> > 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