It is fairly easy to prove that in the Pythagorean equation

x^2 plus y^2 = z^2

i) all three x, y and z cannot be squares themselves.

ii) both x and y cannot be squares themselves.

Can anyone prove that only one of x, y and z can be a square, if any?

take care

kind regards
JB

At least one of the three can definitely be a square; just take any triple and scale it by a number that makes one of the three a square, like 9, 12, 15. As for proving that there can only be one square, the generating equations
a = n^2 - m^2
b = 2nm
c = n^2 + m^2
seem promising. I believe these equations generate all pythagorean triples for non-negative integer values of n and m.

I guess I should have said;

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

<<Can anyone prove that only one of x, y and z can be a square, if any?>>
Example : x=9 ; y=40 ; z=41
(3*3)*(3*3) + 40*40 = 41*41
More generally :
given A and B any different odd numbers :
x = (A*B)*(A*B) is square
y = (A*A*A*A - B*B*B*B)/2
z = (A*A*A*A + B*B*B*B)/2

yes but does that imply that y and z cannot be squares ?

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

I believe that it's a standard result going back to Fermat himself that

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.

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

take care

Hi, neat-maths,

>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

To clarify,

one (1) cannot be regarded as a square.

take care