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## Warming up process

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• Dear Andy, Phil and Paul, Thank you for your comments. Andy, you can be excused as you are new to the F.I.D.N concept. As you correctly point out, there are
Message 1 of 4 , Dec 5, 2003
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Dear Andy, Phil and Paul,
Thank you for your comments. Andy, you can be excused as you
are new to the F.I.D.N concept. As you correctly point out, there
are trivial solutions. The idea being that if no non-trivial
solutions can be obtained then you can always answer the question.
Phil, to atone for your cognitive lapse into triviality, I would
like you to find some non-trivial solutions a la Paul Leyland and
possibly even a formula to generate the solutions. (or even several)
and so set a good example for your juniors.

Regards,
Paul Mills
Kenilworth,
England.
• ... But that condescending answer doesn t defeat my point... You ask for three integers, A,B,C, which are in arithmetic progression, and for which xA^3, yB^3
Message 2 of 4 , Dec 5, 2003
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On Fri, Dec 05, 2003 at 10:14:29AM -0000, paulmillscv wrote:
> Dear Andy, Phil and Paul,
> Thank you for your comments. Andy, you can be excused as you
> are new to the F.I.D.N concept. As you correctly point out, there
> are trivial solutions. The idea being that if no non-trivial
> solutions can be obtained then you can always answer the question.
> Phil, to atone for your cognitive lapse into triviality, I would
> like you to find some non-trivial solutions a la Paul Leyland and
> possibly even a formula to generate the solutions. (or even several)
> and so set a good example for your juniors.

But that condescending answer doesn't defeat my point...

You ask for three integers, A,B,C, which are in arithmetic progression,
and for which xA^3, yB^3 and zC^3 are all fourth powers. There's nothing

Take *any* three A,B,C in arithmetic progression. Then there are
infinitely many triples of integers x,y,z satisfying the requirements
with these A,B,C. That much is painfully obvious, and anyone bored
enough could write down some sort of general formula for these x,y,z.
Where is the 'problem'?

Or perhaps do you mean x,y,z to be in arithmetic progression as well? You
don't explicitly say this, but it would be a reasonably realistic
possibility, and make the problem harder. Or various further
restrictions on the integers involved perhaps. I don't know. But the
problem as stated seems trivial and uninteresting, at least to me. Maybe
the 'seniors' find it harder than the 'juniors'?

All good fun...

Andy
• From: paulmillscv [mailto:paulmillscv@yahoo.co.uk] ... Is it time to set phasers to moderate? __________________________________________________ Virus checked
Message 3 of 4 , Dec 5, 2003
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From: paulmillscv [mailto:paulmillscv@...]
> Sent: 05 December 2003 10:14
> ....

Is it time to set phasers to moderate?

__________________________________________________
Virus checked by MessageLabs Virus Control Centre.
• ... I would claim that my solution was completely trivial also. It took me well under five seconds to find it. The arithmetic progression consisted of the
Message 4 of 4 , Dec 5, 2003
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> > like you to find some non-trivial solutions a la Paul Leyland and
> > possibly even a formula to generate the solutions. (or even
...
> You ask for three integers, A,B,C, which are in arithmetic
> progression,
> and for which xA^3, yB^3 and zC^3 are all fourth powers.
> There's nothing
> magical about it.
>
> Take *any* three A,B,C in arithmetic progression. Then there are

I would claim that my solution was completely trivial also. It took me
well under five seconds to find it.

The arithmetic progression consisted of the values -1, 0, 1. Multiplying
these respectively a negative 4th power, any integer and a positive 4th power
yields a solution to the problem as posed.

Yes, we have undoubtedly been trolled.

Paul
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