- 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. - On Fri, Dec 05, 2003 at 10:14:29AM -0000, paulmillscv wrote:
> Dear Andy, Phil and Paul,

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

> 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.

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

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@...]
> Sent: 05 December 2003 10:14

Is it time to set phasers to moderate?

> ....

__________________________________________________

Virus checked by MessageLabs Virus Control Centre. > > 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

I would claim that my solution was completely trivial also. It took me

> 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

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