# Generalization of Pythagoras equation

Standard

About 3 years ago I discussed following two Diophantine equations of degree 2:

In this post, we will see a slight generalization of the result involving Pythagorean triplets. Unlike Pythagoras equation, $x^2+y^2-z^2=0$, we will work with a little bit more general equation, namely: $ax^2+by^2+cz^2=0$, where $a,b,c\in \mathbb{Z}$. For proofs, one can refer to section 5.5 of Niven-Zuckerman-Montgomery’s An introduction to the theory of numbers.

Theorem: Let $a,b,c\in \mathbb{Z}$ be non-zero integers such that the product is square free. Then $ax^2+by^2+cz^2=0$ have a non-trivial solution in integers if and only if $a,b,c$ do not have same sign, and that $-bc, -ac, -ab$ are quadratic residues modulo $a,b,c$ respectively.

In fact, this result helps us determine the existence of a non-trivial solution of any degree 2 homogeneous equation in three variables, $f(X,Y,Z)=\alpha_1 X^2 +\alpha_2Y^2+\alpha_3Z^2+\alpha_4XY+\alpha_5YZ+\alpha_6ZX$ due to the following lemma:

Lemma: There exists a sequence of changes of variables (linear transformations) so that $f(X,Y,Z)$ can be written as an equation of the form $g(x,y,z)=ax^2+by^2+cz^2$ with $\gcd(a,b,c)=1$.

Now let’s consider the example. Let $f(x,y,z)=3x^2+5y^2+7z^2+9xy+11yz+13zx$, and we want to determine whether this $f(x,y,z)=0$ has a non-trivial solution. Firstly, we will do change of variables:

$\displaystyle{f(x,y,z)=3\left(x+\frac{3}{2}y +\frac{13}{6}z\right)^2 - \frac{7}{4}y^2 - \frac{85}{12}z^2 - \frac{17}{2}yz = g(x',y',z')}$

where $x' = x+\frac{3}{2}y +\frac{13}{6}z$, $y'=y$ and $z'=z$. Thus

$\displaystyle{12g(x',y',z')=36x'^2 - 21 y'^2 - 85z'^2 - 102y'z' = 36x'^2 - 21\left(y'+\frac{17}{7}z'\right)^2+\frac{272}{7}z'^2=h(x'',y'',z'')}$

where $x'' = x'$,$y'' = y'+\frac{17}{7}z'$ and $z''=z'$. Thus

$\displaystyle{7h(x''',y'',z'') = 252x''^2 - 147y''^2+272z''^2=7(6x'')^2-3(7y'')^2 + 17(4z'')^2 = F(X,Y,Z)}$

where $X=6x''$, $Y=7y''$ and $Z=4z''$. Now we apply the theorem to $7X^2-3Y^2+17Z^2=0$. Since all the coefficients are prime numbers, we can use quadratic reciprocity to conclude that the given equation has non-trivial solution (only non trivial thing to note that $-7\times 17$ is quadratic residue mod -3, is same as $-7\times 17$ is quadratic residue mod 3).