Find all solutions $(x, y, z)$ ∈ $\mathbb N^+ ×\mathbb N^+ ×\mathbb N^+$ for the following equations in $\mathbb Z$ 
Find all solutions $(x, y, z)$ ∈ $\mathbb N^+ ×\mathbb N^+ ×\mathbb N^+$ for the following equations in $\mathbb Z$:
(a) $3x^2 + 4y^2 = z^2$
(b) $3x^2 + 6y^2 = z^2$

My attempt:
a) If I divide the equation by $z^2$, I will get: $3(\frac xz)^2$+$4(\frac yz)^2$ = $1$
If I let A = $\frac xz$ and B = $\frac yz$ then $3A^2+4B^2=1$. But then how do I continue?
and what about part b? Any help please?
 A: 
Proposition.  Every solution $(x,y,z)\in\mathbb{Z}_{\geq 0}^3$ to the equation
  $$3x^2+4y^2=z^2$$
  is either of the form
  $$(x,y,z)=d\,\Biggl(2st,\left|\frac{s^2-3t^2}{2}\right|,s^2+3t^2\Biggr)$$
  for some $d\in\mathbb{Z}_{\geq 0}$ and odd coprime $s,t\in\mathbb{Z}_{\geq 0}$ such that $3\nmid s$,
  or the form
  $$(x,y,z)=d\,\Big(4st,\left|s^2-3t^2\right|,2s^2+6t^2\Big)$$
  for some $d\in\mathbb{Z}_{\geq 0}$ and coprime $s,t\in\mathbb{Z}_{>0}$ with different parity such that $3\nmid s$.

Without loss of generality, we may assume that $\gcd(x,y,z)=1$.  First, observe that $x$ is even (otherwise $z^2\equiv -1\pmod{4}$, which is impossible). Note that
$$(z-2y)(z+2y)=3x^2\,.$$
Since $x$ is even, $z$ is also even.  Then, write $x=2u$ and $z=2w$ for some integers $u,w\geq 0$, which gives
$$(w-y)(w+y)=3u^2\,.$$
We have two cases as $\gcd(w-y,w+y)\in\{1,2\}$.  
If $\gcd(w-y,w+y)=1$, then either
$$w-y=s^2\text{ and }w+y=3t^2\,,$$
or
$$w-y=3t^2\text{ and }w+y=s^2\,,$$
for some odd $s,t\in\mathbb{Z}_{\geq 0}$ such that $\gcd(s,t)=1$.  That is,
$$(x,y,z)=(2u,y,2w)=\left(2st,\frac{3t^2-s^2}{2},s^2+3t^2\right)\,,$$
or
$$(x,y,z)=(2u,y,2w)=\left(2st,\frac{s^2-3t^2}{2},s^2+3t^2\right)\,.$$
Hence, we can rewrite these families of solutions as
$$(x,y,z)=\Biggl(2st,\left|\frac{s^2-3t^2}{2}\right|,s^2+3t^2\Biggr)\,,$$
where $s$ and $t$ are odd coprime positive integers.  Note that $3\nmid s$ must hold.
If $\gcd(w-y,w+y)=2$, then either
$$w-y=2s^2\text{ and }w+y=6t^2\,,$$
or
$$w-y=6t^2\text{ and }w+y=2s^2\,,$$
for some $s,t\in\mathbb{Z}_{>0}$ with different parity such that $\gcd(s,t)=1$.  Solving this in a similar manner, we get
$$(x,y,z)=\Big(4st,\left|s^2-3t^2\right|,2s^2+6t^2\Big)$$
for some coprime $s,t\in\mathbb{Z}_{>0}$ with different parity.  Note that $3\nmid s$ must hold.
A: 
Proposition.  Every solution $(x,y,z)\in\mathbb{Z}_{\geq 0}^3$ to the equation $$3x^2+6y^2=z^2$$
  is of the form
  $$(x,y,z)=d\,\Big(\left|s^2-4st-2t^2\right|,\left|s^2+2st-2t^2\right|,3(s^2+2t^2)\Big)\,,$$
  for some $d\in\mathbb{Z}_{\geq0}$ and coprime $s,t\in\mathbb{Z}$ such that $s$ is odd and $3\nmid s+t$.

Without loss of generality, we assume that $\gcd(x,y,z)=1$.    Observe that $x$ must be odd, $z$ is odd, and $3\mid z$.  That is, $z=3w$ for some odd integer $w$.  The equation now becomes
$$x^2+2y^2=3w^2\,.$$
Rewrite the equation as
$$(x-\alpha\,y)(x+\alpha\,y)=(1+\alpha)(1-\alpha)\,w^2\,,\tag{*}$$
where $\alpha:=\sqrt{-2}$.  
The ring $R:=\mathbb{Z}[\alpha]$ is a unique factorization domain whose units are $\pm1$.  Note that
$$\gcd(x-\alpha\,y,x+\alpha\,y)\in\{1,\alpha,2\}\,.$$
If $\gcd(x-\alpha\,y,x+\alpha\,y)\neq 1$, then $2\mid w^2$, which is a contradiction.  That is, $\gcd(x-\alpha\,y,x+\alpha\,y)=1$.
From this observation, (*) implies that either
$$x+\alpha\,y=\pm(1+\alpha)(s+\alpha\,t)^2$$
or
$$x+\alpha\,y=\pm(1-\alpha)(s+\alpha\,t)^2$$
for some coprime $s,t\in\mathbb{Z}$, as $1\pm\alpha$ are prime elements of $R$.  Hence,
$$(x,y,z)=\pm\big(s^2-4st-2t^2,s^2+2st-2t^2,3(s^2+2t^2)\big)$$
or
$$(x,y,z)=\pm\big(s^2+4st-2t^2,-s^2+2st+2t^2,3(s^2+2t^2)\big)\,.$$
We can combine both results into a single family:
$$(x,y,z)=\Big(\left|s^2-4st-2t^2\right|,\left|s^2+2st-2t^2\right|,3(s^2+2t^2)\Big)\,,$$
where $s$ and $t$ are coprime integers.  Since $x$ is odd, $s$ must be odd.  Note also that $3\nmid s+t$.
