# [SOLVED] Diophantine equations

• May 19th 2009, 02:46 PM
curiousmuch
[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 PM
TheAbstractionist
Quote:

Originally Posted by curiousmuch
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!

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

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

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

This contradiction means that there are no positive solutions to the original equation.
• May 19th 2009, 07:29 PM
curiousmuch
Thanks for the help! The previous equation simply refers to x^4 - 4y^4 = z^2.
• May 19th 2009, 07:52 PM
curiousmuch
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?