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        riddles >> hard >> 8-digit squares
        
(Message started by: fatball on Jan 22nd, 2006, 8:49pm)
Title: 8-digit squares
Post by fatball on Jan 22nd, 2006, 8:49pm
Find all 8-digit natural numbers n such that n2 ends in the same 8 digits as n.  Numbers are written in standard decimal notation, with no leading zeroes.
Title: Re: 8-digit squares
Post by Eigenray on Jan 22nd, 2006, 10:01pm
[hideb]> chrem([0,1],[1,0],[2^8,5^8]);
                            [87109376, 12890625][/hideb]
Title: Re: 8-digit squares
Post by fatball on Jan 23rd, 2006, 10:45am
[hide]Very neatly answered.[/hide] Numbers with this property are called Automorphic Numbers and a brief discussion can be found here (http://en.wikipedia.org/wiki/Automorphic_number) or there (http://mathworld.wolfram.com/AutomorphicNumber.html).
Title: Re: 8-digit squares
Post by Eigenray on Jan 24th, 2006, 6:05am
In fact, [xn,yn] := [hide]chrem([0,1],[1,0],[2n,5n])[/hide] is a pair of orthogonal idempotents of the ring R=(Z/10n), the only pair other than [0, 1].  This gives the decomposition
R = Rxn [oplus] Ryn,
giving an explicit inverse to the ring isomorphism
(Z/10n) ~= (Z/5n) x (Z/2n),
(r mod 10n) -> (r mod 5n, r mod 2n)
(axn + byn mod 10n) <- (a mod 5n, b mod 2n).

Moreover, we have xn+1 = xn mod 10n, so that, viewing xn as an integer between 0 and 10n,
x = lim xn = x1 + (x2-x1) + (x3-x2) + ...
= ...109376
exists as a 10-adic integer.  Defining also
y = lim yn = ...890625,
this gives a non-trivial pair of zero-divisors in the 10-adics Z10 ~= Z2 x Z5.


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