ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 24 (2024) #P1.04
https://doi.org/10.26493/1855-3974.2903.9ca
(Also available at http://amc-journal.eu)
Intersecting families of graphs of functions over
a finite field*
Angela Aguglia , Bence Csajbók †
Dipartimento di Meccanica, Matematica e Management, Politecnico di Bari,
Via Orabona 4, I-70125 Bari, Italy
Zsuzsa Weiner
ELKH–ELTE Geometric and Algebraic Combinatorics Research Group,
1117 Budapest, Pázmány P. stny. 1/C, Hungary and
Prezi.com, H-1065 Budapest, Nagymező utca 54-56, Hungary
Received 10 June 2022, accepted 30 December 2022, published online 9 August 2023
Abstract
Let U be a set of polynomials of degree at most k over Fq , the finite field of q elements.
Assume that U is an intersecting family, that is, the graphs of any two of the polynomials in
U share a common point. Adriaensen proved that the size of U is at most qk with equality
if and only if U is the set of all polynomials of degree at most k passing through a common
point. In this manuscript, using a different, polynomial approach, we prove a stability
version of this result, that is, the same conclusion holds if |U | > qk − qk−1. We prove a
stronger result when k = 2.
For our purposes, we also prove the following results. If the set of directions determined
by the graph of f is contained in an additive subgroup of Fq , then the graph of f is a line.
If the set of directions determined by at least q −√q/2 affine points is contained in the set
of squares/non-squares plus the common point of either the vertical or the horizontal lines,
then up to an affinity the point set is contained in the graph of some polynomial of the form
αxp
k
.
Keywords: Direction problem, Erdős-Ko-Rado, finite field, polynomial.
Math. Subj. Class. (2020): 11T06
*We are extremely grateful for the reviewer’s thorough reading and valuable comments. An inaccuracy spotted
out by the reviewer led us to the discovery of Theorem 2.13. The second and the third author acknowledge the
support of the National Research, Development and Innovation Office – NKFIH, grant no. K 124950. This
work was supported by the Italian National Group for Algebraic and Geometric Structures and their Applications
(GNSAGA– INdAM).
†Corresponding author.
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
2 Ars Math. Contemp. 24 (2024) #P1.04
1 Introduction
In 1961, Erdős et al. [6] proved that if F is a k-uniform intersecting family of subsets of
an n-element set X , then |F | ≤
(
n−1
k−1
)
when 2k ≤ n. Furthermore, they proved that if
2k + 1 ≤ n, then equality holds if and only if F is the family of all subsets containing a
fixed element x ∈ X . There are several versions of the Erdős-Ko-Rado theorem. For a
survey of this type of results, see [7, 13] or [5].
In this manuscript, we investigate an Erdős-Ko-Rado type problem for graphs of func-
tions over a finite field. The idea of this work comes from the recent manuscript [1] by
Adriaensen, where the author studies intersecting families of ovoidal circle geometries and,
as a consequence, of graphs of functions over a finite field.
Definition 1.1. If f is an Fq → Fq function, then the graph of f is the affine q-set:
Uf = {(x, f(x)) : x ∈ Fq}.
The set of directions determined by (the graph of) f is
Df =
{
f(x)− f(y)
x− y
: x, y ∈ Fq, x ̸= y
}
.
Definition 1.2. For a family of polynomials, U , we say that U is t-intersecting if for any
two polynomials f1, f2 ∈ U , the graphs of f1 and f2 share at least t points, that is,
|{(x, f1(x)) : x ∈ Fq} ∩ {(x, f2(x)) : x ∈ Fq}| ≥ t.
Instead of 1-intersecting, we will also use the term “intersecting”.
Note that if U is a t-intersecting family of polynomials of degree at most k, then also
|{(x, f1(x)) : x ∈ Fq} ∩ {(x, f2(x)) : x ∈ Fq}| ≤ k
holds for any pairs f1, f2 ∈ U , since (x, f1(x)) = (x, f2(x)) implies that x is a root of
f1 − f2 which has degree at most k.
In this note, we improve a result due to Adriaensen [1, Theorem 6.2] by using different
techniques. Adriaensen’s proof goes through association schemes and circle geometries,
our proof does not use these, as we rely on two classical results (Results 2.3 and 2.8) about
polynomials over finite fields.
To be more precise, our main results are the following theorems.
Theorem 1.3. Let U be a set of intersecting polynomials of degree k ≤ 2 over Fq . Assume
that q ≥ 53, when q is odd and q ≥ 8 when q is even. If
|U | > q2 −
q
√
q
4
+
cq
8
+
√
q
8
,
where c = 1 for q even and c = 3 for q odd, then the graphs of the functions in U share a
common point.
E-mail addresses: angela.aguglia@poliba.it (Angela Aguglia), bence.csajbok@poliba.it (Bence Csajbók),
zsuzsa.weiner@gmail.com (Zsuzsa Weiner)
A. Aguglia et al.: Intersecting families of graphs of functions over a finite field 3
We will prove Theorem 1.3 separately for q odd (Theorem 3.11) and for q even (Theo-
rem 3.15). For k > 2, the proof can be finished by induction as in [1, page 33]. We obtain
the following result.
Theorem 1.4. If U is a set of more than qk − qk−1 intersecting polynomials over Fq ,
q ≥ 53 when q is odd and q ≥ 8 when q is even, and of degree at most k, k > 1, then
there exist α, β ∈ Fq such that g(α) = β for all g ∈ U . Furthermore, U can be uniquely
extended to a family of qk intersecting polynomials over Fq and of degree at most k.
While finalizing our manuscript, a stronger stability version of the above mentioned
result for k = 2 was published by Adriaensen; see [2]. Our proof is different, based on
polynomials and hence might be of independent interest.
2 Preliminaries
Throughout this paper, q = pn for some prime p and a positive integer n. The algebraic
closure of the finite field Fq will be denoted by Fq .
The absolute trace function is defined as Trq/p : Fq → Fp, Trq/p(x) = x+ xp + . . .+
xp
n−1
. Recall that for q even and c ̸= 0, a+ bx+ cx2 ∈ Fq[x] has a root in Fq if and only
if b = 0, or b ̸= 0 and
Trq/2
(ac
b2
)
= 0. (2.1)
When q is a square, we will also use the notation N: Fq → F√q , x 7→ x
√
q+1, which is the
norm of x over F√q .
We will frequently need the following result of Ball, Blokhuis, Brouwer, Storme, Szőnyi
and Ball.
Result 2.1 (Part of [3, 4]). Let f be an Fq → Fq function such that |Df | ≤ (q + 1)/2.
Then Uf is affinely equivalent to the graph of a linearised polynomial, that is, a polynomial
of the form
∑n−1
i=0 aix
pi ∈ Fq[x].
Theorem 2.2. LetU denote a proper Fp-subspace of Fq , q = pn > 2, p prime and consider
a function σ : Fq → Fq . If the set of directions
Dσ =
{
σ(x)− σ(y)
x− y
: x, y ∈ Fq, x ̸= y
}
is contained in U , then σ(x) = ax+ b for some a, b ∈ Fq .
Proof. First note that U is contained in an (n− 1)-dimensional Fp-subspace V and hence
|Dσ| ≤ pn−1. Then by Result 2.1, σ(x) = α + g(x), where α ∈ Fq and g(x) =∑n−1
i=0 bix
pi ∈ Fq[x], thus
Dσ =
{
g(x)
x
: x ∈ Fq \ {0}
}
.
