Diophantine equation
Parameterization

Let now A = (a 1, …, a n) {\displaystyle A=\left(a_{1},\ldots,a_{n}\right)} be an integer solution of the equation Q (x 1, …, x n) = 0. {\displaystyle Q(x_{1},\ldots,x_{n})=0.} As Q is a polynomial of degree two, a line passing through 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 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 a n ≠ 0. {\displaystyle a_{n}\neq 0.} Then one may pass to the affine case by considering the affine hypersurface defined by

which has the rational point

If this rational point is a singular point, that is if all partial derivatives are zero at R, all lines passing through R are contained in the hypersurface, and one has a cone. The change of variables

does not change the rational points, and transforms q into a homogeneous polynomial in 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 q is a product of linear polynomials (possibly with non-rational coefficients), then it defines two hyperplanes. The intersection of these hyperplanes is a rational 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 R:

Substituting this in q, one gets a polynomial of degree two in x, that is zero for x = r. It is thus divisible by x – r. The quotient is linear in x, and may be solved for expressing x as a quotient of two polynomials of degree at most two in t 2, …, t n − 1, {\displaystyle t_{2},\ldots,t_{n-1},} with integer coefficients:

Substituting this in the expressions for x 2, …, x n − 1, {\displaystyle x_{2},\ldots,x_{n-1},} one gets, for i = 1, …, n − 1,

where f 1, …, f n {\displaystyle f_{1},\ldots,f_{n}} are polynomials of degree at most two with integer coefficients.

Then, one can return to the homogeneous case. Let, for i = 1, …, n,

be the homogenization of f i. {\displaystyle f_{i}.} These quadratic polynomials with integer coefficients form a parameterization of the projective hypersurface defined by Q:

A point of the projective hypersurface defined by Q is rational if and only if it may be obtained from rational values of t 1, …, t n − 1. {\displaystyle t_{1},\ldots,t_{n-1}.} As F 1, …, F n {\displaystyle F_{1},\ldots,F_{n}} are homogeneous polynomials, the point is not changed if all t are multiplied by the same rational number. Thus, one may suppose that t 1, …, t n − 1 {\displaystyle t_{1},\ldots,t_{n-1}} are coprime integers. It follows that the integer solutions of the Diophantine equation are exactly the sequences (x 1, …, x n) {\displaystyle (x_{1},\ldots,x_{n})} where, for i = 1,..., n,

where k is an integer, t 1, …, t n − 1 {\displaystyle t_{1},\ldots,t_{n-1}} are coprime integers, and d is the greatest common divisor of the n integers F i (t 1, …, t n − 1). {\displaystyle F_{i}(t_{1},\ldots,t_{n-1}).}

One could hope that the coprimality of the t, could imply that d = 1. Unfortunately this is not the case, as shown in the next section.
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