Editing Diophantine equation (section)

===Degree two===
Homogeneous Diophantine equations of degree two are easier to solve. The standard solving method proceeds in two steps. One has first to find one solution, or to prove that there is no solution. When a solution has been found, all solutions are then deduced.

For proving that there is no solution, one may reduce the equation [[modular arithmetic|modulo {{mvar|p}}]]. For example, the Diophantine equation
:<math>x^2+y^2=3z^2,</math>
does not have any other solution than the trivial solution {{math|(0, 0, 0)}}. In fact, by dividing {{mvar|x, y}}, and {{mvar|z}} by their [[greatest common divisor]], one may suppose that they are [[coprime]]. The squares modulo 4 are congruent to 0 and 1. Thus the left-hand side of the equation is congruent to 0, 1, or 2, and the right-hand side is congruent to 0 or 3. Thus the equality may be obtained only if {{mvar|x, y}}, and {{mvar|z}} are all even, and are thus not coprime. Thus the only solution is the trivial solution {{math|(0, 0, 0)}}. This shows that there is no [[rational point]] on a [[circle]] of radius <math>\sqrt{3}</math>, centered at the origin.

More generally, the [[Hasse principle]] allows deciding whether a homogeneous Diophantine equation of degree two has an integer solution, and computing a solution if there exist. 

If a non-trivial integer solution is known, one may produce all other solutions in the following way.

====Geometric interpretation====
Let 
:<math>Q(x_1, \ldots, x_n)=0</math>
be a homogeneous Diophantine equation, where <math>Q(x_1, \ldots, x_n)</math> is a [[quadratic form]] (that is, a homogeneous polynomial of degree 2), with integer coefficients. The ''trivial solution'' is the solution where all <math>x_i</math> are zero. If <math>(a_1, \ldots, a_n)</math> is a non-trivial integer solution of this equation, then <math>\left(a_1, \ldots, a_n\right)</math> are the [[homogeneous coordinates]] of a [[rational point]] of the hypersurface defined by {{mvar|Q}}. Conversely, if  <math display="inline">\left(\frac {p_1}q, \ldots, \frac {p_n}q \right)</math> are homogeneous coordinates of a rational point of this hypersurface, where <math>q, p_1, \ldots, p_n</math> are integers, then <math>\left(p_1, \ldots, p_n\right)</math> is an integer solution of the Diophantine equation. Moreover, the integer solutions that define a given rational point are all sequences of the form 
:<math>\left(k\frac{p_1}d, \ldots, k\frac{p_n}d\right),</math>
where {{mvar|k}} is any integer, and {{mvar|d}} is the greatest common divisor of the <math>p_i.</math>

It follows that solving the Diophantine equation <math>Q(x_1, \ldots, x_n)=0</math> is completely reduced to finding the rational points of the corresponding projective hypersurface.

====Parameterization====

Let now <math>A=\left(a_1, \ldots, a_n\right)</math> be an integer solution of the equation <math>Q(x_1, \ldots, x_n)=0.</math> As {{mvar|Q}} is a polynomial of degree two, a line passing through {{mvar|A}} crosses the hypersurface at a single other point, which is rational if and only if the line is rational (that is, if the line is defined by rational parameters). This allows parameterizing the hypersurface by the lines passing through {{mvar|A}}, and the rational points are those that are obtained from rational lines, that is, those that correspond to rational values of the parameters.

More precisely, one may proceed as follows. 

By permuting the indices, one may suppose, without loss of generality that <math>a_n\ne 0.</math> Then one may pass to the affine case by considering the [[affine algebraic variety|affine hypersurface]] defined by 
:<math>q(x_1,\ldots,x_{n-1})=Q(x_1, \ldots, x_{n-1},1),</math>
which has the rational point
:<math>R= (r_1, \ldots, r_{n-1})=\left(\frac{a_1}{a_n}, \ldots, \frac{a_{n-1}}{a_n}\right).</math>

If this rational point is a [[singular point of an algebraic variety|singular point]], that is if all [[partial derivative]]s are zero at {{mvar|R}}, all lines passing through {{mvar|R}} are contained in the hypersurface, and one has a [[conical surface|cone]]. The change of variables 
:<math>y_i=x_i-r_i</math>
does not change the rational points, and transforms {{mvar|q}} into a homogeneous polynomial in {{math|''n'' − 1}} variables. In this case, the problem may thus be solved by applying the method to an equation with fewer variables.

