On expressing elements as a sum of squares, where one square is restricted to a subfield

Let q be a prime power and fix a ∈ Fq2 \ Fq. In this note it is proved that, provided q > 5, the set Sa = {a− α : α ∈ Fq} contains both a square and a non-square of Fq2 . In particular, every a ∈ Fq2 \ Fq can be written as a sum of a square of Fq2 and a square of Fq. § 1. The problem and the main result Throughout we use Fq to denote the finite field of q elements and Fq to denote the non-zero elements of the field. We also use q and q to denote the squares and non-squares, respectively, of Fq. In this note we are interested in a specific instance of the following type of problem: Let A be a subset of Fq of order O(q). What is the minimum order of a subset S of Fq so that for all a ∈ A, the set {a− s : s ∈ S} contains both a square and non-square element of Fq? Since there are no non-square elements in a finite field of even characteristic, we assume q is an odd prime power throughout. There are variants of the problem. For example, one can insist that S be a subgroup of Fq instead of simply a subset; or that A = Fq ; or that q be a square, and so on. We now discuss two such variations. In either of these examples, the lower bounds on q avoid degenerate cases. For q ≥ 7, fix A = Fq . It is well known that any non-zero element of Fq can be expressed as the difference of two non-zero squares of the field in many ways. Similar results can be obtained for sums of squares. Together, these show that S = q or S = q suffices. 2 J. math. Ablehnungen 30 (2014) Now fix A = Fq2 \ Fq. Then it is relatively easy to show that for q ≥ 5, S = Fq suffices. Let a ∈ A. Then {1, a} forms a basis for Fq2 over Fq. If {a−s : s ∈ Fq} only contained elements of q2 , then {αa−αs : α, s ∈ Fq} = Fq2 \{αa, α : α ∈ Fq} would only contain elements of q2 as well. Consequently, q2 ⊆ {αa : α ∈ Fq}, implying (q − 1)/2 ≤ q− 1, a contradiction. Thus {a− s : s ∈ Fq} must contain a non-square. Now suppose it contains only non-squares, implying Fq2 \ {αa, α : α ∈ Fq} contains only non-squares. Hence q2 ⊆ {αa, α : α ∈ Fq}, giving the inequality q − 1 2 ≤ { 2q − 1 if a ∈ q2 , q if a ∈ q2 . This yields a contradiction if q ≥ 5. Thus S = Fq suffices when q ≥ 5 and A = Fq2 \ Fq. (Note that it is trivial to refine this argument to obtain exact counts for |{a − s : s ∈ Fq} ∩ q2 | and |{a − s : s ∈ Fq} ∩ q2 |.) In this note we give an improvement of this second variation; our result is somewhat related to the first variation. We shall prove that, if A = Fq2 \ Fq and q ≥ 7, then S = q suffices. More specifically, fix a ∈ Fq2 \ Fq and consider the set Sa = {a− α : α ∈ q}. Our main statement is the following. Theorem 1. For q ≥ 7 and any a ∈ Fq2\Fq, Sa intersects both q2 and q2 non-trivially. More specifically, for q ≥ 5, the following statements hold. (i) Sa ∩ q2 = ∅ if and only if q = 5 and a satisfies one of a = 3, a = 2a+ 1, or a = 3a+ 1. (ii) Sa ∩ q2 = ∅ if and only if q = 5 and a satisfies a = 2. As with the earlier examples, there is a degenerate case. When q = 3, |Sa| = 1, so that Sa ⊂ q2 or Sa ⊂ q2 must hold. The proof is mostly elementary; it relies on the regularity of both the sums and differences of squares in a field, but in one instance we also invoke Weil’s bound on the number of Fq-rational points on an absolutely irreducible curve over a finite field. Theorem 1 also resolves a problem related to Dickson semifields that arose in work of the second author, see [4]. Before proceeding to the proof, we note that the problem considered in the second variation example and Theorem 1 can be restated as a problem concerning the (ir)reducibility of quadratics over Fq2 . Theorem 1 is equivalent to showing that for any a ∈ Fq2 \Fq, there exists elements α, β ∈ q for which a(a−α) ∈ q2 and a(a− β) ∈ q2 . This yields the following corollary. Corollary 2. Let q ≥ 7 be an odd prime power. Then for any a ∈ Fq2 \Fq, there exist elements α, β ∈ q for which X − aX + aα is irreducible over Fq2 and X − aX + aβ is reducible over Fq2 . § 2. Difference sets and sum sets We shall need two results concerning sums and differences of squares. For completeness, we recall the following definitions. Definition 3. Let G be a group of order v and let D ⊂ G with |D| = k. If there exists non-negative integers λ and μ such that every non-identity element of D can be written in precisely λ ways as a quotient in D while every non-identity element of G \ D can be written in μ ways as a quotient in D, then D is called a (v, k, λ, μ) partial difference set. If λ = μ, then D is called a (v, k, λ) difference set. Definition 4. Let G be a group of order v and let D ⊂ G with |D| = k. If there exists non-negative integers λ and μ such that every non-identity element of D can be written in precisely λ ways as a product in D while every non-identity element of G \ D can be written in μ ways as a product in D, then D is called a (v, k, λ, μ) partial sum set. If λ = μ, then D is called a (v, k, λ) sum set. R.S. Coulter and P. Kosick, On expressing elements as a sum of squares 3 While difference sets and partial difference sets have been studied for quite some period of time [1], an in-depth systematic treatment of sum sets and partial sum sets was only initiated recently by Gutekunst [3], see also the papers [2, 5, 6, 8, 10]. Though the two types of objects are similarly defined, their behaviour is generally quite distinct. However, in one particular case, there is a very strong connection. The following result is well known. Lemma 5. If q ≡ 1 mod 4, the set q forms a (q, q−1 2 , q−5 4 , q−1 4 ) partial difference set. If q ≡ 3 mod 4, the set q forms a (q, q−1 2 , q−3 4 ) difference set. Gutekunst noted [3, Lemma 1.5] the following. Lemma 6. If q ≡ 1 mod 4, the set q forms a (q, q−1 2 , q−5 4 , q−1 4 ) partial sum set. If q ≡ 3 mod 4, the set q forms a (q, q−1 2 , q−3 4 , q+1 4 ) partial sum set. Let a ∈ Fq . We denote by sna the number of ways in which a can be written as x − y with x ∈ q and y ∈ q. Similarly, we use nsa to denote the number of ways in which a can be written as y − x with x ∈ q and y ∈ q. One can use the previous two lemmas to prove the following. Lemma 7. If q ≡ 1 mod 4, then sna = nsa = q−1 4 . If q ≡ 3 mod 4, then sna = { q−3 4 if a ∈ ∗ q , q+1 4 if a ∈ q; and nsa = { q+1 4 if a ∈ ∗ q , q−3 4 if a ∈ q. These counts will prove crucial in our proof of Theorem 1 § 3. Proof of Theorem 1 We proceed to establishing our main result. Fix a ∈ Fq2 \ Fq}. Define the set T by T = {βa− α : (α, β ∈ Fq) ∧ (βα ∈ q)} = {β(a− α) : β ∈ Fq ∧ α ∈ q}. Note that if a− α, a− β ∈ Fqb for some b ∈ Fq2 \ Fq and α, β ∈ Fq, then α = β is forced. Consequently, every coset of Fq in Fq2 , apart from F ∗ q itself, contains a unique element a − α with α ∈ Fq. This also means T = ⋃