It is well-known that βq/p
∏
γ∈V (x− γ) = Trq/p(βx) for some β ∈ Fq \ {0}. Next define
f(x) := βg(x). Then Dσ ⊆ V implies
Trq/p
(
f(x)
x
)
= 0
4 Ars Math. Contemp. 24 (2024) #P1.04
for each x ∈ Fq \{0}. To prove the assertion, it is enough to prove that f(x) is linear. With
f(x) =
∑n−1
i=0 aix
pi ∈ Fq[x],
Trq/p
(
f(x)
x
)
= Trq/p
(
n−1∑
i=0
aix
pi−1
)
=
n−1∑
j=0
n−1∑
i=0
ap
j
i x
pi+j−pj ,
and because of our assumption, this polynomial vanishes at every element of Fq \ {0}.
The p > 2 case:
If we multiply this polynomial by x1+p+p
2+...+pn−1 , then we obtain
n−1∑
j=0
n−1∑
i=0
ap
j
i x
1+p+...+pn−1+pi+j−pj
and this polynomial vanishes at every element of Fq . As a function, this polynomial re-
mains the same if we consider it modulo xp
n − x, so it is the same function as the poly-
nomial we obtain when we replace the exponents pi+j with pi+j−n for each i + j ≥ n.
Denote this new polynomial with f̃ . The fact that we multiplied f by x1+p+p
2+...+pn−1
ensures that the exponents of f̃ are larger than 0 and smaller than q. We claim that each
monomial has different degree in f̃ . It is clear that f̃ is the sum of at most n2 monomials
and the set of degrees of these monomials is contained in the set
A := {1 + p+ p2 + . . .+ pn−1 + pc − pd : c, d ∈ {0, 1, . . . , n− 1}}.
Assume that for some c1, c2 ∈ {0, 1, . . . , n − 1}, d1, d2 ∈ {0, 1, . . . , n − 1} and
(c1, d1) ̸= (c2, d2)
1 + p+ p2 + . . .+ pn−1 + pc1 − pd1 = 1 + p+ p2 + . . .+ pn−1 + pc2 − pd2 ,
or equivalently pc1 + pd2 = pc2 + pd1 . Since the base p-digits of an integer are uniquely
determined, this implies {c1, d2} = {c2, d1}, so either c1 = c2 and d1 = d2, or c1 = d1
and c2 = d2. We conclude that two distinct monomials of f̃ have the same degree d if any
only if d = 1 + p+ . . .+ pn−1, that is, when in
1 + p+ . . .+ pn−1 + pi+j − pj
we have i = 0.
Note that the degree of f̃ is at most
m := 1 + p+ . . .+ pn−1 + pn−1 − 1.
Since p > 2, m is clearly smaller than q, but f̃ has q roots (the elements of Fq). Thus it is
the zero polynomial, all of its coefficients are zero. The coefficients of f̃ are the pj-powers
of a1, . . . , an−1 and Trq/p(a0). So f(x) = a0x.
The p = 2 case:
Note that when i ̸= 0, then
2i+j − 2j = 2j + 2j+1 + . . .+ 2i+j−1.
A. Aguglia et al.: Intersecting families of graphs of functions over a finite field 5
If 2i+j − 2j ≥ 2n = q then write this number as
2i+j−2j = 2j+2j+1+ . . .+2i+j−1 = 2j+2j+1+ . . .+2n−1+2n(1+ . . .+2i+j−1−n).
Clearly, as Fq → Fq functions,
x 7→ x2
i+j−2j
are the same as
x 7→ x2
j+2j+1+...+2n−1x1+...+2
i+j−1−n
=
x1+...+2
i+j−1−n+2j+2j+1+...+2n−1 .
When 2i+j − 2j ≥ 2n = q, then substitute these exponents in
Trq/2
(
f(x)
x
)
=
n−1∑
j=0
n−1∑
i=0
a2
j
i x
2i+j−2j
with the exponents
Bij := 1 + . . .+ 2
i+j−1−n + 2j + 2j+1 + . . .+ 2n−1 < q,
and denote this new polynomial with f̃ . Note that in this case i+ j− 1−n < j− 1. When
2i+j − 2j < q, i ̸= 0, then define
Aij := 2
i+j − 2j = 2j + 2j+1 + . . .+ 2i+j−1 < q.
If i = 0 then put A0j = 0. Since base 2-digits of an integer are uniquely determined,
we have the following: (1) If i1 ̸= 0, then Ai1j1 = Ai2j2 iff (i1, j1) = (i2, j2), (2)
A0j1 = A0j2 for any pair (j1, j2), (3) Bi1j1 = Bi2j2 iff (i1, j1) = (i2, j2), finally (4)
Bi1j1 = Ai2j2 iff i1+j1−n = j1 and j2 = 0 and i2 = n (otherwiseBi1j1 in base 2 has the
form 11 . . . 110 . . . 011 . . . 11, whileAi2j2 has the form 11 . . . 1100 . . . 00, a contradiction);
but i1, i2 < n. It follows that the only exponent which appears in more than one monomial
is the 0. The degree of f̃ is at most q − 2 (obtained in xA(n−1) 1 ) and it has q − 1 roots
(the elements of Fq \ {0}), so it is the zero polynomial. Hence all of its coefficients are
zero. These coefficients are the 2j-powers of a1, . . . , an−1 and Trq/2(a0). It follows that
f(x) = a0x.
We will need the following two results regarding functions over finite fields.
Result 2.3 ([9, Theorem 5.41], Weil’s bound). Let ψ be a multiplicative character of Fq
of order m > 1 and let f ∈ Fq[x] be a monic polynomial of positive degree that is not an
m-th power of a polynomial. Let d be the number of distinct roots of f in Fq . Then for
every a ∈ Fq we have ∣∣∣∣∣∣
∑
c∈Fq
ψ(af(c))
∣∣∣∣∣∣ ≤ (d− 1)√q.
We will also consider polynomials of degree
√
q+1 admitting square values for almost
every element of Fq . In this case, the inequality above seems to be useless. In Lemma 2.6,
we show a way how to derive information from Weil’s bound also in this case. When
m = d = 2 then the following, stronger result holds which can be easily proved by counting
Fq-rational points of a conic of PG(2, q):
6 Ars Math. Contemp. 24 (2024) #P1.04
Result 2.4 ([11, Exercise 5.32]). Let q be an odd prime power, f(x) = ax2+bx+c ∈ Fq[x]
with a ̸= 0, and let ψ denote the quadratic character Fq → {−1, 1, 0}. Then∑
x∈Fq
ψ(ax2 + bx+ c)
equals −ψ(a) if b2 − 4ac ̸= 0 and (q − 1)ψ(a) if b2 − 4ac = 0.
To use Result 2.3, we will need the following.
Lemma 2.5. Put f(x) = axp
k+1 + dxp
k
+ bx + c ∈ Fq[x], k ̸= 0. If q is odd and
f(x) = g(x)2, then dp
k
a = bap
k
and dp
k+1a = cap
k+1, or a = b = d = 0.
Proof. If a = 0, then b = d = 0 otherwise the degree of f was odd. Assume a ̸= 0 and
suppose f(x) = g(x)2. Then the roots of f have multiplicities at least 2 and hence they
are also roots of f ′(x) = axp
k
+ b = (ap
−k
x+ bp
−k
)p
k
. It follows that f(x) has a unique
root, −(b/a)p−k , so
f(x) = a(x+ γ)p
k+1 = a(xp
k
+ γp
k
)(x+ γ) = axp
k+1 + γaxp
k
+ aγp
k
x+ γp
k+1a,
with γ = (b/a)p
−k
. It follows that f has the listed properties.
Lemma 2.6. If for some odd, square q > 9 there is a subset D of Fq of size larger than
q −√q/2 + 1/2 such that the Fq → Fq function x 7→ ℓ(x) := ax
√
q+1 + dx
√
q + bx+ c,
a ̸= 0, has the property that ℓ(x) is a square of Fq for each x ∈ D, then a
√
qb = d
√
qa.
