Can anyone prove that in the Pythagorean equation

i) x, y and z cannot all be squares

ii) x and y cannot both be squares

iii) x and z cannot both be squares

iv) y and z cannot both be squares

take care

the same logic would apply for y exchanged with x as the square, but what about if z is a square?
for example I think x=7, y=24 and z=25 is a pythagorean trio and z is a square

take care

x⁴ + y⁴ = z²

has no solution in integers. (Note the square on the right.) This is pp. 9-10 of Edwards Fermat's Last Theorem.

take care

>Yes there is a similar non-result for x^4 - y^4 = z^2 as
>well.
>
>take care

Now you've given away the answer to your puzzle! This, together with alexb's response number 5 provide the complete solution. The fact mentioned by Alex takes care of x and y both being squares, and rearranging the variables and changing their names in the formula you gave here takes care of the x,z and y,z cases. Of course, if no two of them can be squares, it follows automatically that not all three can be squares. (Besides, the case where all three are squares is really a special case of Fermat's Last Theorem, which has for a long time been known to be true for exponent 4.)

Thanks for an interesting question.

--Stuart Anderson