Sums of Cubes in Quaternion Rings

Madison Gamble, Spencer Hamblen, Blake Schildhauer, and Chung Truong

Department of Mathematics and Computer Science McDaniel College Westminster, MD 21157 USA

                mlg011@connections.mcdaniel.edu
                     shamblen@mcdaniel.edu
                  blakeschildhauer@gmail.com
                hanhchung.truong@stella.edu.vn

Abstract We investigate a version of Waring’s problem over quaternion rings, focusing on cubes in quaternion rings with integer coefficients. We determine the global upper and lower bounds for the number of cubes necessary to represent all such quaternions.

1 Introduction and definitions

Theorem 1 (Waring’s problem/Hilbert-Waring theorem). For every integer k ≥ 2 there exists a positive integer g(k) such that every positive integer is the sum of at most g(k) k-th powers of integers.

The idea behind Waring’s problem—examining sums of powers—can be easily extended to any ring. (For example, number fields [7] and polynomial rings over finite fields [5].) For an excellent and thorough exposition of the research on Waring’s problem and its generalizations, see Vaughan and Wooley [8]. We will specifically look at sums of cubes in quaternion rings, extending the previous work on sum of squares begun in Cooke, Hamblen, and Whitfield [4]. Definition 2. Let LQ_{a,b} denote the quaternion ring

{α_{0} + α_{1}i + α_{2}j + α_{3}k | α_{n}, a, b ∈ ℤ, i^{2} = −a, j^{2} = −b, ij = −ji = k}.

Let LQ_{a,b}^{n} denote the additive group generated by all nth powers in LQ_{a,b}.

Note here that k^{2} = −ab, and that if a = b = 1, we have the Lipschitz quaternions. We then have the following analogue of Waring’s problem.

Conjecture 3. For every integer k ≥ 2 and all positive integers a, b there exists a positive integer g_{a,b}(k) such that every element of LQ_{a,b}^{k} can be written as the sum of at most g_{a,b}(k) k-th powers of elements of LQ_{a,b}.

In contrast with the case when k = 2, it is much harder when an element of a ring can be represented as a sum of a small number of cubes. For example, it was only recently determined [1] that 33 is the sum of 3 integer cubes. Our goal in this paper, therefore, is to determine global upper and lower bounds for g_{a,b}(3), the number of cubes necessary to represent all elements of LQ_{a,b}^{3}. We have the following main result.

Theorem 4. Let a, b be positive integers. Then

• if 3 ∤ a or 3 ∤ b, then 3 ≤ g_{a,b}(3) ≤ 6, and

• if 3 | a and 3 | b, then 4 ≤ g_{a,b}(3) ≤ 5.

The upper bounds of Theorem 4 are given in Section 2, following an algorithmic approach based on cubic algebraic identities. The lower bounds are given in Section 3.

It seems quite possible that the lower bounds in Theorem 4 are the actual values for g_{a,b}(3). A number of individual quaternions were tested in SAGE, and all were found to be expressible as the minimum number of cubes. Additionally, the identities of Eqns. (4) and (5), while very useful for our upper bound proof, are by no mean optimal. A search for similar identities involving quaternions was unsuccessful, due to the complications introduced by non-commutativity.

Lastly, it should be noted that Propositions 6 and 12 were both initially proven by checking individual residue classes in SAGE. While we were able to cover all possible cases, more theoretical versions of the proofs are provided here.

2 LQ_{a,b}^{3} and upper bounds

Recall that LQ_{a,b}^{3} is the additive subgroup generated by all cubes in LQ_{a,b}. Our first goal is to determine the shape of elements in LQ_{a,b}^{3}; we therefore first give the general forms of cubes in LQ_{a,b}. If α = α_{0} + α_{1}i + α_{2}j + α_{3}k, we have

α^{3} = α_{0}^{3} − 3aα_{0}α_{1}^{2} − 3bα_{0}α_{2}^{2} − 3abα_{0}α_{3}^{2}  (1)
+ (3α_{0}^{2}α_{1} − aα_{1}^{3} − bα_{1}α_{2}^{2} − abα_{1}α_{3}^{2})i
+ (3α_{0}^{2}α_{2} − aα_{1}^{2}α_{2} − bα_{2}^{3} − abα_{3}α_{3}^{2})j
+ (3α_{0}^{2}α_{3} − aα_{1}^{2}α_{3} − bα_{1}α_{2}^{2} − abα_{3}^{3})k.