If the polynomial {{mvar|q}} is a product of linear polynomials (possibly with non-rational coefficients), then it defines two [[hyperplane]]s. The intersection of these hyperplanes is a rational [[flat (geometry)|flat]], and contains rational singular points. This case is thus a special instance of the preceding case.

In the general case, consider the [[parametric equation]] of a line passing through {{mvar|R}}:
:<math>\begin{align}
x_2 &= r_2 + t_2(x_1-r_1)\\
&\;\;\vdots\\
x_{n-1} &= r_{n-1} + t_{n-1}(x_1-r_1).
\end{align}</math>
Substituting this in {{mvar|q}}, one gets a polynomial of degree two in {{math|''x''{{sub|1}}}}, that is zero for {{math|1=''x''{{sub|1}} = ''r''{{sub|1}}}}. It is thus divisible by {{math|''x''{{sub|1}} − ''r''{{sub|1}}}}. The quotient is linear in {{math|''x''{{sub|1}}}}, and may be solved for expressing {{math|''x''{{sub|1}}}} as a quotient of two polynomials of degree at most two in <math>t_2, \ldots, t_{n-1},</math> with integer coefficients:
:<math>x_1=\frac{f_1(t_2, \ldots, t_{n-1})}{f_n(t_2, \ldots, t_{n-1})}.</math>
Substituting this in the expressions for <math>x_2, \ldots, x_{n-1},</math> one gets, for {{math|1=''i'' = 1, …, ''n'' − 1}},
:<math>x_i=\frac{f_i(t_2, \ldots, t_{n-1})}{f_n(t_2, \ldots, t_{n-1})},</math>
where <math>f_1, \ldots, f_n</math> are polynomials of degree at most two with integer coefficients.

Then, one can return to the homogeneous case. Let, for {{math|1=''i'' = 1, …, ''n''}}, 
:<math>F_i(t_1, \ldots, t_{n-1})=t_1^2 f_i\left(\frac{t_2}{t_1}, \ldots, \frac{t_{n-1}}{t_1} \right),</math>
be the [[homogenization of a polynomial|homogenization]] of <math>f_i.</math> These quadratic polynomials with integer coefficients form a parameterization of the projective hypersurface defined by {{mvar|Q}}:
:<math>\begin{align}
x_1&= F_1(t_1, \ldots, t_{n-1})\\
&\;\;\vdots\\
x_n&= F_n(t_1, \ldots, t_{n-1}).
\end{align}</math>

A point of the projective hypersurface defined by {{mvar|Q}} is rational if and only if it may be obtained from rational values of <math>t_1, \ldots, t_{n-1}.</math> As <math>F_1, \ldots,F_n</math> are homogeneous polynomials, the point is not changed if all {{mvar|t{{sub|i}}}} are multiplied by the same rational number. Thus, one may suppose that <math>t_1, \ldots, t_{n-1}</math> are [[coprime integers]]. It follows that the integer solutions of the Diophantine equation are exactly the sequences <math>(x_1, \ldots, x_n)</math> where, for {{math|1=''i'' = 1, ..., ''n''}},
:<math>x_i= k\,\frac{F_i(t_1, \ldots, t_{n-1})}{d},</math>
where {{mvar|k}} is an integer, <math>t_1, \ldots, t_{n-1}</math> are coprime integers, and {{mvar|d}} is the greatest common divisor of the {{mvar|n}} integers <math>F_i(t_1, \ldots, t_{n-1}).</math>

One could hope that the coprimality of the {{mvar|t{{sub|i}}}}, could imply that {{math|1=''d'' = 1}}. Unfortunately this is not the case, as shown in the next 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}},