Thread: Variation of a Fermat Theorem

I need to prove x^4-y^4=z^2 has no solutions in the nonzero integers. I know and understand the proof for x^4+y^4=z^2 but can't seem to get it to work in the slightly different case.

I have thought of 2 possible ways.
1) In the proof of x^4+y^4=z^2 you say x^2=m^2-n^2 and y^2/4 = mn/2 and from there use the method of infinite descent to show there is no solution. This doesn't work in x^4 = y^4 + z^2 (rewritten x^4-y^4=z^2) because z is only squared, not to the 4th power.

2) Rewriting x^4-y^4=z^2 as (x^2+y^2)(x^2-y^2) = z^2. I just noticed this by observation, and don't really know what to do with it.

Any help is appreciated.

Originally Posted by breadbakingfanatic

I need to prove x^4-y^4=z^2 has no solutions in the nonzero integers. I know and understand the proof for x^4+y^4=z^2 but can't seem to get it to work in the slightly different case.

I have thought of 2 possible ways.
1) In the proof of x^4+y^4=z^2 you say x^2=m^2-n^2 and y^2/4 = mn/2 and from there use the method of infinite descent to show there is no solution. This doesn't work in x^4 = y^4 + z^2 (rewritten x^4-y^4=z^2) because z is only squared, not to the 4th power.

2) Rewriting x^4-y^4=z^2 as (x^2+y^2)(x^2-y^2) = z^2. I just noticed this by observation, and don't really know what to do with it.

Any help is appreciated.

See this.