We can simplify this equation by noting common factors in each of the coefficients on the right side of Eqn. (1). For α = α_{0} + α_{1}i + α_{2}j + α_{3}k, let

P_{α} = aα_{1}^{2} + bα_{2}^{2} + abα_{3}^{2}.  (2)

We then have

α^{3} = (α_{0}^{2} − 3P_{α})α_{0} + (3α_{0}^{2} − P_{α}) (α_{1}i + α_{2}j + α_{3}k) .  (3)

Additionally, we will make frequent use of the following two identities:

6z = (z + 1)^{3} + (z − 1)^{3} + (−z)^{3} + (−z)^{3}  (4)
6z + 3 = (−z − 5)^{3} + (z + 1)^{3} + (−2z − 6)^{3} + (2z + 7)^{3}.  (5)

These two identities, and these proofs, are inspired by Cohn’s results [2, 3] on sums of cubes in quadratic fields: g_{ℤ[i]}(3) = 4 and g_{ℤ[\sqrt{d}]}(3) ≤ 5.

We start by treating the case when 3 ∤ a or 3 ∤ b.

Proposition 5. If 3 ∤ a or 3 ∤ b, then LQ_{a,b}^{3} = LQ_{a,b}.

Note that in the Lipschitz quaternions (a = b = 1), this follows from Pollack’s work on quaternions [6, Theorem 1.1].

Proposition 6. If 3 ∤ a or 3 ∤ b, then every element of LQ_{a,b}^{3} can be written as the sum of at most 6 cubes of elements in LQ_{a,b}.

We will prove that every element of LQ_{a,b} can be written as the sum of at most 6 cubes, which yields both propositions.

Proof. First, note that by Eqns. (4) and (5), we immediately have that every element in LQ_{a,b} that is a multiple of 6, or 3 more than a multiple of 6, can be written as the sum of 4 cubes. It then suffices to restrict our attention to the resulting residue classes, and we need only consider the residue of a, b modulo 6. We will break the problem into two cases, and in each case will need two supporting Lemmas.

Our two cases are as follows:

• Case 1: Suppose 3 ∤ ab, and at least one of a or b is congruent to 2 (mod 3), and

• Case 2: All other cases: either a ≡ b ≡ 1 (mod 3), or exactly one of a and b is divisible by 3.

For the following Lemmas, we let Re(x) be the real part of x and Im(x) be the imaginary or pure part of x. That is, if x = x_{0} + x_{1}i + x_{2}j + x_{3}k, then Re(x) = x_{0} and Im(x) = x_{1}i + x_{2}j + x_{3}k. Additionally, we write Im(x) ≡ Im(y) (mod 6) if 6 divides each of the coefficients of Im(x − y). Lastly, for n ∈ ℤ, we write \overline{n} for the least non-negative residue of n (mod 6); that is \overline{n} ≡ n (mod 6) and \overline{n} ∈ \{0, 1, …, 5\}.

Lemma 7. Suppose we are in Case 1: 3 ∤ ab, and at least one of a or b is congruent to 2 (mod 3), and let

S = {α ∈ LQ_{a,b} | 2 ∤ α_{0} and 3 ∤ α_{1}α_{2}α_{3}}.

Then, for all α ∈ S, there exists x ∈ LQ_{a,b} such that Re(x^{3}) ≡ Re(α) (mod 3) and Im(x^{3}) ≡ Im(α) (mod 6).

Note that as an immediately corollary of Lemma 7 and Eqns. (4) and (5), every element of S can be written as the sum of at most 5 cubes.

