Editing Diophantine equation (section)

====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.