Proof. Suppose a
√
qb ̸= d
√
qa. Then the value set of ℓ clearly does not change if we
replace x with g(y) = ((b/a)
√
q − (d/a))y − d/a, since g is a permutation polynomial.
Also, C := g−1(D), will have the properties that |C| > q − √q/2 + 1/2 and for each
y ∈ C,
f(y) := ℓ(g(y))
is a square of Fq . One can easily verify f(y) = ℓ(g(y)) = βy
√
q+1 + βy + α, where
β =
(a
√
qb− ad
√
q)
√
q+1
a2
√
q+1
and α = c− bd/a. We will show that this is not possible.
Since the norm x 7→ N(x) takes (√q − 1) distinct non-zero values in Fq , and
(
√
q − 1)− (√q/2− 1/2) ≥ 2 (here we use q > 9 square), we may take t1, t2 ∈ Fq \ {0}
such that N(t1) ̸= N(t2) and N(t1),N(t2) /∈ {N(d) : d ∈ Fq \ C}. We show that if f(x)
is a square for each x ∈ C then also the polyomials
f(t1y
√
q−1) = N(t1)βy
q−1 + βt1y
√
q−1 + α ∈ Fq[y],
f(t2y
√
q−1) = N(t2)βy
q−1 + βt2y
√
q−1 + α ∈ Fq[y],
have only square values for each y ∈ C. Indeed, this follows from the fact that N(tiy
√
q−1) /∈
{N(d) : d ∈ Fq \ C} and hence tiy
√
q−1 ∈ C for i = 1, 2.
A. Aguglia et al.: Intersecting families of graphs of functions over a finite field 7
Then the polynomials
G1(y) := N(t1)β + βt1y
√
q−1 + α ∈ Fq[y],
G2(y) := N(t2)β + βt2y
√
q−1 + α ∈ Fq[y],
take only square values on the non-zero elements of C. Denote by ψ the multiplicative
character of Fq of order two. The polynomialGi has at most
√
q−1 roots, and ψ(Gi(x)) =
1 for every element x of C \ {0} if x is not a root of Gi. Define ε to be ψ(Gi(0)) if 0 ∈ C
and to be 0 otherwise. Then |C \ {0}| − (√q − 1) + ε ≤
∑
x∈C ψ(Gi(x)). On the other
hand, −(q − |C|) ≤
∑
x∈Fq\C ψ(Gi(x)), and hence
2|C| − q −√q − 1 ≤
∣∣∣∣∣∣
∑
y∈Fq
ψ(Gi(y))
∣∣∣∣∣∣ .
Since
(
√
q − 2)√q < 2|C| − q −√q − 1,
by Result 2.3 (with m = 2) this can only happen if Gi = g2i for some polynomials gi,
i = 1, 2. Then the roots of Gi (in the algebraic closure of Fq) are multiple roots of Gi
and hence also roots of gcd(Gi, G′i). The only root of G
′
i is 0, thus Gi(0) = 0 and hence
N(ti)β + α = 0. Since a
√
qb ̸= d
√
qa, we have β ̸= 0. It follows that N(ti) = −α/β for
i = 1, 2, a contradiction because of the choice of t1 and t2.
The next example shows that ℓ(x) = ax
√
q+1 + dx
√
q + bx + c can have only square
values if a
√
qb = d
√
qa holds.
Example 2.7. For t, r ∈ Fq , the polynomial
f(x) = r
√
q+1x
√
q+1 + r
√
qtx
√
q + rt
√
qx+ t
√
q+1 = (t+ rx)
√
q+1
has only square values in Fq .
We will need a generalisation of the following result by Göloğlu and McGuire.
Result 2.8 ([8, Theorem 1.2]). Let q be odd and consider a non zero polynomial L(x) =∑n−1
i=0 aix
pi ∈ Fq[x]. Denote by □q the set of non-zero squares in Fq . Then
Im
(
L(x)
x
)
⊆ □q ∪ {0}
if and only if L(x) = axp
d
for some a ∈ □q and 0 ≤ d ≤ n.
Definition 2.9. If U is a point set of AG(2, q), then the set of directions defined by U is
DU =
{(
a− b
c− d
)
: (a, b), (c, d) ∈ U, (a, b) ̸= (c, d)
}
.
(If the denominator is zero then
(
a−b
0
)
= (∞), the ideal point of vertical lines.)
8 Ars Math. Contemp. 24 (2024) #P1.04
In the proof of Theorem 2.13 the following result of Szőnyi will be crucial.
Result 2.10 ([10, Theorem 4 and Proposition 6]). Let U be a point set of AG(2, q) of size
at least q −√q/2 and let DU be the set of directions determined by U .
1. If U determines less than (q + 1)/2 directions, then U can be extended to a q-set
determining the same set of directions as U .
2. If U determines exactly (q + 1)/2 directions, one of them is (∞) and there is no
point P ∈ AG(2, q) \ U such that U ∪ {P} determines the same set of directions as
U , then the (q + 1)/2-set
{d ∈ Fq, (d) /∈ DU}
is the set of Y coordinates of the points of an irreducible conic C of AG(2, q) and the
direction (0) is an internal point of C.
Remark 2.11. By [12, Remark 3.3] a blocking set of size at most 2q contains a unique
minimal blocking set. Let U denote an affine point set of size at least q−√q/2 such that U
determines less than (q + 1)/2 directions. Assume that P and P ′ are two affine point sets
of size q−|U | which extend U to a q-set determining the same set of directions as U . Then
B := U ∪P ∪P ′ ∪DU is a blocking set of size at most ⌊q+
√
q/2+ (q+1)/2⌋ ≤ 2q and
hence B contains a unique minimal blocking set. But both U ∪ P ∪DU and U ∪ P ′ ∪DU
are minimal blocking sets and this proves P = P ′, that is, the unicity of the extension of
U in Result 2.10.
Lemma 2.12. Let S denote the set of non-zero squares or non-squares in GF(q). If the
set of Y coordinates of the points of an irreducible conic C of AG(2, q), q ≥ 53 odd, is
contained in S ∪ {0} then C is a parabola with equation
Y = α(a′X + b′Y + c′)2,
where α ∈ S.
Proof. Note that horizontal translations of C does not affect the properties that we are
examining, so after substituting X by X − β for a suitable β ∈ Fq we may assume that
(0, 0) is not a point of C and hence the equation of the conic is
aX2 + bXY + cY 2 + dX + eY + 1 = 0.
The direction (0) cannot be a point of the projective extension of C since otherwise we
would get at least q − 1 > (q + 1)/2 different Y coordinates. It follows that there are at
most 2 horizontal lines meeting C in 1 point and at least (q− 3)/2 horizontal lines meeting
C in 2 points. Fix some α ∈ S. At least (q − 5)/2 horizontal lines meet C in 2 points
(Ai, αB
2
i ) and (A
′
i, αB
2
i ) with Bi ̸= 0; and C has at most 2 points on the X axis. Next
define the quartic Q (which might as well be of smaller degree if c = 0):
aX2 + αbXY 2 + α2cY 4 + dX + αeY 2 + 1 = 0.
Points of C on the X axis are points of Q as well, and if (Ai, αB2i ), (A′i, αB2i ), Bi ̸= 0,
were two points of C then (Ai,±Bi), (A′i,±Bi) are 4 points of Q. It follows that
A = {(x, y), (x,−y) : (x, αy2) ∈ C}
A. Aguglia et al.: Intersecting families of graphs of functions over a finite field 9
is a subset of the point set of Q of size at least 2 + 2(q − 5) = 2q − 8. Note that 2q − 8 >
q + 1 + 6
√
q (here we use q ≥ 53). It follows instantly from the Hasse-Weil bound that Q
cannot be an irreducible cubic or an irreducible quartic.
