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<DIV><BR>Richard:</DIV>
<DIV> </DIV>
<DIV>Clearly, in</DIV>
<DIV> </DIV>
<DIV>[1] n! = (x+y)(x^2-xy+y^2),</DIV>
<DIV> </DIV>
<DIV>x+y is the minority factor. Indeed, assuming nonnegative<BR>x and y,
we can show that</DIV>
<DIV> </DIV>
<DIV>x+y <= (4n!)^(1/3).</DIV>
<DIV> </DIV>
<DIV>I assume your thoughts then proceed that all the prime<BR>divisors of
x^2-xy+y^2 are congruent to 1 mod 3, so that<BR>all primes not congruent to 1
mod 3 must divide x+y, and<BR>that for sufficient n, there will be enough such
prime<BR>divisors to force x+y > (4n!)^(1/3).</DIV>
<DIV> </DIV>
<DIV>Unfortunately, the argument is not that simple. The<BR>observation
that all the prime divisors of x^2-xy+y^2<BR>are congruent to 1 mod 3 rests on
the assumption that<BR>gcd(x,y) = 1, and we do not have that luxury. For
example,<BR>if x and y are both even, so is x^2-xy+y^2.</DIV>
<DIV> </DIV>
<DIV>Letting g = gcd(x,y), x = gx', y = gy', we have</DIV>
<DIV> </DIV>
<DIV>[2] n! = g^3(x'+y')(x'^2+x'y'+y'^2); (x',y') = 1.</DIV>
<DIV> </DIV>
<DIV>Now we can assert that all prime divisors of x'^2+x'y'+y'^2<BR>are
congruent to 1 mod 3, but unfortunately, g^3(x'+y') in<BR>[2] is not such a
minority factor of n! as is x+y in [1].<BR>I don't know if the idea can be
salvaged.</DIV>
<DIV></FONT> </DIV>
<DIV style="FONT: 10pt arial">----- Original Message -----
<DIV style="BACKGROUND: #e4e4e4; font-color: black"><B>From:</B> <A
title=rkg@cpsc.ucalgary.ca href="mailto:rkg@cpsc.ucalgary.ca">Richard Guy</A>
</DIV>
<DIV><B>To:</B> <A title=parabola@paradise.net.nz
href="mailto:parabola@paradise.net.nz">Don McDonald</A> </DIV>
<DIV><B>Cc:</B> <A title=seqfan@ext.jussieu.fr
href="mailto:seqfan@ext.jussieu.fr">seqfan@ext.jussieu.fr</A> </DIV>
<DIV><B>Sent:</B> Tuesday, June 24, 2003 10:46 AM</DIV>
<DIV><B>Subject:</B> Re: Fwd: Diophantine equation with factorial</DIV></DIV>
<DIV><BR></DIV>Hand-waving ... Most of the primes
dividing<BR>n! do so only to the first power. Half of<BR>these don't
divide x^2 - xy + y^2 and so<BR>would have to divide x +
y. But x + y is small<BR>compared with the quadratic ...
Maybe you<BR>are allowing y to be negative, in which case<BR>this
`proof' fails ?? R.</BODY></HTML>