Fwd: Diophantine equation with factorial

[Richard Guy]

Hand-waving ...   Most of the primes dividing
n!  do so only to the first power.  Half of
these don't divide  x^2 - xy + y^2  and so
would have to divide  x + y.  But  x + y  is small
compared with the quadratic ...  Maybe you
are allowing  y  to be negative, in which case
this `proof' fails ??   R.

On Wed, 25 Jun 2003, Don McDonald wrote:

> In message <Pine.LNX.4.44.0306221449380.16191-100000 at csl> you write:
> Richard Guy:
> 
> > Where angels fear to tread ?
>   (Fools rush in??  )
> > 
> > x^3 + y^3 = (x + y)(x^2 - xy + y^2)   and all squarefree
> > factors of the latter (paren) are  3  or of form  6k + 1.  R.
> >
> Sorry, I do not follow. Please Richard.
> So intermediate factors  ..., 6k+5,... may be very thin, sparse?
> 
> Could you kindly say some more please? Thank you.
> 
> seqfan, diophantine equation with ... factorial
> 
>  x^3 + y^3 = n!,         24.6.03  date.
> solution 1^3 + 1^3= 2!.
> test don.mcdonald. bas64 calc circle FACTScube3
> 
> 
> try x^3+y^3 mod 30, (also Richard Guy.)
> ?
> For all integers,   k>=0, m>0, and 0 <= x <= 30.
> 
> (30k+x)^3 + (30m-x)^3 >= 2*15^3 is divisible by 2.3.5=30.
> 
> We desire factor 4 for 4!, followed by factor 6 for 6!.
> 
> /  don.mcdonald
> 
> CAUTION Working ..  [delete.]
> 
> 
> 0^3 == 0 mod 30.       0^3 +  0^3 ==  0  mod 30.
>                        0^3 +  30^3 ==  0  mod 30.
>                         
> 1^3 == 1 mod 30.       1^3 +  29^3 ==  0  mod 30. 
> 2^3 == 8 mod 30.       2^3 +  28^3 ==  0  mod 30. 
> 3^3 == 27 mod 30.       3^3 +  27^3 ==  0  mod 30. 
> 4^3 == 4 mod 30.       4^3 +  26^3 ==  0  mod 30. 
> 5^3 == 5 mod 30.       5^3 +  25^3 ==  0  mod 30. 
> 6^3 == 6 mod 30.       6^3 +  24^3 ==  0  mod 30. 
> 7^3 == 13 mod 30.       7^3 +  23^3 ==  0  mod 30. 
> 8^3 == 2 mod 30.       8^3 +  22^3 ==  0  mod 30. 
> 9^3 == 9 mod 30.       9^3 +  21^3 ==  0  mod 30. 
> 10^3 == 10 mod 30.       10^3 +  20^3 ==  0  mod 30. 
> 11^3 == 11 mod 30.       11^3 +  19^3 ==  0  mod 30. 
> 12^3 == 18 mod 30.       12^3 +  18^3 ==  0  mod 30. 
> 13^3 == 7 mod 30.       13^3 +  17^3 ==  0  mod 30. 
> 14^3 == 14 mod 30.       14^3 +  16^3 ==  0  mod 30. 
> 15^3 == 15 mod 30.       15^3 +  15^3 ==  0  mod 30. 
> 16^3 == 16 mod 30.    
> 17^3 == 23 mod 30.    
> 18^3 == 12 mod 30.    
> end close *ram 
> 
> To: all at abouthugo.de,seqfan
> Subject: Re: Fwd: Diophantine equation with factorial
> 
> In message <26800538$10562687033ef5619f4a1cd4.97542386 at config3.schlund.de> you write:
> 
> > SeqFans,
> >                  ...
> > 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 cuberoot 2.   Until x = (n!)^(1/3.)
> 
> y decreases rapidly.
> This greatly reduces the number of trials.
> 
> The next (and better) idea is try x^3+y^3 mod 21.
> 
> E.g. does in contain factorial 7.?
> (Later..  Experimented modulo 30 above/ top.)
> 
> Only 200 easy trials needed.
> 
> don.mcdonald
> 23.06.03  01:12
> 
> > > -------- 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?
> > 
>