First we show that Q cannot be a quartic curve which is the product of an irreducible
cubic and a line. Vertical and horizontal lines meet A in at most 2 points and hence if such
a line would be a factor of Q then the remaining at least 2q − 10 points of A should lie
on the cubic, a contradiction again by the Hasse-Weil bound now applied to cubic curves.
Similarly, if Y = mX + n was a factor of Q, for some m ̸= 0, then
aX2 + αbX(mX + n)2 + α2c(mX + n)4 + dX + αe(mX + n)2 + 1
was the zero polynomial. The coefficient of X4 is α2cm4, so c has to be zero but then Q is
not a quartic curve, a contradiction.
From now on we may assume that
aX2 + αbXY 2 + α2cY 4 + dX + αeY 2 + 1 = F ·G,
where F and G are of degree at most 2. Put
F = (a1Y
2 + b1X + c1Y + 1 + d1X
2 + e1XY ),
G = (a2Y
2 + b2X + c2Y + 1 + d2X
2 + e2XY ).
In F · G the coefficient of Y is c1 + c2, while it is 0 in the equation of Q, so clearly
c2 = −c1 and we will use this from now on. The coefficient of X4 is d1d2 and it has to be
zero, so from now on we may assume d1 = 0. Then the coefficient of X3 is d2b1 and it has
to be zero.
In this paragraph assume d2 ̸= 0 , then b1 = 0 and the coefficient of X3Y is d2e1 so
e1 = 0. The coefficient of Y X2 is c1d2, so c1 = 0. Then the coefficient of XY is e2,
so e2 = 0. The coefficient of X2Y 2 is d2a1, so a1 = 0. We arrived to the conclusion
that the equation of Q is a2Y 2 + b2X + 1 + d2X2. It follows that the equation of C is
Y = −α(b2X+1+d2X2)/a2. Then by Result 2.4, −(b2X+1+d2X2)/a2 = (a′X+b′)2
for some a′, b′ ∈ Fq and this finishes the proof of the d2 ̸= 0 case.
Now assume d2 = 0 (recall also d1 = 0). Then the coefficient of X2Y 2 is e1e2. We
may assume e1 = 0. The coefficient ofX2Y is e2b1. First assume e2 ̸= 0, so b1 = 0. Then
the coefficient of Y 3X is a1e2, so a1 = 0. Then the coefficient of Y 3 is c1a2. We cannot
have c1 = 0 since then the coefficient of XY would be e2 ̸= 0, so a2 = 0. The coefficient
of XY is c1b2 + e2 = 0 and hence the equation of Q is: (1 + c1Y )(1 − c1Y )(1 + b2X),
a contradiction since vertical and horizontal lines contain at most 2 points of A. So e2 = 0
and from now on we may assume that Q has equation
(a1Y
2 + b1X + c1Y + 1)(a2Y
2 + b2X − c1Y + 1) = 0.
The fact that Y 3 and XY should have zero coefficient yields a1 = a2 and b1 = b2, or
c1 = 0. In the former case the equation of Q is
(a1Y
2 + b1X + c1Y + 1)(a1Y
2 + b1X − c1Y + 1) = 0,
so C had equation
(α−1a1Y + b1X + 1)
2 − α−1c21Y = 0,
10 Ars Math. Contemp. 24 (2024) #P1.04
which proves the assertion.
In the latter case the equation of Q is
(a1Y
2 + b1X + 1)(a2Y
2 + b2X + 1) = 0,
so C had equation
(a1Y/α+ b1X + 1)(a2Y/α+ b2X + 1) = 0,
a contradiction since C was irreducible.
The following can be considered as a generalisation of Result 2.8.
Theorem 2.13. Let U denote a point set of AG(2, q), q ≥ 53 odd, of size at least q−√q/2.
Let S denote the set of non-zero squares or non-squares in Fq and let (d) denote one of
the directions (0) or (∞). If DU is contained in {(s) : s ∈ S} ∪ {(d)}, then U is affinely
equivalent to a subset of the graph of a function of the form
f(x) = αxp
k
,
where α ∈ S.
Proof. If |DU | < (q + 1)/2 then by Result 2.10 U can be extended to a q-set U ′ deter-
mining the same set of directions. According to Result 2.1 U ′ is affinely equivalent to the
graph of a linearised polynomial f . Then Result 2.8 shows that f has the requested form.
Now assume |DU | = (q + 1)/2. If (d) = (0), then apply the affinity φ : (x : y : z) 7→
(y : x : z). Clearly, DUφ = (DU )φ and U can be extended if and only if Uφ can be
extended. We have (0)φ = (∞) and if m ̸= 0 then (m)φ = (1/m), so
{(s)φ : s ∈ S} = {(s) : s ∈ S}.
By Result 2.10, if Uφ (or U , if (d) = (∞)) cannot be extended, then the set of non-
zero squares or non-squares together with the zero equals the set of Y coordinates of an
irreducible affine conic C and (0) is an internal point of C. Then the line at infinity is not
a tangent to C, thus C is not a parabola (and not a hyperbola because then the size of the
set of Y coordinates would be (q − 1)/2; but we don’t need this) but this is not possible
because of the Lemma 2.12. It follows that U can be extended to a q-set determining the
same set of directions as U and the proof can be finished as in the previous paragraph.
3 On intersecting families of graphs of functions
Our first aim is to prove Theorem 1.3 which we will do separately in the odd and even case.
Lemma 3.1. If U is a set of t-intersecting polynomials of degree at most k over Fq , then
the (k− t+1)-ple of coefficients of the monomials xt, . . . , xk in elements of U are distinct
elements of Fk−t+1q .
Proof. If the coefficients of xt, . . . , xk coincide in f1, f2 ∈ U , then f1 − f2 would have
degree at most t − 1, and hence at most t − 1 roots, thus the graphs of f1 and f2 would
share at most t− 1 points, a contradiction.
A. Aguglia et al.: Intersecting families of graphs of functions over a finite field 11
Next, we report below Lemma 6.1 from [1] with an alternative proof.
Lemma 3.2 ([1, Lemma 6.1]). Assume t ≤ k < q. Let U be a set of polynomials of degree
at most k over Fq .
(1) If for any f, g ∈ U there exist x1, . . . , xt ∈ Fq such that f(xi) = g(xi) for i =
1, . . . , t, then |U | ≤ qk−t+1.
(2) If for any f, g ∈ U there are no x1, . . . , xt ∈ Fq such that f(xi) = g(xi) for
i = 1, . . . , t, then |U | ≤ qt.
Proof. Proof of Part (1): It is a direct consequence of Lemma 3.1.
Proof of Part (2): Take any t distinct field elements, say, x1, . . . , xt. For any polynomial
f over Fq , (f(x1), . . . , f(xt)) can take at most qt distinct values of Ftq and hence if |U | >
qt then there will be at least 2 polynomials in U which have the same values on the set
{x1, . . . , xt}.
Lemma 3.3. LetU be a set of intersecting polynomials of degree at most 2 over Fq . Assume
that there are more than ⌊(q+1)/2⌋ polynomials hi in U , so that their x2 coefficients are c,
for some fixed c ∈ Fq and suppose also that there exist values α and β so that hi(α) = β.
Then for every polynomial f ∈ U , whose coefficient in x2 is not c, f(α) = β.
Proof. First assume that α = 0. Then the constant term in the polynomials hi is always β
and we want to show that for any polynomial f ∈ U , if the coefficient of x2 is not c, the
constant term must be β. Assume to the contrary, that there is a polynomial g ∈ U , whose
constant term is not β. Consider the polynomials:
{g − hi}.