Proof. Take α = α_{0} + α_{1}i + α_{2}j + α_{3}k ∈ S. Then let x = x_{0} + x_{1}i + x_{2}j + x_{3}k, where x_{ℓ} = \overline{α_{ℓ}} for ℓ ∈ \{1, 2, 3\} and x_{0} = \overline{α_{0}} − 3δ_{α}, where

δ_{α} = \begin{cases}1, & if P_{α} is odd; \\ 0, & otherwise. \end{cases}

By Eqn. (3), it suffices to show that x_{0}^{3} −3x_{0}P_{x} ≡ α_{0} (mod 3), and x_{ℓ}(3x_{0}^{2} −P_{x}) ≡ α_{ℓ} (mod 6) for ℓ ∈ {1, 2, 3}.

We then have

x_{0}^{3} − 3x_{0}P_{x} = (\overline{α_{0}} − 3δ_{α})^{3} − 3(\overline{α_{0}} − 3δ_{α})P_{x} ≡ α_{0}^{3} ≡ α_{0} (mod 3),  (6)

so Re(x^{3}) ≡ Re(α) (mod 3). Then, note that in this case we have α ∈ S, α_{1}^{2} ≡ α_{2}^{2} ≡ α_{3}^{2} ≡ 1 (mod 3), so

P_{α} ≡ a · 1 + b · 1 + ab · 1 (mod 3)
≡ (a + 1)(b + 1) − 1 (mod 3).

Since at least one of a or b is congruent to 2 (mod 3), we must have that P_{α} ≡ 2 (mod 3). Therefore if δ_{α} = 1, then P_{α} ≡ 5 (mod 6), and if δ_{α} = 0, then P_{α} ≡ 2 (mod 6); in either case, 3δ_{α} − P_{α} ≡ −2 (mod 6).

Then note that since P_{x} ≡ P_{α} (mod 6) (since by definition, Im(x) ≡ Im(α) (mod 6)) and α_{0} is odd, we have

3x_{0}^{2} − P_{x} = 3(\overline{α_{0}} − 3δ_{α})^{2} − P_{x} ≡ 3α_{0}^{2} + 3δ_{α} − P_{α} (mod 6)
≡ 3 − 2 = 1 (mod 6).

Therefore x_{ℓ}(3x_{0}^{2} − P_{x}) ≡ α_{ℓ} (mod 6) for ℓ ∈ {1, 2, 3}, so Im(x^{3}) ≡ Im(α) (mod 6), which completes the proof.

Lemma 8. Suppose 3 ∤ ab, and at least one of a or b is congruent to 2 (mod 3), and let S be defined as in Lemma 7. Then, for all α ∈ LQ_{a,b}, there exists α′, α′′ ∈ S such that Re(α′ + α′′) ≡ Re(α) (mod 3) and Im(α′ + α′′) ≡ Im(α) (mod 6).

Proof. Notice that elements of S can have real coefficient equivalent to 1, 3, or 5 (mod 6), and can have imaginary coefficients equivalent to 1, 2, 4, or 5 (mod 6). The first conclusion then follows since the real coefficients cover all residue classes mod 3, and the second follows from the fact that in ℤ_{6}, we have {1, 2, 4, 5} + {1, 2, 4, 5} = ℤ_{6}.

As a consequence of Lemmas 7 and 8, for all α ∈ LQ_{a,b}, there exists x_{1}, x_{2} ∈ LQ_{a,b} such that α − x_{1}^{3} + x_{2}^{3} is either a multiple of 6, or 3 more than a multiple of 6; Eqns. (4) and (5) then imply that under the hypotheses of Case 1, every element of LQ_{a,b} can be written as the sum of at most 6 cubes. We have therefore proven Propositions 5 and 6 in the case when 3 ∤ ab, and at least one of a or b is congruent to 2 (mod 3).

We then move to Case 2, where we suppose that we are in one of the following cases:

• Case 2a: a ≡ b ≡ 1 (mod 3).

