A homogeneous Diophantine equation is a Diophantine equation that is defined by a
homogeneous polynomial. A typical such equation is the equation of
Fermat's Last Theorem :x^d+y^d -z^d=0. As a homogeneous polynomial in indeterminates defines a
hypersurface in the
projective space of dimension , solving a homogeneous Diophantine equation is the same as finding the
rational points of a projective hypersurface. Solving a homogeneous Diophantine equation is generally a very difficult problem, even in the simplest non-trivial case of three indeterminates (in the case of two indeterminates the problem is equivalent with testing if a
rational number is the th power of another rational number). A witness of the difficulty of the problem is Fermat's Last Theorem (for , there is no integer solution of the above equation), which needed more than three centuries of mathematicians' efforts before being solved. For degrees higher than three, most known results are theorems asserting that there are no solutions (for example Fermat's Last Theorem) or that the number of solutions is finite (for example
Faltings' theorem). For the degree three, there are general solving methods, which work on almost all equations that are encountered in practice, but no algorithm is known that works for every cubic equation.
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
modulo. For example, the Diophantine equation :x^2+y^2=3z^2, does not have any other solution than the trivial solution . In fact, by dividing , and 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 , and are all even, and are thus not coprime. Thus the only solution is the trivial solution . This shows that there is no
rational point on a
circle of radius \sqrt{3}, 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 :Q(x_1, \ldots, x_n)=0 be a homogeneous Diophantine equation, where Q(x_1, \ldots, x_n) is a
quadratic form (that is, a homogeneous polynomial of degree 2), with integer coefficients. The
trivial solution is the solution where all x_i are zero. If (a_1, \ldots, a_n) is a non-trivial integer solution of this equation, then \left(a_1, \ldots, a_n\right) are the
homogeneous coordinates of a
rational point of the hypersurface defined by . Conversely, if \left(\frac {p_1}q, \ldots, \frac {p_n}q \right) are homogeneous coordinates of a rational point of this hypersurface, where q, p_1, \ldots, p_n are integers, then \left(p_1, \ldots, p_n\right) is an integer solution of the Diophantine equation. Moreover, the integer solutions that define a given rational point are all sequences of the form :\left(k\frac{p_1}d, \ldots, k\frac{p_n}d\right), where is any integer, and is the greatest common divisor of the p_i. It follows that solving the Diophantine equation Q(x_1, \ldots, x_n)=0 is completely reduced to finding the rational points of the corresponding projective hypersurface.
Parameterization Let now A=\left(a_1, \ldots, a_n\right) be an integer solution of the equation Q(x_1, \ldots, x_n)=0. As is a polynomial of degree two, a line passing through 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 , 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\ne 0. Then one may pass to the affine case by considering the
affine hypersurface defined by :q(x_1,\ldots,x_{n-1})=Q(x_1, \ldots, x_{n-1},1), which has the rational point :R= (r_1, \ldots, r_{n-1})=\left(\frac{a_1}{a_n}, \ldots, \frac{a_{n-1}}{a_n}\right). If this rational point is a
singular point, that is if all
partial derivatives are zero at , all lines passing through are contained in the hypersurface, and one has a
cone. The change of variables :y_i=x_i-r_i does not change the rational points, and transforms into a homogeneous polynomial in variables. In this case, the problem may thus be solved by applying the method to an equation with fewer variables. If the polynomial 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 : :\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} Substituting this in , one gets a polynomial of degree two in , that is zero for . It is thus divisible by . The quotient is linear in , and may be solved for expressing as a quotient of two polynomials of degree at most two in t_2, \ldots, t_{n-1}, with integer coefficients: :x_1=\frac{f_1(t_2, \ldots, t_{n-1})}{f_n(t_2, \ldots, t_{n-1})}. Substituting this in the expressions for x_2, \ldots, x_{n-1}, one gets, for , :x_i=\frac{f_i(t_2, \ldots, t_{n-1})}{f_n(t_2, \ldots, t_{n-1})}, where 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 , :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), be the
homogenization of f_i. These quadratic polynomials with integer coefficients form a parameterization of the projective hypersurface defined by : :\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} A point of the projective hypersurface defined by is rational if and only if it may be obtained from rational values of t_1, \ldots, t_{n-1}. As F_1, \ldots,F_n are homogeneous polynomials, the point is not changed if all are multiplied by the same rational number. Thus, one may suppose that 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, \ldots, x_n) where, for , :x_i= k\,\frac{F_i(t_1, \ldots, t_{n-1})}{d}, where is an integer, t_1, \ldots, t_{n-1} are coprime integers, and is the greatest common divisor of the integers F_i(t_1, \ldots, t_{n-1}). One could hope that the coprimality of the , could imply that . Unfortunately this is not the case, as shown in the next section.
Example of Pythagorean triples The equation :x^2+y^2-z^2=0 is probably the first homogeneous Diophantine equation of degree two that has been studied. Its solutions are the
Pythagorean triples. 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 , corresponding to the point of the unit circle. A line passing through this point may be parameterized by its slope: :y=t(x+1). Putting this in the circle equation :x^2+y^2-1=0, one gets :x^2-1 +t^2(x+1)^2=0. Dividing by , results in :x-1+t^2(x+1)=0, which is easy to solve in : :x=\frac{1-t^2}{1+t^2}. It follows :y=t(x+1) = \frac{2t}{1+t^2}. Homogenizing as described above one gets all solutions as :\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} where is any integer, and are coprime integers, and is the greatest common divisor of the three numerators. In fact, if and are both odd, and if one is odd and the other is even. The
primitive triples are the solutions where and . This description of the solutions differs slightly from Euclid's formula because Euclid's formula considers only the solutions such that , and are all positive, and does not distinguish between two triples that differ by the exchange of and , ==Diophantine analysis==