Fwd: Diophantine equation with factorial

[Richard Guy]

More hand-waving!  The factor  g^3  comes only
 g  from  x+y  and  g^2  from  x^2 - xy + y^2
co their relative sizes are still similar.

I'm not sure about your inequality, but
it would imply that

x'^2 - x'y' + y'^2  is of order (n/e)^(2n/3) 

(which can surely be proved by other means?)

On the other hand, this quantity can only come
from primes of shape  6k+1  which are less than
n,  or possibly their squares.  The product of
these is less than  e^n.

Most such problems get out of computer reach
because we have such poor estimates about the
exact behavior of primes, but here,  n!  is
so huge, that we can see it by hand.

32! = 2^31.3^14.5^7.7^4.11^2.13^2.17.19.23.29.31 =

(2^10.3^4.5^2.7)^3 * (3.11.13)^2 * 2.7.17.19.23.29.31

x'^2 - x'y' + y'^2 = 7.13^2.19.31, x' = 846, y' = 23,

which are too small for the job.    R.

On Wed, 25 Jun 2003, David Wilson wrote:

> 
> Richard:
> 
> Clearly, in
> 
> [1] n! = (x+y)(x^2-xy+y^2),
> 
> x+y is the minority factor.  Indeed, assuming nonnegative
> x and y, we can show that
> 
> x+y <= (4n!)^(1/3).
> 
> I assume your thoughts then proceed that all the prime
> divisors of x^2-xy+y^2 are congruent to 1 mod 3, so that
> all primes not congruent to 1 mod 3 must divide x+y, and
> that for sufficient n, there will be enough such prime
> divisors to force x+y > (4n!)^(1/3).
> 
> Unfortunately, the argument is not that simple.  The
> observation that all the prime divisors of x^2-xy+y^2
> are congruent to 1 mod 3 rests on the assumption that
> gcd(x,y) = 1, and we do not have that luxury.  For example,
> if x and y are both even, so is x^2-xy+y^2.
> 
> Letting g = gcd(x,y), x = gx', y = gy', we have
> 
> [2] n! = g^3(x'+y')(x'^2+x'y'+y'^2); (x',y') = 1.
> 
> Now we can assert that all prime divisors of x'^2+x'y'+y'^2
> are congruent to 1 mod 3, but unfortunately, g^3(x'+y') in
> [2] is not such a minority factor of n! as is x+y in [1].
> I don't know if the idea can be salvaged.
> 
> ----- Original Message ----- 
> From: Richard Guy 
> To: Don McDonald 
> Cc: seqfan at ext.jussieu.fr 
> Sent: Tuesday, June 24, 2003 10:46 AM
> Subject: Re: Fwd: Diophantine equation with factorial
> 
> 
> 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.