# 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.

.

**Follow-Ups**:**Re: Cubic Diophantine Equations***From:*Wolfgang Rave

**References**:**Re: Cubic Diophantine Equations***From:*Larry Hammick

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