(1) Assume that 
. We know that any
polynomial in one variable over 
can be written as 
where 
are pairwise coprime and squarefree polynomials, and
essentially in a unique manner (up to multiplication of each 
by suitable
constants). This result can be homogenized and transformed into an identical
one for homogeneous polynomials in 2 variables.
Since 
, by definition of the discriminant this means that
F has at least a double root in 
, in other words in the
decomposition above there exists i>1 such that 
is not equal to a
constant. Since F is of degree 3, this means that 
or else
with 
, 
and 
of degree 1. It follows in
particular that all the roots of F modulo p are in 
itself.
(2) We have just seen that if 
then all the roots of F are
in 
and there is at least a double root. Hence write
Then one finds that
Now since F is primitive, 
and 
cannot both be zero modulo
p, hence
On the other hand, if 
then we cannot have 
otherwise 
so 
, absurd.
(3) From now on we assume that 
. Replacing F by an equivalent
form G, we may assume that the triple root of G modulo p is at 
,
i.e that 
for some 
. This
implies that 
. Since G is primitive,
we have 
.
Assume first 
. We thus have
by definition of 
.
We could also have written 
for an integral form 
, from which we obtain
, which
immediately implies the result.
Now assume p=3 and that 
, or equivalently that 
.
Then 
, hence 
, so
, hence 
, so finally
.
(4) Assume p=3 and that 
. Then since
, we see that 
and 
. It follows that
and since F is
primitive we cannot have 
and 
, so 
is a root of F
modulo 3, hence the root of F modulo 3. Hence 
if and
only if 
. Since b and c are divisible by 3,
the value of 
modulo 9 depends only on x and y modulo 3, hence
and the result follows.
