Solve $a^2 + 2b^2 - 3c^2 - 6d^2 = 1$ and $ab=3cd$ over integers I'd like to solve this set of equations over $\mathbb{Z}$.  Consider this pair of conic sections:
\begin{eqnarray*}
a^2 + 2b^2 - 3c^2 - 6d^2 &=& 1 \\
ab - 3 cd&=& 0
\end{eqnarray*}
If these two equations were over $\mathbb{R}$ , this should define the intersection of a pair of quadrics in four-dimensional space.
One possible starting point is to try that $\frac{a}{c} \times \frac{b}{d} = 3 $ so these numbers are proportional.

From $ab=3cd$ it follows that $3 \mid ab$. If $3\mid a$ then we get that $1=a^2+2b^2-3c^2-6d^2 \equiv 2b^2 \pmod 3$. Multiplying by $2$ we get $2 \equiv 4b^2\equiv b^2 \pmod 3$ which is a contradiction because there are no squares of integers congruent to $2$ modulo $3$.
Therefore $3 \mid b$. Write $b=3b'$. Then $ab'=cd$. This means that $a=kl, b'=mn, c=km, d=ln$ for some integers $k,l,m,n$. Substituting this to the first equation we get
$$(k^2-6n^2)(l^2-3m^2)=1.$$
It follows that $k^2-6n^2=1=l^2-3m^2$ or $k^2-6n^2=-1=l^2-3m^2$. The second case is impossible, as then $l^2 \equiv 2 \pmod 3$. Therefore we have $k^2-6n^2=1=l^2-3m^2$.
These equations are known as Pell equations and all their solutions are known. See e.g. here.
So, all solutions to the OP's question are quadruples $(a,b,c,d)=(kl,3mn,km,ln)$, where $(k,n)$ is a solution to the Pell equation $k^2-6n^2=1$, and $(l,m)$ is a solution to the Pell equation $l^2-3m^2=1$.