Since g and hi are intersecting, g − hi must have a root in Fq . Also, by the assumptions
of the lemma, (g − hi)(x) = dx2 + vx + w, where d ̸= 0 and w ̸= 0 are fixed. We
claim that there are at most ⌊(q + 1)/2⌋ such polynomials, hence a contradiction. Indeed,
if (g− hi)(x) has a root in Fq , then it can be written as d(x− u)(x− wdu ). So to bound the
number of possible polynomials, we have to bound the number of different (u, wdu ) pairs,
where the order does not matter.
First assume q to be odd. If w/d is not a square, then we get (q − 1)/2 such pairs. If it
is a square, then we see 2 + (q − 3)/2 pairs, which is (q + 1)/2.
Now, assume q to be even. In this case the number of different pairs (u, wdu ) is
(q − 2)/2 + 1, that is, q/2.
Finally, if α ̸= 0, then instead of the polynomials f in U , consider the polynomials
f̄(x) := f(x + α). This new family is clearly an intersecting family, the h̄i polynomials
will still have the same leading coefficients and h̄i(0) = β, so we are in the previous
case.
Lemma 3.4. Let q be even and U be a set of intersecting polynomials of degree at most 2
over Fq . Assume that |U | > q
2+q
2 and assume also that H is a subset of U with more than
q2
2 polynomials hi, so that there exist values α and β for which hi(α) = β. Then for every
polynomial f ∈ U , f(α) = β.
12 Ars Math. Contemp. 24 (2024) #P1.04
Proof. By the pigeon hole principle, there exists a value c such that there are more than
q
2 polynomials in H with x
2 coefficient c. Let U c denote the polynomials in U with x2
coefficient c. Then Lemma 3.3 implies that for any polynomial f ∈ (U \ U c), f(α) = β.
Note that |U \ U c| > q
2+q
2 − q. Again by the pigeon hole principle, there exists a value
c′ ̸= c so that there are more than q
2−q
2(q−1) =
q
2 polynomials in (U \ U
c) with x2 coefficient
c′. Lemma 3.3 yields that for any polynomial g ∈ U c, g(α) = β.
The next result can be proved in exactly the same way as Lemma 3.4.
Lemma 3.5. Let q be odd and U be a set of intersecting polynomials of degree at most 2
over Fq . Assume that |U | > q
2+2q−1
2 and suppose also that H is a subset of U with more
than q
2+q
2 polynomials hi, so that there exist values α and β for which hi(α) = β. Then
for every polynomial f ∈ U , f(α) = β.
Lemma 3.6. Let U be a set of intersecting polynomials of degree at most k > 1 over
Fq . Assume that there are more than (q − 1)qk−2 polynomials hi in U , so that their xk
coefficients are c, for some fixed c ∈ Fq and suppose also that there exist values α and β
so that hi(α) = β. Then for every polynomial f ∈ U , whose coefficient of xk is not c, it
holds that f(α) = β.
Proof. First assume that α = 0. Then the constant term in the polynomials hi is always β
and we want to show that for any polynomial f ∈ U , if the coefficient of xk is not c, the
constant term must be β. Assume to the contrary, that there is a polynomial g ∈ U , whose
constant term is not β. Consider the polynomials:
{g − hi}.
Since g and hi are intersecting, g − hi must have a root in Fq . Also, by the assumptions of
the lemma, (g − hi)(x) = dxk + v1xk−1 + v2xk−2 + . . .+ vk−1x+w, where d ̸= 0 and
w ̸= 0 are fixed. We claim that there are at most (q − 1)qk−2 such polynomials, hence a
contradiction. Indeed, such polynomials can be written in the form (x− u)(dxk−1 + . . .−
w/u). Note that u ̸= 0, because w ̸= 0, hence u can take q − 1 values. The second term
is a polynomial of degree k − 1, its coefficient in xk−1 and its constant term are fixed, so
there are at most qk−2 different such polynomials.
As before, if α ̸= 0, then instead of the polynomials f in U , consider the polynomials
f̄(x) := f(x+α). This new family is clearly an intersecting family, the h̄i polynomials will
still have the same leading coefficients and h̄i(0) = β, so we are in the previous case.
3.1 Intersecting families of polynomials of degree at most 2, over Fq , q odd
According to Lemma 3.1, the members of an intersecting family of polynomials of degree
at most 2 are of the form f(b, c) + bx+ cx2 for some function f . More precisely:
Definition 3.7. Suppose that U is a set of intersecting polynomials. Put D = {(b, c) ∈
Fq × Fq : a+ bx+ cx2 ∈ U} and define f : D → Fq as f(b, c) = a, where a ∈ Fq is the
unique field element such that a+ bx+ cx2 ∈ U .
Lemma 3.8. Let U be a set of intersecting polynomials of degree at most 2 and for b ∈ Fq ,
q ≥ 53 odd, define
Cb := {c ∈ Fq : f(b, c) + bx+ cx2 ∈ U}
A. Aguglia et al.: Intersecting families of graphs of functions over a finite field 13
and
domo := {b ∈ Fq : |Cb| > q −
√
q/2 + 1/2}.
There exist functions s, t : domo → Fq and h : domo → {0, 1, . . . , n− 1} (where q = pn)
such that for every c ∈ Cb
f(b, c) = s(b)cp
h(b)
+ t(b),
and −s(b) is square in Fq .
Proof. If f(b, c)+ bx+ cx2 and f(d, e)+ dx+ ex2 are members of U , then the difference
of the two polynomials must have a root in Fq and hence
F (b, c, d, e) := (b− d)2 − 4(f(b, c)− f(d, e))(c− e)
is a square. For c ∈ Cb put fb(c) for f(b, c).
For each b ∈ Fq and C,E ∈ Cb, C ̸= E, consider F (b, C, b, E) = −4(fb(C) −
fb(E))(C − E), which has to be a square, or, equivalently, after dividing by (C − E)2,
−fb(C)− fb(E)
C − E
is in □q ∪ {0} for each C,E ∈ Cb.
If b ∈ domo, then by Theorem 2.13, fb can be uniquely extended to a function f̃b : Fq →
Fq determining the same set of directions as fb and for each c ∈ Fq f̃b(c) = s(b)cp
h(b)
+t(b)
for some domo → Fq functions s, t such that −s(b) is a square, and a function h : domo →
{0, 1, . . . , n− 1}. Then for c ∈ Cb we have fb(c) = s(b)cp
h(b)
+ t(b).
Lemma 3.9. If q ≥ 53 is odd, U is a set of intersecting polynomials of degree at most 2
such that |domo| > 1, then for b, d ∈ domo and c ∈ Cb, e ∈ Cd recall that
F (b, c, d, e) = (b− d)2 − 4(f(b, c)− f(d, e))(c− e)
= (b− d)2 − 4(s(b)cp
h(b)
+ t(b)− s(d)ep
h(d)
− t(d))(c− e).
For b ∈ domo, one of the following holds
1. s(b) = s(d) = 0 and t(d) = t(b) for each d ∈ domo,
2. s(b) = s(d) ̸= 0, h(d) = h(b) = 0 and (t(b) − t(d))2 = −s(b)(b − d)2 for each
d ∈ domo,
3. s(b) = s(d) ̸= 0, h(d) = h(b) = n/2 and t(b) = t(d) for each d ∈ domo.
Proof. For b, d ∈ domo and c ∈ Cb, e ∈ Cd recall that F (b, c, d, e) is a square in Fq .
