How to conclude the congruence $2\cdot \text{disc}(F)\equiv 2b^2c^2\in\mathbb{Z}$

Question

I am studying a proof which shows that the cubic form $F(x,y)=ax^3+bx^2y+cxy^2+dy^3$ is integral (i.e has integer coefficients.) So far I have the following facts:

$a,d\in\mathbb{Z}$,

$b+c\in\mathbb{Z}$,

$b-c\in\mathbb{Z}$, and

$b,c\in\frac{1}{2}\mathbb{Z}$ with $b\equiv c \pmod 1$

Now the proof says: ''by the explicit formula for $\text{disc}(F)$ and using that $b,c\in\frac{1}{2}\mathbb{Z}$ we get $2\cdot disc(F)\equiv 2b^2c^2\in\mathbb{Z}$

where, $\text{disc}(F)=b^2c^2-27a^2d^2+18abcd-4ac^3-4b^3d$

I would like to see how this congruence can be worked out but don't know where to begin.

Any tips greatly appreciated

Answer 1

All the terms of 2·disc(F) are integers (because $c^3$ and $b^3 \in 1/8 \mathbb Z$) except $2b^2c^2$. Is that what you need?