=3n^2-2n^3 mod 10^(2k)

The original relation can be written as n^2 - n = a 10^k for some
integer a. With a little algebra,
n'^2 - n'
= (n^2 - n)^2 (2n-3) (2n+1)
= a^2 10^(2k) (2n-3) (2n+1).

So
n'^2 - n' = 0 mod 10^(2k).


> I am prompted to look for other bases than 10 due to an Euler
> Project problem (which I have solved) and if I understand where it
> comes from it will help.

The derivation above is not specific to base 10.


- Tim
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