Find all solutions $(x, y, z)$ ∈ $\mathbb N^+ ×\mathbb N^+ ×\mathbb N^+$ for the following equations in $\mathbb Z$

Question

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?

Answer 1

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.

Answer 2

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 $3\nmid s+t$.