# Re: Cubic Diophantine Equations

• From: Wolfgang Rave <info@xxxxxxxxxxx>
• Date: Sat, 30 Aug 2008 21:52:35 -0700 (PDT)

On Aug 31, 4:43 am, "Larry Hammick" <larryhamm...@xxxxxxxxx> wrote:
"Wolfgang Rave"

hello there,

Has anybody of you got ideas on how to solve some special cubic
Diophantine Equations like [rewritten in ascii]:
(1) a^3 pm b^3 = c^2 where pm means plus or minus
(2) a^2 pm b^2 = c^3

aa - bb can be factored (a-b)(a+b) and we can set each factor to a cube, say
d^3 and e^3, both odd or both even, and we get a parametric solution:
c = de
a = (d+e)/2
b = (e-d)/2.
where d and e are arbitrary. The same approach works for the equation aa pm
bb = c^s for any s>1.

aa + bb can be factored in the ring Z[i]: (a+bi)(a-bi) = aa + bb, and that
ring has the unique factorization property, so the continuation is similar.

I can't say much about (1).

LH

hello Larry,

thanks for the translation again. I used the Alt Gr 2 and 3 key
combination for square and cube, which is inconvenient here - sorry
again.

And thanks for your hints. Equation (2) may be regarded as solved
then.

best wishes, Wolfgang.
.

## Relevant Pages

• Re: Cubic Diophantine Equations
... "Wolfgang Rave" ... Diophantine Equations like [rewritten in ascii]: ... ring has the unique factorization property, ...
(sci.math)
• Re: A known factorization?
... then this difference is  a multiple of their sum. ... dozen upon dozens of Diophantine equations are available, ... useless to apply it for real integer factorization. ...
(sci.math)