Prove that there are no positive solutions to

x^4 - 4y^4 = z^2.

(Hint: Use the previous equation to first compute z^4 + (2xy)^4)

You assume that there are positive solutions, but I still don't understand the hint. Any help would be great. Thanks!

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- May 19th 2009, 03:46 PMcuriousmuch[SOLVED] Diophantine equations
Prove that there are no positive solutions to

x^4 - 4y^4 = z^2.

(Hint: Use the previous equation to first compute z^4 + (2xy)^4)

You assume that there are positive solutions, but I still don't understand the hint. Any help would be great. Thanks! - May 19th 2009, 08:27 PMTheAbstractionist
What’s your previous equation?

Suppose there are positive solutions to the equation. We may assume WLOG that For if for some prime then so if we write we end up where we started.

If is an odd prime dividing both and then would divide both their sum and their difference It would follow that divides both and contrary to Hence and are either 1 or powers of 2. It is impossible for both of them to be 1 as it would imply that Neither is it possible for one of them to be 1 and the other to be a power of 2, since the parity of would then be indeterminate. Hence both and are powers of 2. This implies that is even.

But note that form a Pythagorean triple. This would imply that must be odd instead.

This contradiction means that there are no positive solutions to the original equation. - May 19th 2009, 08:29 PMcuriousmuch
Thanks for the help! The previous equation simply refers to x^4 - 4y^4 = z^2.

- May 19th 2009, 08:52 PMcuriousmuch
Also, on a completely different topic: what is the sum of the series (-10)^n

where n goes from 0 to infinity when looked at 2-adically and 5-adically?

2-adically, I got that the sum of the series is 2

5-adically, I got that the sum of the series is 5/4.

Are these correct?