• Case 2b: Exactly one of a and b is divisible by 3, and the other is 2 (mod 3). Without loss of generality, in this case we assume a ≡ 2 (mod 3) and b ≡ 0 (mod 3).

• Case 2c: Exactly one of a and b is divisible by 3, and the other is 1 (mod 3). Without loss of generality, in this case we assume a ≡ 1 (mod 3) and b ≡ 0 (mod 3).

Lemma 9. Given a and b satisfying one of the cases above, let

T_{2} = {α ∈ LQ_{a,b} | 2 ∤ α_{0} and 3 ∤ α_{1}α_{3} and 3 | α_{2}},
T_{3} = {α ∈ LQ_{a,b} | 2 ∤ α_{0} and 3 ∤ α_{1}α_{2} and 3 | α_{3}},

and T = T_{2} ∪ T_{3}. Then, for all α ∈ T, there exists x ∈ LQ_{a,b} such that Re(x^{3}) ≡ Re(α) (mod 3) and Im(x^{3}) ≡ Im(α) (mod 6).

Proof. The proofs in each subcase are very similar to that of Lemma 7; we will only highlight where the definitions and calculations differ.

Take α = α_{0} + α_{1}i + α_{2}j + α_{3}k ∈ S, let x_{0} = \overline{α_{0}} − 3δ_{α} as defined in Lemma 7, and let

x_{ℓ} = \begin{cases}\overline{α_{ℓ}}, & in Cases 2a and 2b; \\ 6 − \overline{α_{ℓ}}, & in Case 2c. \end{cases}

Immediately by Eqn. (6) in Lemma 7, we have that Re(x^{3}) ≡ Re(α) = α_{0} (mod 3).

Then, for α ∈ T_{2}, we have α_{1}^{2} ≡ α_{3}^{2} ≡ 1 (mod 3) and α_{2}^{2} ≡ 0 (mod 3), so from Eqn. (2) we get:

P_{α} ≡ \begin{cases}2 ≡ 1 · 1 + 1 · 0 + 1 · 1 (mod 3), & in Case 2a; \\ 2 ≡ 2 · 1 + 0 · 0 + 0 · 1 (mod 3), & in Case 2b; \\ 1 ≡ 1 · 1 + 0 · 0 + 0 · 1 (mod 3), & in Case 2c. \end{cases}

Note that in all of these cases, b ≡ ab (mod 3), so for α ∈ T_{3}, the values of P_{α} mod 3 are the same as for α ∈ T_{2}.

Therefore, in Cases 2a and 2b, if δ_{α} = 1, then P_{α} ≡ 5 (mod 6), and if δ_{α} = 0, then P_{α} ≡ 2 (mod 6); either way, 3δ_{α} − P_{α} ≡ −2 (mod 6). Since P_{x} ≡ P_{α} (mod 6) and α_{0} is odd, we have

3x_{0}^{2} − P_{x} = 3(\overline{α_{0}} − 3δ_{α})^{2} − P_{x} ≡ 3α_{0}^{2} + 3δ_{α} − P_{α} (mod 6)
≡ 3 − 2 = 1 (mod 6).

Therefore Im(x^{3}) ≡ Im(α) (mod 6), which completes the proof for Cases 2a and 2b.

In Case 2c, if δ_{α} = 1, then P_{α} ≡ 1 (mod 6), and if δ_{α} = 0, then P_{α} ≡ 4 (mod 6); either way, 3δ_{α} − P_{α} ≡ 2 (mod 6). The same calculation as above then yields

3x_{0}^{2} − P_{x} ≡ 3 + 2 ≡ −1 (mod 6).

But, as we have defined x_{ℓ} = 6 − \overline{α_{ℓ}} in this case, we have

x_{ℓ}(3x_{0}^{2} − P_{x}) ≡ (6 − \overline{α_{ℓ}})(−1) ≡ α_{ℓ} (mod 6)

for ℓ ∈ {1, 2, 3}, which implies Im(x^{3}) ≡ Im(α) (mod 6), completing the proof for Case 2c.

