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How can we find all the primitive solutions of the diophantine equation $x^2+3y^2=z^2$ ?

Some solutions are $(\pm n , 0 , \pm n)$ for $n\in \mathbb{N}$. How can we find also the other ones?

Mary Star
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    I haven't worked this out (hence the comment), but if you divide by $z^2$ and work in $\Bbb{Q}(\sqrt{-3})$ you should be able to apply the same method from this answer. – A.P. Dec 17 '18 at 23:15
  • $$x^2+ay^2=z^2$$ $$x=p^2-as^2$$ $$y=2ps$$ $$z=p^2+as^2$$ – individ Dec 18 '18 at 04:42
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    As mentioned above by @individ, this can be easily transformed into a Pell equation by looking at it in the form $x^2 = z^2 - 3y^2$ and dividing by $x^2$, giving the equation $Z^2 - 3Y^2 = 1$ in $Z,Y \in \Bbb{Q}$. These solutions are in bijection with primitive solutions to your equation, since given $(Z,Y)$ there is a unique way to write $Z = z/x$ and $Y = y/x$ with $\gcd(x,y,z) = 1$. – A.P. Dec 18 '18 at 14:19
  • @A.P. The slightly nontrivial work that you did warrants a more general duplicate for both which address $x^2+Dy^2=z^2$ for all $D \in \Bbb Q$. – Lord_Farin Dec 18 '18 at 18:31
  • @Lord_Farin I'm happy to post such a duplicate for $D \in \Bbb{Z}$, but I'm not entirely sure that $D \in \Bbb{Q}$ admits the same kind of solution, as it would lead to an equation of the form $ax^2 - by^2 = a$ with $a,b \in \Bbb{Z}$. – A.P. Dec 18 '18 at 19:50
  • @A.P. You're right, my bad. $D \in \Bbb Z$ would be great. – Lord_Farin Dec 18 '18 at 21:23

1 Answers1

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We can follow the stereographic projection method used to find Pythagorean triples. We start by dividing through by $z^2$, giving us the equation $$\left(\frac{x}{z}\right)^2 + 3\left(\frac{y}{z}\right)^2 = 1.$$ Or, in other words, we search for rational points on the ellipse $$x^2 + 3y^2 = 1.$$ We'll take the point $(1, 0)$ and stereographically project onto the $y$-axis. Consider an arbitrary point $(0, y)$ on the $y$-axis, and consider the line $$r = (1, 0) + t(-1, y)$$ between $(1, 0)$ and $(0, y)$. This line passes through the ellipse at $(1, 0)$ and a second point, which will be the point that stereographically projects onto $(0, y)$. The $t$ value for this point must satisfy, $$(1 - t)^2 + 3(ty)^2 = 1 \iff t((3y^2 + 1)t - 2) = 0.$$ The $t = 0$ solution produces $(1, 0)$, so we discard it. The other solution is $$t = \frac{2}{3y^2 + 1},$$ which yields the point on the ellipse, $$\left(\frac{3y^2 - 1}{3y^2 + 1}, \frac{2y}{3y^2 + 1}\right).$$ Now, when we have a rational point on the ellipse, the line will be of rational slope, and hence will stereographically project to a point $(y, 0)$ where $y$ is rational. That is, the rational point on the ellipse must take the above form where $y \in \mathbb{Q}$. If we take $y = \frac{m}{n}$, where $n \neq 0$, and $m, n \in \mathbb{Z}$, then this becomes $$\left(\frac{3\left(\frac{m}{n}\right)^2 - 1}{3\left(\frac{m}{n}\right)^2 + 1}, \frac{2\frac{m}{n}}{3\left(\frac{m}{n}\right)^2 + 1}\right) = \left(\frac{3m^2 - n^2}{3m^2 + n^2}, \frac{2mn}{3m^2 + n^2}\right).$$

To wrap this up, suppose we have an integer solution $(x, y, z)$ to the Diophantine equation $x^2 + 3y^2 = z^2$. If $z = 0$, then $x = y = 0$, which is one solution. Otherwise, $z \neq 0$ and $\left(\frac{x}{z}, \frac{y}{z}\right)$ must be a (rational) point on the ellipse. This point may be $(1, 0)$, in which case $(x, y, z) = k(1, 0, 1)$ (the points you've mentioned). The other solutions must take the form $$\left(\frac{x}{z}, \frac{y}{z}\right) = \left(\frac{3m^2 - n^2}{3m^2 + n^2}, \frac{2mn}{3m^2 + n^2}\right).$$ Equating the fractions, we must have, for some integers $k, l$, \begin{align*} x &= k(3m^2 - n^2) \\ z &= k(3m^2 + n^2) \\ y &= l(2mn) \\ z &= l(3m^2 + n^2). \end{align*} As such, $k = l$, so our solution is of the form $$(x, y, z) = k(3m^2 - n^2, 2mn, 3m^2 + n^2)$$ for integers $m, n, k$. Importantly, each of these are a solution, as $$(k(3m^2 - n^2))^2 + 3(k(2mn))^2 = k^2(9m^4 + n^4 - 6m^2 n^2 + 12m^2 n^2) = (k(3m^2 + n^2))^2.$$ Note that the $(0, 0, 0)$ solution can be obtained from $k = 0$, but there is no way to obtain the $k(1, 0, 1)$ solutions. Hence, our general solution is $$(x, y, z) = k(1, 0, 1) \text{ or } k(3m^2 - n^2, 2mn, 3m^2 + n^2) \text{ for } k, m, n \in \mathbb{Z}.$$

Theo Bendit
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  • from your final line, if your $z$ is even so is $x.$ When $v$ is odd we can use $$ x = 2 u^2 + 2 uv - v^2, ; y = 2uv + v^2, ; z = 2 u^2 + 2uv + 2 v^2$$ to get $z$ even but the others odd. In short, the rational solutions produce a finite set of parametrizations of the same type as Pythagorean triples, when writing out all integer solutions. – Will Jagy Dec 18 '18 at 00:38
  • for some recent examples https://math.stackexchange.com/questions/3036094/determine-whether-the-following-quadratic-forms-are-isotropic-over-mathbbq/3036132#3036132 – Will Jagy Dec 18 '18 at 00:45