Find all solutions to the Diophantine equation x^2+y^2 = 7z^2.

Please teach me how to solve it . Thank you.

Results 1 to 3 of 3

- Oct 24th 2006, 08:00 AM #1

- Joined
- Sep 2006
- Posts
- 83

- Oct 24th 2006, 08:45 AM #2

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

There are no solutions to this diophantine equation.

7 is a number of the form 4k+3

$\displaystyle z^2$ can have only two form, $\displaystyle 4k,4k+1$.

Therefore,

$\displaystyle 7z^2$ has the form of 4k+3.

The important thing here is that the sum of two squares,

$\displaystyle x^2+y^2$ can never take the form 4k+3.

Thus, there are no solutions to this equation.

- Oct 24th 2006, 09:04 AM #3

- Joined
- Sep 2006
- Posts
- 83

Hi Perfecthacker,

I got it . Thank you very much.

============================

Also, how should I prove the below question?

Prove Fermat's last theorem for exponent three; i.e. prove that if

x^3 + y^3 = z^3

where x, y, and z are rational integers, then x, y, or z is 0.

Hint: show that x^3 + y^3 = (epsilon)z^3 , where x, y, and z are quadratic integers in Q[-3], and epsilon is a unit in Q[-3] , then x, y, or z is 0.