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!

Printable View

- May 19th 2009, 02:46 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, 07:27 PMTheAbstractionist
What’s your previous equation?

Suppose there are positive solutions $\displaystyle x,y,z$ to the equation. We may assume WLOG that $\displaystyle \gcd(x,y)=1.$ For if $\displaystyle x=kx_0,\,y=ky_0$ for some prime $\displaystyle k,$ then $\displaystyle z^2=k^4(x_0^2-4y_0^2)$ so if we write $\displaystyle z=k^2z_0$ we end up where we started.

If $\displaystyle k$ is an odd prime dividing both $\displaystyle x^2-2y^2$ and $\displaystyle x^2+2y^2,$ then $\displaystyle k$ would divide both their sum $\displaystyle 2x^2$ and their difference $\displaystyle 4y^2.$ It would follow that $\displaystyle k$ divides both $\displaystyle x$ and $\displaystyle y,$ contrary to $\displaystyle \gcd(x,y)=1.$ Hence $\displaystyle x^2-2y^2$ and $\displaystyle x^2+2y^2$ are either 1 or powers of 2. It is impossible for both of them to be 1 as it would imply that $\displaystyle y=0.$ Neither is it possible for one of them to be 1 and the other to be a power of 2, since the parity of $\displaystyle x$ would then be indeterminate. Hence both $\displaystyle x^2-2y^2$ and $\displaystyle x^2+2y^2$ are powers of 2. This implies that $\displaystyle x$ is even.

But note that $\displaystyle z^2+4y^4=x^4\ \implies\ z,\ 2y^2,\ x^2$ form a Pythagorean triple. This would imply that $\displaystyle x$ must be odd instead.

This contradiction means that there are no positive solutions to the original equation. - May 19th 2009, 07:29 PMcuriousmuch
Thanks for the help! The previous equation simply refers to x^4 - 4y^4 = z^2.

- May 19th 2009, 07:52 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?