Thread: x^2+y^2 = 7z^2

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

Please teach me how to solve it . Thank you.

Originally Posted by beta12

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

Please teach me how to solve it . Thank you.

There are no solutions to this diophantine equation.

7 is a number of the form 4k+3

can have only two form, .

Therefore,
has the form of 4k+3.

The important thing here is that the sum of two squares,
can never take the form 4k+3.

Thus, there are no solutions to this equation.

Hi Perfecthacker,

I got it . Thank you very much.
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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.