Define the function Gb,d,e : Fq → Fq , as
c 7→ (b− d)2 − 4(s(b)cp
h(b)
+ t(b)− s(d)ep
h(d)
− t(d))(c− e) =
−4s(b)cp
h(b)+1 + 4es(b)cp
h(b)
− 4(t(b)− s(d)ep
h(d)
− t(d))c+
(b− d)2 + 4e(t(b)− s(d)ep
h(d)
− t(d)).
14 Ars Math. Contemp. 24 (2024) #P1.04
First assume 0 < h(b) < n/2 for some b ∈ domo.
Denote by ψ the quadratic character of Fq and apply Result 2.3 to the function Gb,d,e.
Then we have
q −√q − ph(b) < −(q − |Cb|) + (|Cb| − ph(b) − 1) ≤
∑
c∈Fq
ψ(Gb,d,e(c)),
as Gb,d,e is a polynomial of degree ph(b) + 1 in c, so the number of its roots is at most
ph(b) + 1.
Thus, we cannot have ∣∣∣∣∣∣
∑
c∈Fq
ψ(Gb,d,e(c))
∣∣∣∣∣∣ ≤ ph(b)√q.
It follows that Gb,d,e is the square of a polynomial in c. And hence by Lemma 2.5, one of
the following holds:
(i) s(b) = 0 and t(b) − s(d)eph(d) − t(d) = 0 for each d ∈ domo, e ∈ Cd. If we fix d
as well and let e run through Cd then we obtain s(d) = 0 and t(b) = t(d), for each
d ∈ domo.
(ii) s(b) ̸= 0 and
(4es(b))p
h(b)
(−4s(b)) = −4(t(b)− s(d)ep
h(d)
− t(d))(−4s(b))p
h(b)
, (3.1)
(4es(b))p
h(b)+1(−4s(b)) = ((b−d)2+4e(t(b)−s(d)ep
h(d)
−t(d)))(−4s(b))p
h(b)+1.
(3.2)
Then (3.1) yields s(d)ep
h(d) − s(b)eph(b) = t(b)− t(d), for each d ∈ domo, e ∈ Cd.
Fix d as well and let e run through Cd. PutK for the dimension over Fp of the kernel
of the Fp-linear Fq → Fq function e 7→ s(d)ep
h(d) − s(b)eph(b) . Then
|Cd|
pK
≤
∣∣∣{s(d)eph(d) − s(b)eph(b) : e ∈ Cd}∣∣∣ = |{t(b)− t(d)}| = 1,
thus K = n (q = pn) and hence s(d) = s(b), h(d) = h(b) and t(d) = t(b) for each
d ∈ domo.
Then (3.2) reads as 0 = (b − d)2 for each d ∈ domo, a contradiction since
|domo| > 1.
We proved that 0 < h(b) < n/2 implies s(d) = 0 and t(b) = t(d) for each d ∈ domo.
Next assume n/2 < h(b) < n for some b ∈ domo.
Apply Result 2.3 to the function c 7→ (Gb,d,e(c))p
n−h(b)
(mod cq − c) and continue as
above. It turns out that s(d) = 0 and t(b) = t(d) for each d ∈ domo also in this case.
A. Aguglia et al.: Intersecting families of graphs of functions over a finite field 15
Now assume h(b) = n/2 for some b ∈ domo.
If s(b) = 0, then
−4(t(b)− s(d)ep
h(d)
− t(d))c+ (b− d)2 + 4e(t(b)− s(d)ep
h(d)
− t(d))
is a square for each d ∈ domo and c ∈ Cb, e ∈ Cd. If we consider d and e fixed as well,
then it follows that as a function of c it has to be a constant, so t(b)− s(d)eph(d) − t(d) = 0
for each e ∈ Cd and hence s(d) = 0 and t(b) = t(d) for each d ∈ domo.
If s(b) ̸= 0, then Lemma 2.6 applied to Gb,d,e gives (3.1) and hence, as before,
s(d) = s(b), t(d) = t(b) and h(d) = h(b) for each d ∈ domo.
Finally, consider the case when h(b) = 0 for some b ∈ domo.
Then again from Result 2.3, one obtains Gb,d,e(c) = (b − d)2 − 4(s(b)c + t(b) −
s(d)ep
h(d) − t(d))(c− e) = (α+ βc)2, for some α, β ∈ Fq .
If s(b) = 0, that is, when Gb,d,e is a constant, then t(b) − s(d)ep
h(d) − t(d) = 0 for
each d ∈ domo, e ∈ Cd, so s(d) = 0 and t(b) = t(d) for each d ∈ domo.
If s(b) ̸= 0, that is, when Gb,d,e is of degree two, then the discriminant of Gb,d,e has to
be zero, i.e.
s(b)(b− d)2 + (s(b)e− s(d)ep
h(d)
+ t(b)− t(d))2 = 0.
For d ∈ domo let εd be an element of Fq for which ε2d = −s(b)(b − d)2. Consider
d ∈ domo fixed as well, then for e ∈ Cd:
s(b)e− s(d)ep
h(d)
∈ {εd + t(d)− t(b),−εd + t(d)− t(b)}.
Put K for the dimension over Fp of the kernel of the Fp-linear Fq → Fq function e 7→
s(b)e− s(d)eph(b) . Then
|Cd|
pK
≤
∣∣∣{s(b)e− s(d)eph(d) : e ∈ Cd}∣∣∣ ≤ 2,
which is a contradiction, unless K = n. It follows that s(b)e − s(d)eph(b) = 0 for each
e ∈ Fq , so h(d) = 0, s(d) = s(b) and t(d)− t(b) is one of εd and −εd.
Lemma 3.10. If q ≥ 53 is odd, U is a set of intersecting polynomials of degree at most 2
such that |domo| > (q+1)/2, then there exist α, β ∈ Fq such that g(α) = β for all g ∈ U
with g = f(b, c) + bx+ cx2 where b ∈ domo.
Proof. According to Lemma 3.9, we consider the following two cases.
Suppose that there exists some b′ ∈ domo such that s(b′) = 0.
Then s(d) = 0 and t(d) = t(b) for each d ∈ domo. Put T for t(b). It follows
that for b ∈ domo and c ∈ Cb the polynomials f(b, c) + bx + cx2 ∈ U have the shape
T + bx+ cx2 ∈ U and hence (0, T ) is a common point of their graphs.
Suppose that s(b) ̸= 0 for each b ∈ domo.
Then s(b) = s(d) and h(d) = h(b) for each d ∈ domo. We will denote these values by
S and h, respectively. Note that h = 0, or h = n/2.
16 Ars Math. Contemp. 24 (2024) #P1.04
When h = 0.
Then (t(b)− t(d))2 = −S(b− d)2 for each b, d ∈ domo, and so{
t(b)− t(d)
b− d
: b, d ∈ domo, b ̸= d
}
⊆ {s,−s},
where s2 = −S.
It follows that the point set {(b, t(b)) : b ∈ domo} determines at most two directions.
But there is no point set determining exactly two directions, thus t determines a unique
direction, i.e.
t(d) = γd+ T, for d ∈ domo,
where γ is a constant satisfying γ2 = s2 = −S. Then for b ∈ domo and c ∈ Cb the
polynomials f(b, c) + bx+ cx2 ∈ U have the shape Sc+ γb+ T + bx+ cx2, so (−γ, T )
is the common point of their graphs.
When h = n/2.
Then also t(b) = t(d) for each d ∈ domo. Then
(b− d)2 − 4S(c− e)
√
q+1
has to be a square for each b, d ∈ domo and c ∈ Cb, e ∈ Cd.
