I've been tacking a problem in my abstract algebra assignment which wants me to show that the ideal $(4 + \sqrt{-5}, 1 + 2\sqrt{-5})$ is not principle in $\mathbb{Z}[\sqrt{-5}]$ and I've reduced it to a problem of finding that there does not exist a solution to the linear Diophantine system:
$$ \begin{cases} 4a - 5b + c - 10d = 1 \\ a + 4b + 2c + d = 0 \end{cases} $$
However, I just can't seem to find a contradiction. I know that it's probably not true given solving it numerically, but I would appreciate a hint in what direction to take this. I've tried reducing this modulo some number, but $2$ just gives me a parity argument on $a,d$ and $b,c$ and $3$ actually solves properly. $4,5$ seemed useless since a variable would get eliminated.
Thanks!
Here's my solution to the overall problem so far:
Let $\mathfrak{a} = (4+\sqrt{-5},1+2\sqrt{-5})$. Recall that the norm in $\mathbb{Z}[\sqrt{-5}]$ is $N(a + b\sqrt{-5}) = a^2 + 5b^2$. Notice that $N(4 + \sqrt{-5}) = 16 + 5(1)^2 = 21$ and $N(1 + 2\sqrt{-5}) = 1 + 5(2)^2 = 21$. Therefore, if $\mathfrak{a}$ were principle generated by $\gamma \implies N(\gamma) \mid 21 \implies N(\gamma) = 1,3$ or $7$. However, the values $3$ and $7$ are not in the range of $N$, which tells us that $N(\gamma) = 1 \implies \gamma$ a unit of $\mathbb{Z}[\sqrt{-5}]$, i.e. is $\pm 1$. So then, it should be true that there exist $a + b\sqrt{-5}, c + d\sqrt{-5} \in \mathbb{Z}[\sqrt{-5}]$ such that:
\begin{gather*} (a + b\sqrt{-5})(4 + \sqrt{-5}) + (c + d\sqrt{-5})(1 + 2\sqrt{-5}) = 1 \\ 4a + (a+4b)\sqrt{-5} - 5b + c + (2c+d)\sqrt{-5} - 10d = 1 \\ \begin{cases} 4a - 5b + c - 10d = 1 \\ a + 4b + 2c + d = 0 \end{cases} \end{gather*}
EDIT: Easy solution pointed out by quasi, following his steps we end up with $14a + 35b + 21c = 1 \implies 7(\dots) = 1$ which is a contradiction. I can't believe I began to look up general solvability of systems of diophantine equations before trying this, probably a side effect of a having the internet at my fingertips.
Solve the second equation for $d$.
Substitute the result for $d$ in the first equation.
Look at the coefficients of the new equation.