All solutions $(x, y, z) \in (\mathbb{N},\mathbb{N},\mathbb{N})$ for $3x^2 + 6y^2 = z^2$

Question

My attempt:

$z^2 = 3(x^2 + 2y^2)$

so $3|z^2$ and thus $3|z$

letting $z = 3k$, we reduce the equation to $x^2+2y^2=3k^2$

Obviously $(n,n,n) $ represents infinitely many solutions, but not all since $(5,1,3)$ is a solution for example.

Also, if $x$ and $k$ are even then $4|2y^2$, then $y$ is even

If $x$ and $k$ are odd, then $x^2$ and $k^2$ are congruent to 1 mod 8 so $2y^2$ is congruent to 2 mod 8 so y is also odd.Therefore, it's sufficient to find the odd solutions.

But I don't know how to proceed from here. Any hints on how to parametrize the odd solutions or a suitable mod $n$ to consider to simplify it further would be appreciated.

Answer 1

There is a systematic way to solve such equation: the idea is to parametrize the ellipse $C: 3X^2+6Y^2=1$ by a family of lines through a fixed rational point. (in other words, an ellipse is a rational curve).

Here $A=({1\over 3}, {1\over 3})$ is a rational point on this ellipse. Note that a line through $A$ has equation $L_t:(X-1/3)=t(Y-1/3)$ must cut the ellipse in exactly one other point which is a rational function of $t$.

It is preferable to write $C= (3(X-1/3)^2+6(Y-1/3)^2+6(X-1/3)+12(Y-1/3)=0$

Let $U=X-1/3, V=Y-1/3$ The intersection of $L_t\cap C$ is $U=tV, 3U^2+6V^2+6U+12V=0$

Or $3t^2V^2+6V^2+6tV+12V=0$

If we exclude the point $A$ (where $V=0$) we get; $V(3t^2+6)=-6t-12$, $V=-{t+4\over t^2+3}, U=t V$

Now $t={a\over b}\in \bf Q$, ${x\over z}=U+1/3=-t{t+4\over t^2+3}+1/3$ and ${y\over z}=V+1/3=-{t+4\over t^2+3}$

${x\over z}=-{a^2+4ab\over a^2+3b^2}+1/3={-2a^2+3b^2-12ab\over a^2+3b^2}$

${y\over z}=-{a/b+4\over (a/b)^2+3}+1/3=$ $-{ab+4b^2\over a^2+3b^2}+1/3=$ $a^2-3ab-9b^2\over 3(a^2+b^2)$

So $x=-2a^2+3b^2-12ab$, $y=a^2-3ab-9b^2$, $z= 3(a^2+b^2)$ is a solution if $(a,b,c)\in \bf Z$.

Conversely, for every solution, the point $ P=(x/z, y/z)$ is on the ellipse, so $(x,y,z)$ is proportional to the triple $(-2a^2+3b^2-12ab , a^2-3ab-9b^2,3(a^2+b^2))$, for $(a,b)$ the slope of the line $(A,P)$.

Note that a different choice for $A$ yields a different parametrization of the set of solution...

Answer 2

It is easy enough to get all rational solutions by stereographic projection around a single point. The trouble is that we want all integer solutions with $\gcd(x,y,k) = 1.$

There is a method for producing recipes of the Pythagorean Triple type. As far as I know, first written down in Fricke and Klein (1897).

$$ x = u^2 + 4 u v - 2 v^2 \; , \; \; y = u^2 - 2 uv - 2 v^2 \; , \; \; k = u^2 + 2 v^2 $$

In order to keep both $u,v$ positive, I just added in the effect of negating $v$ as:

$$ x = u^2 - 4 u v - 2 v^2 \; , \; \; y = u^2 + 2 uv - 2 v^2 \; , \; \; k = u^2 + 2 v^2 $$

The proof that this one recipe suffices is simply that forcing all to have gcd 2 or 3 leads to a new version of the same thing, when divided by that 2 or 3. We stick with the first description and allow negative variables $u,v.$

If we demand $u = -2t,$ we arrive at $$ \frac{-x}{2} = v^2 + 4tv - 2t^2 \; , \; \frac{-y}{2} = v^2 - 2tv - 2t^2 \; , \; \frac{k}{2} = v^2 + 2t^2 $$

If we demand $u = s +2t, v = s - t,$ so that $u \equiv v \pmod 3,$ we arrive at $$ \frac{x}{3} = s^2 + 4st - 2t^2 \; , \; \frac{-y}{3} = s^2 - 2st - 2t^2 \; , \; \frac{k}{3} = s^2 + 2t^2 $$

Thu Oct 29 08:38:25 PDT 2020
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