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• Hello, Thanks for your response to the last question. For those of you who cannot wait until Xmas here is another question. There may be others as well. Find
Message 1 of 6 , Oct 31, 2003
Hello,
Thanks for your response to the last question. For those of
you who cannot wait until Xmas here is another question. There may be
others as well.

Find the least 3 integer cubes so that X^3 = b, Y^3 = a + b and
Z^3 = 2a + b. I.e. 3 cubes in arithmetic progression. If that is
too hard , try 2 cubes in AP (!) or too easy, try 4 cubes in AP etc.

Regards,
Paul Mills
Kenilworth,
England.
• -1 , 0 , 1
Message 2 of 6 , Oct 31, 2003
-1 , 0 , 1

On Fri, 31 Oct 2003, paulmillscv wrote:

> Hello,
>       Thanks for your response to the last question. For those of
> you who cannot wait until Xmas here is another question. There may be
> others as well.
>
> Find the least  3 integer  cubes so that X^3 = b,  Y^3 = a + b and
> Z^3 = 2a + b.  I.e. 3 cubes in arithmetic progression.  If that is
> too hard , try 2 cubes in AP (!) or too easy, try 4 cubes in AP etc.
>
> Regards,
> Paul Mills
> Kenilworth,
> England.
>
>
>
>
> [rand=130868432]
> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> The Prime Pages : http://www.primepages.org/
>
>
>
>
• ... Then of course, any positive integer n gives -n,0,n as an answer. Is this a clever solution type question? Or a boring computer search type
Message 3 of 6 , Oct 31, 2003
--- In primenumbers@yahoogroups.com, Lawrence Hon <lh2072@c...>
wrote:
>
> -1 , 0 , 1

Then of course, any positive integer n gives -n,0,n as an answer.

Is this a 'clever solution' type question? Or a boring 'computer
search' type question...? Cuz looking at the 'clever solution'
approach, it doesn't seem to go too far...

Andy
• ... Am I misreading this? Seems to be you are asking for solutions to 2Y^3 = X^3 + Z^3 If we let r be a solution to r^2 + r + 1 = 0, then we have the
Message 4 of 6 , Oct 31, 2003
At 12:11 PM 10/31/2003 +0000, you wrote:
>Hello,
> Thanks for your response to the last question. For those of
>you who cannot wait until Xmas here is another question. There may be
>others as well.
>
>Find the least 3 integer cubes so that X^3 = b, Y^3 = a + b and
>Z^3 = 2a + b. I.e. 3 cubes in arithmetic progression. If that is
>too hard , try 2 cubes in AP (!) or too easy, try 4 cubes in AP etc.

Am I misreading this? Seems to be you are asking for solutions to

2Y^3 = X^3 + Z^3

If we let r be a solution to r^2 + r + 1 = 0, then we have the solutions

Y = 0, X = r^n*Z where n = 0, 1 or 2

and

X = r^n*Y = r^m*Z

but sadly that is it in regular integers or Q[r] (Phil's field). The same
is true (sans the last solution) if we replace 2 by any prime that is
congruent to 2 or 5 modulo 9 (or those primes squared).

Short answer, Lawrence was right: -1,0,1,

Chris
• ... of ... be ... etc. ... solutions ... The same ... is ... I expect you re right, Chris (if you correct your typo and say X = - r^n*Z ), and both the
Message 5 of 6 , Nov 1, 2003
--- In primenumbers@yahoogroups.com, Chris Caldwell <caldwell@u...>
wrote:
> At 12:11 PM 10/31/2003 +0000, you wrote:
> >Hello,
> > Thanks for your response to the last question. For those
of
> >you who cannot wait until Xmas here is another question. There may
be
> >others as well.
> >
> >Find the least 3 integer cubes so that X^3 = b, Y^3 = a + b and
> >Z^3 = 2a + b. I.e. 3 cubes in arithmetic progression. If that is
> >too hard , try 2 cubes in AP (!) or too easy, try 4 cubes in AP
etc.
>
> Am I misreading this? Seems to be you are asking for solutions to
>
> 2Y^3 = X^3 + Z^3
>
> If we let r be a solution to r^2 + r + 1 = 0, then we have the
solutions
>
> Y = 0, X = r^n*Z where n = 0, 1 or 2
>
> and
>
> X = r^n*Y = r^m*Z
>
> but sadly that is it in regular integers or Q[r] (Phil's field).
The same
> is true (sans the last solution) if we replace 2 by any prime that
is
> congruent to 2 or 5 modulo 9 (or those primes squared).
>
>
> Short answer, Lawrence was right: -1,0,1,
>
> Chris

I expect you're right, Chris (if you correct your typo and say X = -
r^n*Z ), and both the equation and your assertions have a familiar
ring to them, but I can neither prove that no other solutions exist
nor lay hands on a relevant reference.

Mike
• ... The wonderful text: L. J. Mordell Diophantine Equations Academic Press, 1969, page 124-130 includes things like: Theorem 3: Let a = p or p^2 where p =
Message 6 of 6 , Nov 1, 2003
At 01:12 PM 11/1/2003 +0000, you wrote:
>I expect you're right, Chris (if you correct your typo and say X = -
>r^n*Z ), and both the equation and your assertions have a familiar
>ring to them, but I can neither prove that no other solutions exist
>nor lay hands on a relevant reference.

The wonderful text: L. J. Mordell "Diophantine Equations"
Academic Press, 1969, page 124-130 includes things like:

Theorem 3: Let a = p or p^2 where p = 2, 5 (mod 9) is a prime,
and let e be a unit in Q(r) (r^2 + r + 1 = 0). Then the equation

x^3 + y^3 + eaz^3 = 0

has no solutions (x,y,z,e) in Q(r) except

z = 0, x = -y, -ry, -r^2y

unless a = 2, when the equation also has the solutions

x^3 = y^3 = -e z^3, e = +/-1.

Proof is about one page and is done by choosing x, y, z
pairwise relatively prime and N(xyz) minimal. Then do
descent...

But we are wandering away from primes...

Chris
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