Fwd: Diophantine equation with factorial

[David Wilson]

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.
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