Lemma 10. Given a and b satisfying Case 2, let T be defined as in Lemma 9. Then, for all α ∈ LQ_{a,b}, there exists α′, α′′ ∈ T such that Re(α′ + α′′) ≡ Re(α) (mod 3) and Im(α′ + α′′) ≡ Im(α) (mod 6).

Proof. If 3 | α_{2} or 3 | α_{3}, we can satisfy the conclusions by choosing α′ and α′′ both to be in T_{2} or T_{3}, respectively, and following the reasoning in Lemma 8. If 3 ∤ α_{2}α_{3}, then there exists α′ ∈ T_{2} and α′′ ∈ T_{3} satisfying the conclusions.

This completes the proofs of Propositions 5 and 6: as in Case 1, Lemmas 9 and 10 imply that in Case 2, every element of LQ_{a,b} can be written as the sum of at most 6 cubes.

If 3 | a and 3 | b, there is slightly more work to do, as not all elements of the ring can be written as the sum of cubes.

Proposition 11. Let 3 | a and 3 | b, and define

C_{a,b} = {α_{0} + 3α_{1}i + 3α_{2}j + 3α_{3}k | i^{2} = −a, j^{2} = −b, = ij = −ji = k, α_{n} ∈ ℤ}.

Then LQ_{a,b}^{3} = C_{a,b}.

Proof. Note that if 3 | a and 3 | b, then for all α ∈ LQ_{a,b}, we have 3 | P_{α} from Eqn. (2). Then by Eqn. (3), we have that each of the imaginary coefficients (the coefficients of i, j, k) are each divisible by 3, showing that LQ_{a,b}^{3} ⊆ C_{a,b}.

The proposition then follows from Proposition 12, which shows that every element in C_{a,b} can be written as the sum of at most 5 cubes.

Proposition 12. If 3 | a and 3 | b, then every element of C_{a,b} can be written as the sum of at most 5 cubes of elements in LQ_{a,b}.

Proof. In light of Eqns. (4) and (5), it suffices to show that for all elements α ∈ C_{a,b}, there exists x ∈ LQ_{a,b} such that Re(x^{3}) ≡ Re(α) (mod 3) and Im(x^{3}) ≡ Im(α) (mod 6).

Take α = α_{0} +α_{1}i+α_{2}j+α_{3}k ∈ C_{a,b}. Then let x_{ℓ} = \overline{α_{ℓ}} for ℓ ∈ \{1, 2, 3\} and x_{0} = \overline{α_{0}} −3δ_{α}, where

δ_{α} = \begin{cases}1, & if P_{α} ≡ α_{0} (mod 2); \\ 0, & otherwise. \end{cases}

We immediately get Re(x^{3}) ≡ Re(α) (mod 3) by the calculations in Lemma 7.

For α ∈ C_{a,b}, since 3 | a and 3 | b, we have P_{α} ≡ 0 (mod 3). Therefore if δ_{α} = 1, then α_{0} is odd and P_{α} ≡ 3 (mod 6), or α_{0} is even and P_{α} ≡ 0 (mod 6). If δ_{α} = 0, then α_{0} is odd and P_{α} ≡ 0 (mod 6), or α_{0} is even and P_{α} ≡ 3 (mod 6). Specifically, an odd number of α_{0}, δ_{α}, and P_{α} will be odd. We then have

3x_{0}^{2} − P_{x} = 3(\overline{α_{0}} − 3δ_{α})^{2} − P_{x} ≡ 3α_{0}^{2} + 3δ_{α} − P_{α} (mod 6)
≡ 3 (mod 6).

Then, since α ∈ C_{a,b}, we have α_{ℓ} is a multiple of 3 for ℓ ∈ {1, 2, 3}, so 3α_{ℓ} ≡ α_{ℓ} (mod 6). But these are now exactly the mod 6 imaginary coefficients of x^{3}.

