Looking for proof of no solution to 4-variable quadratic diophantine equation Show there are no integers $a,b,c,d$ such that $$\begin{cases}1=9ac+3ad+3bc-16bd \\ 1=3ad+3bc+2bd \end{cases}$$
Motivation: The ideal $I=(3,1+\sqrt{-17})$ in $R=\mathbb{Z}[\sqrt{-17}]$ has the property that that $\omega=1+\sqrt{-17} \in I^2$ but is not a product of two elements of $I$. Writing out the claim (that $\omega$ is a product of two elements of $I$) as a diophantine equation (and noting the integral basis of $I$) gives the equation.
Mild progress: Subtracting the equations gives $0=9(ac-2bd)$ so that $ac=2bd$ and so we get an equivalent system:
$$\begin{cases}0=ac-2bd \\ 1=3ad+3bc+2bd \end{cases}$$
Another method: The norm $N(a+b\sqrt{-17}) = a^2+17b^2$ is multiplicative, and the minimum value of $N$ on $I\setminus \{0\}$ is $9$, so the minimum value of a nonzero product of two elements of $I$ is $81$, while the value on $\omega$ is only $18$. However, I'm not familiar with low-tech methods of finding the minimum nonzero norm of a set. In particular how does one show:
$$\min\left(\{ 3a^2+2ab+6b^2 : a,b\in\mathbb{Z}; (a,b) \neq 0\}\right) = 3$$
or at least that it is strictly greater than $1$? [ I've shown it is at least equal to 1. ]
 A: Suppose we want to make $3a^2+2ab+6b^2$ small, where neither $a$ nor $b$ is $0$. It is clearly best to choose $a$ and $b$ of opposite sign. So we want to make $3x^2-2xy+6y^2$ small, with $x$ and $y$ positive. Note that 
$$3x^2-2xy+6y^2=3(x-\sqrt{2}y)^2+(6\sqrt{2}-2)xy.$$
For positive $x$ and $y$, the right-hand side is greater than $6$.
A: We have $3a^2+2ab+6b^2=(a+b)^2+2a^2+5b^2$, which is as a sum of non-negative squares at least $5$, if not $b=0$. Hence we may assume that $b=0$. Then obviously $3a^2\ge 3$. The general estimates can be found in the nice answer of Will.
A: You only have to show it doesn't equal $1$. Solving the equation for $a$ gives
$$
b=\frac{-a\pm \sqrt{6-17a^2}}6
$$
If $a\neq 0$, the root is imaginary, so $a=1$. But then, $b$ is not an integer, so this doesn't have any solutions.
A: Not sure you need all this high power machinery. Here is my analysis
From $ac = 2 bd$ we have
$$ a/b = 2d /c ~\text{ (say) } = s/t$$
Then
$$
a = u s \\ b=u t \\ d =v s/2 \\ c= v t$$
Substitute in the second and solve for $u$
$$
\frac{2}{\left( 6\,{t}^{2}+2\,s\,t+3\,{s}^{2}\right) \,v}$$
So 
$$ v = \pm 1 ~\text{ or } \pm 2$$
and 
$$
6\,{t}^{2}+2\,s\,t+3\,{s}^{2} = \pm1 ~\text{ or } \pm 2$$
Argue that $6\,{t}^{2}+2\,s\,t+3\,{s}^{2} $ can never be $\pm 1$ or $\pm 2$ by setting it to $\pm 1$ and $\pm 2$ and solving for $t$.
