Fwd: Diophantine equation with factorial

[Don McDonald]

In message <26800538$10562687033ef5619f4a1cd4.97542386 at config3.schlund.de> you write:

> 
> SeqFans,
> 
> a few days ago the following was posted to
> NG sci.math.research. No reply so far.
> At least there don't seem too many solutions....
> Any ideas?
> 
> Hugo 

I tested up to 20! = 2.4E18.
But my Basic is only good for 4.5E15. (by trial.)

Tip:
start at x = (n! /2) ^(1/3).
x increases by a factor of cubert 2.

y decreases rapidly.
This greatly reduces the number of trials.

The next (and better) idea is try x^3+y^3 mod 21.
Only 200 easy trials needed.

don.mcdonald
23.06.03  01:12

..Calc.Profile.eisintegsq.Seqfan.x3+y3=nfac
> >
> > -------- Original Message --------
> > Betreff: Diophantine equation with factorial
> > Datum: Thu, 19 Jun 2003 15:26:48 GMT
> > Von: Alex Liu <mrlwk at netvigator.com>
> > Foren: sci.math.research
> > 
> > It is clear that (x,y,n)=(1,1,2) is [a] positive integer solution to
> > the equation x^3+y^3=(n!). Any other positive integer solution to this
> > equation?
>