Fwd: Diophantine equation with factorial

[Don McDonald]

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?
>