An explicit computation show that 
is a square modulo 4.
Hence, it can always be written as 
. Assume first 
. Then
and so we can write
with 
integral.
Assume 
. A computation shows that
Since 
we have 
hence if 
, either
, i.e 
, or 
,
i.e 
.
Assume now that p=2. If 
, we have
Since 
, 
. Since
the square of an integer is congruent to 0 or 1 modulo 4, it follows that
in other words 
, hence
once again either 
or 
as before.
Conversely, assume that either 
or
. If 
, then since F is
primitive, by Proposition 3.3 (2), we have 
.
Assume first p>3. Then 
, hence
, so 
.
Assume now that p=2. Then 
for some other quadratic form 
,
hence 
, hence the same is true for
and hence 
.
Finally, assume that p=3. From the explicit formulas, 
is equivalent to 
and 
, from which it follows by the
formula for the discriminant that 
, and in particular
.
Assume now that 
,
and 
. We may assume that
, otherwise we are in the case
which we have just dealt with.
By the same reasoning as before, writing
, we have
Hence 
is equivalent to
which implies 
,
hence if p>2 we have 
.
For p=2 and 
, we have as before
and since
, we deduce that
and hence 
, thus finishing the proof of the proposition.
