A quick count shows that $A$ has 36 points.
Those more than half a square count as whole ones.
We shall thus avoid doubly counting the same contribution.
At the core of our proof of Theorem 1 is a simple counting argument.
the number of zeros of $f$ counted according to their multiplicities $\langle$counted with multiplicity$\rangle$
Go to the list of words starting with: a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
s
t
u
v
w
y
z