Fix b, c, d. Note that for k ∈ F√q \ {0} there are
√
q + 1 elements x in Fq such that
x
√
q+1 = k. Since e runs through more than q −√q/2 + 1/2 values, we have
F√q \ {0} ⊆ {(c− e)
√
q+1 : e ∈ Cd}
so
(b− d)2 − Sk = b2 − 2db+ d2 − Sk (3.3)
has to be a square in Fq for each b, d ∈ domo, k ∈ F√q \ {0}. As a polynomial in b, the
discriminant of (3.3) is
4d2 − 4(d2 − Sk) = Sk.
Recall S = s(b) ̸= 0, so this discriminant cannot be zero. By Result 2.4, for fixed d ∈
domo and k ∈ F√q \ {0} and for the character ψ of order 2,∑
b∈Fq
ψ(b2 − 2db+ d2 − Sk) = −ψ(1) = −1.
On the other hand, a lower bound for this sum is
|domo| − 2− |Fq \ domo|,
which is at least 2|domo| − q − 2, a contradiction when |domo| > (q + 1)/2.
Next, we prove Theorem 1.3 when q is odd.
Theorem 3.11. If q ≥ 53 is odd and U is a set of q2 − ε, ε < q
√
q
4 −
3q
8 −
√
q
8 , intersecting
polynomials of degree at most 2 over Fq , then the graphs of the functions in U share a
common point.
A. Aguglia et al.: Intersecting families of graphs of functions over a finite field 17
Proof. Let domo denote the set of values as before and let us call a polynomial f(b, c) +
bx+ cx2 ∈ U good if b ∈ domo. According to the previous lemma, if |domo| > (q+1)/2
the graphs of the good polynomials share a common point. If there are more than q
2+q
2
good polynomials then Lemma 3.5 finishes the proof.
Clearly,
|U | ≤ |domo|q + (q − |domo|)(q −
√
q/2 + 1/2) = q2 − (q − |domo|)(
√
q/2− 1/2).
Assume to the contrary that the number of good polynomials is at most q
2+q
2 , then |domo| <
q2+q
2(q−√q/2+1/2) <
q
2 +
√
q
4 +
1
2 . Hence:
|U | ≤ q2 −
(
q
√
q
4
− 3q
8
−
√
q
8
+
1
4
)
,
which is a contradiction.
3.2 Intersecting families of polynomials of degree at most 2, over Fq , q even
Lemma 3.12. Let U be a set of intersecting polynomials of degree at most 2 and for t ∈ Fq ,
q > 2 even, define
Bt := {b ∈ Fq : f(b, b+ t) + bx+ (b+ t)x2 ∈ U}
and
dome := {t ∈ Fq : |Bt| ≥ q −
√
q/2}.
There exist functions A,B : dome → Fq such that for every b ∈ Bt
f(b, b+ t) = A(t)b+B(t),
and A(t) ∈ kerTrq/2.
Proof. Consider F (x) = f(b, c) + bx + cx2 and G(x) = f(d, e) + dx + ex2. Then the
graphs of F and G share a common point if and only if F − G has a root in Fq , that is,
b = d or
H(b, c, d, e) := Trq/2
(
(c+ e)(f(b, c) + f(d, e))
(b+ d)2
)
= 0.
Then for each t ∈ Fq , b, d ∈ Bt, b ̸= d,
H(b, b+ t, d, d+ t) = Trq/2
(
(b+ d)(f(b, b+ t) + f(d, d+ t))
(b+ d)2
)
= 0.
Simplifying by b+ d yields
Trq/2
(
f(b, b+ t) + f(d, d+ t)
b+ d
)
= 0.
Define Rt : Bt → Fq as Rt(x) = f(x, x+ t). For each x, y ∈ Bt, x ̸= y, it holds that
Trq/2
(
Rt(x) +Rt(y)
x+ y
)
= 0.
18 Ars Math. Contemp. 24 (2024) #P1.04
In particular, the set of directions determined by the graph of Rt is contained in kerTrq/2,
and hence it has size at most q/2.
From now on assume t ∈ dome and hence |Bt| ≥ q −
√
q/2. By results of Szőnyi
[10], there exists a unique extension R̃t : Fq → Fq of Rt such that the set of direc-
tions determined by R̃t is the same as the set of directions determined by the point set
{(x,Rt(x)) : x ∈ Bt} ⊆ AG(2, q). So the set of directions determined by R̃t is
contained in kerTrq/2 and hence by Theorem 2.2 there exist A(t), B(t) ∈ Fq such that
R̃t(x) = A(t)x+B(t) with Trq/2(A(t)) = 0. It follows that for b ∈ Bt we have
Rt(b) = f(b, b+ t) = A(t)b+B(t).
Lemma 3.13. Let U be a set of intersecting polynomials of degree at most 2 and define
Bt, dome and the functions A and B as in the previous lemma. Then there exist α, β ∈ Fq ,
q ≥ 8, such that A(t) = αq/2 + α and B(t) = αt+ β for each t ∈ dome.
Proof. If |dome| = 1, then the assertion is trivial, so assume |dome| ≥ 2 and take any
s, t ∈ dome. Fix some b ∈ Bs. Then for each d ∈ Bt \ {b},
H(b, b+ s, d, d+ t) = Trq/2
(
(b+ s+ d+ t)(f(b, b+ s) + f(d, d+ t))
(b+ d)2
)
= 0,
that is,
Trq/2
(
(b+ s+ d+ t)(f(b, b+ s) +A(t)d+B(t))
(b+ d)2
)
= 0,
i.e.,
Trq/2
(
A(t) +
f(b, b+ s) +B(t) +A(t)b+A(t)(s+ t)
b+ d
+
(s+ t)
f(b, b+ s) +B(t) +A(t)b
(b+ d)2
)
= 0.
Applying Trq/2(A(t)) = 0 and Trq/2(z) = Trq/2(z2) for each z ∈ Fq , we obtain for each
d ∈ Bt \ {b}, d ̸= b
Trq/2
(
f2(b, b+ s) +B2(t) +A2(t)b2 +A2(t)(s+ t)2+
(b+ d)2
(s+ t)(f(b, b+ s) +B(t) +A(t)b)
(b+ d)2
)
= 0.
The numerator does not depend on d, while the denominator ranges over a subset of
F∗q of size |Bt \ {b}| > degTrq/2 = q/2 and hence this is possible only if
f2(b, b+ s)+B2(t)+A2(t)b2+A2(t)(s+ t)2+(s+ t)(f(b, b+ s)+B(t)+A(t)b) = 0.
Since f(b, b+ s) = A(s)b+B(s), this is equivalent to
(A(s)b+B(s))2 +B2(t) +A2(t)b2 +A2(t)(s+ t)2+
(s+ t)(A(s)b+B(s) +B(t) +A(t)b) = 0,
A. Aguglia et al.: Intersecting families of graphs of functions over a finite field 19
that is,
b2(A2(s) +A2(t)) + b(A(s) +A(t))(s+ t)+
(B(s) +B(t))(B(s) +B(t) + s+ t) +A2(t)(t+ s)2 = 0.
