Re: x^3 + y^3 = 5z^3 has no soln not div. by 3

On Mar 1, 5:07 pm, Robert Israel
<isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
Timothy Murphy <gayle...@xxxxxxxxxx> writes:
nullgraph wrote:

Hi, I'm trying to solve this problem: show that x^3 + y^3 = 5z^3 has
no integer solution not divisible by 3. The hint is to consider Z[w]
where w is a primitive cube root of unity.
Any suggestion highly appreciated!

What do you mean by "solution divisible by 3"?

I would have assumed it meant x,y,z all divisible by 3.  But then there's
x=1, y=-1, z=0.  

I think "solution divisible by 3" means xyz = 0 (mod 3). At least this
is what other problems that I've solved meant.


But I am sure the hint means you are to consider
the quadratic number field Z[w];
This is a unique factorisation domain;
and the equation reads
(x+y)(x+wy)(x+w^2y) = 5z^3

--
Robert Israel              isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics        http://www.math.ubc.ca/~israel
University of British Columbia            Vancouver, BC, Canada

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