<s>[INST] Solve the following equation in integers $x,y:$ $x^2+6xy+8y^2+3x+6y=2.$ Question: Solve the following equation in integers $x,y:$ $$x^2+6xy+8y^2+3x+6y=2.$$
Solution: For some $x,y\in\mathbb{Z}$ $$x^2+6xy+8y^2+3x+6y=2\\\iff x^2+2xy+4xy+8y^2+3x+6y=2\\\iff x(x+2y)+4y(x+2y)+3(x+2y)=2\\\iff(x+4y+3)(x+2y)=2.$$ 
Now if $(x+4y+3)(x+2y)=2$, then either $$\begin{cases} x+4y+3=1\\ x+2y=2\end{cases}\text{ or }\begin{cases} x+4y+3=2\\ x+2y=1\end{cases}\text{ or }\begin{cases} x+4y+3=-1\\ x+2y=-2\end{cases}\text{ or }\begin{cases} x+4y+3=-2\\ x+2y=-1\end{cases}.$$ 
We have $$\begin{cases} x+4y+3=1\\ x+2y=2\end{cases}\iff (x,y)=(6,-2), \\\begin{cases} x+4y+3=2\\ x+2y=1\end{cases}\iff (x,y)=(3,-1), \\\begin{cases} x+4y+3=-1\\ x+2y=-2\end{cases}\iff (x,y)=(0,-1),\\\begin{cases} x+4y+3=-2\\ x+2y=-1\end{cases}\iff (x,y)=(3,-2).$$ 
Now since, all the four pairs $(6,-2),(3,-1),(0,-1),(3,-2)$ satisfies the integer equation $(x+4y+3)(x+2y)=2$, thus we can conclude that $(x+4y+3)(x+2y)=2\iff (x,y)=(6,-2),(3,-1),(0,-1),(3,-2).$ 
Hence, we can conclude that the integer equation $x^2+6xy+8y^2+3x+6x=2$ is satisfied if and only if $(x,y)=(6,-2),(3,-1),(0,-1),(3,-2)$, and we are done. 
Is the solution correct and rigorous enough? And, I am always confused while solving equations regarding the usage of the if and only if arguments, which I feel is very necessary in order to have a complete and rigorous solution, but I rarely find it's usage in any book while solving equations of any kind. So, is it necessary? Also, is there a better solution than this? 
 [/INST] Another solution is to let $k=2y$ and get
 $$x^2+3xk+2k^2+3x+3k=2$$
$$\iff (x^2+2xk+k^2)+(k^2+3xk)+3(x+k)=2\iff (x+k)^2+(x+k)(k+3)=2$$
So now we can let $m=x+k$ and get
$$m(m+k+3)=2 \implies (m,k) \in \{(1,-2),(-1,-4),(2,-4),(-2,-2)\}$$
Noting that $k=\frac{2}{m}-m-3$ in our calculations. This gives (after dividing the solutions of $k$ by $2$ and putting $x=m-k$)
$$(x,y) \in \{(3,-1),(3,-2),(6,-2),(0,-1)\}$$
This is much simpler.


Or else, I guess we can stop at the point where
$$(x+4y+3)(x+2y)=2$$
and let $a$ be an integer such that $a=x+2y$, this simplifies a lot:
$$a(a+2y+3)=2$$
And we note that $a \in \{1,-1,2,-2\}$ and $y=\frac{1}{a}-\frac{a}{2}-\frac{3}{2}$
$$\implies (a,y) \in \{(1,-1),(-1,-2),(2,-2),(-2,-1)\}$$
$$x=a-2y\implies (x,y) \in \{(3,-1),(3,-2),(6,-2),(0,-1)\}$$
That makes stuff a lot easier.
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