Since this holds for every b ∈ Bs, and |Bs| > 2, it follows that as a polynomial of b, this is
the zero polynomial, so A(s) = A(t) and
(B(s) +B(t))(B(s) +B(t) + s+ t) +A2(t)(t+ s)2 = 0. (3.4)
Since Trq/2(A(t)) = 0, and A(t) is a constant function, this proves the existence of α′ ∈
Fq such thatA(x) = α′q/2+α′ for each x ∈ dome. If |dome| = 2, then clearlyB is linear,
so assume |dome| ≥ 3 and take some t′ ∈ dome \ {s, t}. The same arguments show
(B(s) +B(t′))(B(s) +B(t′) + s+ t′) +A2(t′)(t′ + s)2 = 0. (3.5)
Summing up (3.4) and (3.5) we obtain
B(s)(t+t′)+s(B(t)+B(t′))+B2(t)+B2(t′)+B(t)t+B(t′)t′+(α′2+α′)(t+t′)2 = 0,
so for x ∈ dome \ {t, t′} it holds that
B(x) = x
B(t) +B(t′)
t+ t′
+
B2(t) +B2(t′) +B(t)t+B(t′)t′
t+ t′
+ (α′2 + α′)(t+ t′),
and from (3.4) (with s = t′) one obtains the same for x ∈ {t, t′}, so B is linear. Put
B(x) = γx+ β, then from (3.4)
γ(s+ t)(γ(s+ t) + (s+ t)) = (α′2 + α′)(s+ t)2,
so γ2 + γ = α′2 + α′ which proves γ = α′ or γ = α′ + 1. Now, if γ = α′ then we set
α := α′ whereas if γ = α′+1 we set α := α′+1. Since α′q/2+α′ = (α′+1)q/2+α′+1,
our lemma follows.
Corollary 3.14. If g(x) = f(b, c) + bx+ cx2 ∈ U and b+ c ∈ dome, then g(αq/2) = β.
Proof. Put t = b+ c. Then t ∈ dome and hence
f(b, c) = f(b, b+ t) = A(t)b+B(t) = αq/2b+ αb+ αt+ β,
hence
g(αq/2) = αq/2b+ αb+ αt+ β + bαq/2 + (b+ t)α = β.
Finally, we prove Theorem 1.3 for q even.
Theorem 3.15. If q ≥ 8 is even and U is a set of q2 − ε, ε < q
√
q
4 −
q
8 −
√
q
8 , intersecting
polynomials of degree at most 2 over Fq , then the graphs of the functions in U share a
common point.
20 Ars Math. Contemp. 24 (2024) #P1.04
Proof. Let dome denote the set of values as before and let us call a polynomial f(b, c) +
bx + cx2 ∈ U good if b + c ∈ dome. According to the previous corollary, the graphs of
the good polynomials share a common point. If there are more than q
2
2 good polynomials
then Lemma 3.4 finishes the proof.
Clearly,
|U | ≤ |dome|q + (q − |dome|)(q −
√
q/2) = q2 − (q − |dome|)
√
q/2.
Assume to the contrary that the number of good polynomials is at most q
2
2 , then |dome| ≤
q2
2(q−√q/2) <
q
2 +
√
q
4 +
1
4 . Hence:
|U | < q2 −
(
q
√
q
4
− q
8
−
√
q
8
)
,
which is a contradiction.
3.3 Intersecting families of polynomials of degree at most k > 2
Theorem 1.4. If U is a set of more than qk − qk−1 intersecting polynomials over Fq ,
q ≥ 53 when q is odd and q ≥ 8 when q is even, and of degree at most k, k > 1, then
there exist α, β ∈ Fq such that g(α) = β for all g ∈ U . Furthermore, U can be uniquely
extended to a family of qk intersecting polynomials of degree at most k over Fq .
Proof. Let U be a set of more than qk − qk−1 intersecting polynomials over Fq and of
degree at most k, k > 1. First we show that there exist α, β ∈ Fq such that g(α) = β for
all g ∈ U . We prove this by induction.
For k = 2, this is true by Theorem 1.3. Now assume that it is true for k − 1 and we
want to prove it for k. By the pigeon hole principle there must be a value c, such that there
are more than qk−1−qk−2 polynomials hi in U whose coefficient in xk is c. Now consider
the family of polynomials in the form of {hi − cxk}. Clearly, this is an intersecting family
of polynomials of degree at most k − 1. So by the induction hypothesis, there are values α
and β so that for every i, (hi − cxk)(α) = β and hence of course hi(α) = β + cαk and so
Lemma 3.6 finishes the proof of the first part.
Next, we will prove that U can be uniquely extended to a family of qk intersecting
polynomials of degree at most k over Fq . Hence, let F and F ′ be two intersecting families
of size qk, both of them containing U . Then, there exist α, α′, β, β′ ∈ Fq such that g(α) =
β for all g ∈ F and g(α′) = β′ for all g ∈ F ′. The polynomials in U are in F ∩ F ′, a
contradiction unless (α, β) = (α′, β′), since there are at most qk−1 < |U | polynomials of
degree at most k, whose graph contains two distinct, fixed points. Theorem 1.4 follows.
4 Large intersecting families whose graphs do not share a common
point
The following construction was drawn to our attention in a talk by Sam Adriaensen. Note
that it shows the sharpness of the lower bound on |U | in Lemma 3.4.
Example 4.1 (Hilton-Milner type). Pick a point P := (α, β) and a line e := {(x, vx+w) :
x ∈ Fq} in AG(2, q), so that β ̸= vα + w. Let U ′ be the set of those polynomials over
Fq , which are of the form h(x) = cx2 + bx + a and for which h(α) = β and there exist
A. Aguglia et al.: Intersecting families of graphs of functions over a finite field 21
values α′ and β′ so that h(α′) = β′ and β′ = vα′ + w. The set U = U ′ ∪ {e} is a set
of intersecting polynomials of degree at most 2 over Fq . The size of U is q
2+q
2 and clearly
there exist no values s, t ∈ Fq so that for every polynomial f ∈ U , f(s) = t.
Proof. Clearly, we may assume that P = (0, 1) and e = {(x, 0) : x ∈ Fq}. Then a = 1
for the polynomials in U ′. Pick a point R := (u, 0) from e. The number of polynomials
g in U ′, so that g(u) = 0 is 0 if u = 0, q otherwise. Hence if we count the polynomials
of U ′ corresponding to R when R runs on the points of e, we see q(q − 1) polynomials.
But most of the polynomials in U ′ will correspond to two different points R and R′ of e.
Actually, only the polynomials which are of the form bx + 1 (b ∈ F∗q) and polynomials
of the form c−2(x + c)2 (c ∈ F∗q) in U ′ will correspond to exactly one point in e. Hence
|U ′| = q(q−1)−2(q−1)2 + 2(q − 1) =
q2+q
2 − 1 and so |U | =
q2+q
2 .
Example 4.2. Let q be odd. There is a family M of intersecting polynomials of degree at
most 2 such that |M| = q
2−q+1
2 and there exists f ∈ M with the property that |Uf ∩Ug| =
1 for each g ∈ M, g ̸= f .
Proof. Choose a polynomial f(x) = Ax2 + Bx + C and let □q be the set of non-zero
squares in Fq . Let
P =
{
aix
2 + bix+ C −
(B − bi)2
4(A− ai)
: A− ai ∈ □q, bi ∈ Fq
}
.
Note that |P| = q(q − 1)/2.
If (ai, bi) and (aj , bj) correspond to two elements of P then the graphs of the corre-
sponding polynomials meet each other if and only if
(bi − bj)2 − 4(ai − aj)
(
(B − bj)2
4(A− aj)
− (B − bi)
2
4(A− ai)
)
=
(aiB − ajB −Abi + ajbi +Abj − aibj)2
(A− ai)(A− aj)
is a (possibly zero) square in Fq . This certainly holds since both (A − ai) and (A − aj)
are squares. Hence P is an intersecting family. Finally, we prove that M = P ∪ {f} is
also an intersecting family. We will do this by proving that for each g ∈ P , Ug meets Uf
in a unique point. So assume g(x) = ax2 + bx + C − (B−b)
2
4(A−a) . It is easy to see that the
discriminant of f − g is zero and hence the result follows.
ORCID iDs
Angela Aguglia https://orcid.org/0000-0001-6854-8679
Bence Csajbók https://orcid.org/0000-0003-2801-452X
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