Editing Diophantine equation (section)

====Example of Pythagorean triples====
The equation 
:<math>x^2+y^2-z^2=0</math>
is probably the first homogeneous Diophantine equation of degree two that has been studied. Its solutions are the [[Pythagorean triple]]s. This is also the homogeneous equation of the [[unit circle]]. In this section, we show how the above method allows retrieving [[Euclid's formula]] for generating Pythagorean triples.

For retrieving exactly Euclid's formula, we start from the solution {{math|(−1, 0, 1)}}, corresponding to the point {{math|(−1, 0)}} of the unit circle. A line passing through this point may be parameterized by its slope:
:<math>y=t(x+1).</math>
Putting this in the circle equation 
:<math>x^2+y^2-1=0,</math>
one gets 
:<math>x^2-1 +t^2(x+1)^2=0.</math>
Dividing by {{math|''x'' + 1}}, results in
:<math>x-1+t^2(x+1)=0,</math>
which is easy to solve in {{mvar|x}}:
:<math>x=\frac{1-t^2}{1+t^2}.</math>
It follows
:<math>y=t(x+1) = \frac{2t}{1+t^2}.</math>
Homogenizing as described above one gets all solutions as 
:<math>\begin{align}
x&=k\,\frac{s^2-t^2}{d}\\
y&=k\,\frac{2st}{d}\\
z&=k\,\frac{s^2+t^2}{d},
\end{align}</math>
where {{mvar|k}} is any integer, {{mvar|s}} and {{mvar|t}} are coprime integers, and {{mvar|d}} is the greatest common divisor of the three numerators. In fact, {{math|1=''d'' = 2}} if {{mvar|s}} and {{mvar|t}} are both odd, and {{math|1=''d'' = 1}} if one is odd and the other is even.

The ''primitive triples'' are the solutions where {{math|1=''k'' = 1}} and {{math|''s'' > ''t'' > 0}}.

This description of the solutions differs slightly from Euclid's formula because Euclid's formula considers only the solutions such that {{mvar|x, y}}, and {{mvar|z}} are all positive, and does not distinguish between two triples that differ by the exchange of {{mvar|x}} and {{mvar|y}},