Solving $a^2+3b^2=c^2$

I’m looking for how to solve the equation $a^2+3b^2=c^2$ where $a,b,c$ are integers and $b$ is even, I’m looking for the algorithm used to solve this kind of equations, not just the solution.

Regards

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The argument below is much like the argument for generating the Pythagorean triples, and the answer is structurally very similar. There is the trivial solution $(0,0,0)$.

We now will look for solutions with $a$ and $c$ relatively prime. The other solutions can then be obtained by multiplying by an integer scaling factor.

Rewrite our equation as $3b^2=c^2-a^2=(c-a)(c+a)$. Note that $a$ cannot be even. For if $a$ is even, then by relative primality $b$ is odd, and therefore $a^2+3b^2$ is of the form $4k+3$. But $c^2$ cannot be of that form.

It follows that the
gcd of $c-a$ and $c+a$ is $2$ or $1$. Since $3$ divides $(c-a)(c+a)$, we get that $3$ divides $c-a$ or $3$ divides $c+a$. By changing the sign of $a$ if necessary, we can assume that $3$ divides $c+a$.

We first deal with the case where $c-a$ and $c+a$ are both even. In that case, we have $c-a=\pm2 s^2$ and $c+a=\pm 6t^2$ for some non-negative (and relatively prime, of opposite parity) integers $s$ and $t$.
If we are looking for non-negative solutions, then solving for $c$ and $a$ we get
$c=3t^2+s^2$, $a=|3t^2-s^2|$, and therefore $b=2st$. And of course we can change signs at will.

Finally, we deal with the case $c-a$ and $c+a$ both odd. Then $c-a=s^2$ and $c+a=3t^2$ for odd $s$ and $t$. Again, solve for $c$ and $a$.

Another way to think about such problems is to convert this to
$$\left(\frac{a}{c}\right)^2+3\left(\frac{b}{c}\right)^2=1,$$
and then think about rational points (both coordinates rational) on the curve
$$x^2+3y^2=1.$$
One obvious point is $(1,0)$. Let us consider a line that passes through this point $y=m(x-1)$ with rational slope $m$. Observe that all the points of intersection of this line and the curve (given above) will be rational. To find the intersection we solve
$$x^2+3m^2(x-1)^2=1.$$
This is equivalent to
$$x^2(3m^2+1)-6m^2x+(3m^2-1)=0.$$
Since one of the roots of this equation has to be $x=1$ (our trivial rational point) so the other root will be
$$\alpha=\frac{3m^2-1}{3m^2+1}.$$
Thus we obtain the following point as the other point of intersection:
$$\left(\frac{3m^2-1}{3m^2+1}, \, \frac{-2m}{3m^2+1}\right).$$

So if we take $a=3m^2-1, b=-2m$ and $c=3m^2+1$ (with $m \in \mathbb{Z}$) then we can obtain integer solutions to the equation $a^2+3b^2=c^2$.

More generally, if we take $m=\frac{r}{s}$ (with $r,s \in \mathbb{Z}$ and $s \neq 0$) then we can have $a=3r^2-s^2, b=-2rs$ and $c=3r^2+s^2$ as the solutions.

Here are stubborn! There is a formula framed and immediately the answer! What is the problem?

In this equation:

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

After substituting in the formula of the solution obtained:

$$x=2p^2+2ps-s^2$$

$$y=(2p+s)s$$

$$z=2(p^2+ps+s^2)$$

And more:

$$x=2p^2+6ps+3s^2$$

$$y=(2p+3s)s$$

$$z=2(p^2+3ps+3s^2)$$

And more:

$$x=2p^2-10ps-s^2$$

$$y=8p^2+2ps-s^2$$

$$z=2(7p^2+ps+s^2)$$

$p,s$ – integers of any sign.

To obtain a primitive solutions may be necessary to divide all the numbers on common divisor.