Therefore Im(x^{3}) ≡ Im(α) (mod 6), which completes the proof.

3 Lower bounds

We now prove the lower bounds of Theorem 4 via example.

Proposition 13. If 3 ∤ a or 3 ∤ b, then 3 + 3i cannot be written as the sum of 2 cubes in LQ_{a,b}.

Proof. Suppose x, y ∈ LQ_{a,b} are such that

3 + 3i = x^{3} + y^{3},  (7)

and write x = x_{0} + x_{1}i + x_{2}j + x_{3}k, y = y_{0} + y_{1}i + y_{2}j + y_{3}k with x_{n}, y_{n} ∈ ℤ. We then have the following four equations from the coefficients of Eqn. (7):

x_{0}^{3} − 3x_{0}P_{x} + y_{0}^{3} − 3y_{0}P_{y} = 3  (real coefficient);  (8)
3x_{0}^{2}x_{1} − x_{1}P_{x} + 3y_{0}^{2}y_{1} − y_{1}P_{y} = 3  (i coefficient);  (9)
3x_{0}^{2}x_{2} − x_{2}P_{x} + 3y_{0}^{2}y_{2} − y_{2}P_{y} = 0  (j coefficient);  (10)
3x_{0}^{2}x_{3} − x_{3}P_{x} + 3y_{0}^{2}y_{3} − y_{3}P_{y} = 0  (k coefficient).  (11)

From Eqn. (8), we get x_{0}^{3} + y_{0}^{3} ≡ 0 (mod 3); as the only cubes mod 9 are 0, 1, and 8, we immediately get x_{0}^{3} + y_{0}^{3} ≡ 0 (mod 9). Since x_{0}^{3} ≡ x_{0} (mod 3), we also get

x_{0} + y_{0} ≡ 0 (mod 3).  (12)

We can then examine Eqn. (8) (mod 9) and simplify:

x_{0}^{3} − 3x_{0}P_{x} + y_{0}^{3} − 3y_{0}P_{y} ≡ 3 (mod 9)
−3x_{0}P_{x} − 3y_{0}P_{y} ≡ 3 (mod 9)
−x_{0}P_{x} − y_{0}P_{y} ≡ 1 (mod 3)
−x_{0}P_{x} − y_{0}P_{y} ≡ 1 (mod 3)
y_{0}(P_{x} − P_{y}) ≡ 1 (mod 3).  (13)

We first assume (without loss of generality) that P_{x} ≡ 0 (mod 3). Then P_{y} ≢ 0 (mod 3), and Eqns. (9), (10), (11) reduce to

−y_{1}P_{y} ≡ 0 (mod 3)
−y_{2}P_{y} ≡ 0 (mod 3)
−y_{3}P_{y} ≡ 0 (mod 3).

Therefore y_{1} ≡ y_{2} ≡ y_{3} ≡ 0 (mod 3), which implies that P_{y} ≡ 0 (mod 3), a contradiction. Therefore P_{x}, P_{y} ≢ 0 (mod 3).

We additionally have from Eqn. (13) that P_{x} ≢ P_{y} (mod 3), so assume P_{x} ≡ 1 (mod 3) and P_{y} ≡ 2 (mod 3). From Eqns. (9), (10), and (11) we have x_{n} ≡ 2y_{n} (mod 3) for n ∈ {1, 2, 3}, which implies x_{n}^{2} ≡ y_{n}^{2} (mod 3). We then have

1 ≡ P_{y} − P_{x} ≡ (ay_{1}^{2} + by_{2}^{2} + aby_{3}^{2}) − (ax_{1}^{2} + bx_{2}^{2} + abx_{3}^{2}) (mod 3)
≡ a(y_{1}^{2} − x_{1}^{2}) + b(y_{1}^{2} − x_{1}^{2}) + ab(y_{3}^{2} − x_{3}^{2}) (mod 3)
≡ 0 (mod 3).

We therefore have the contradiction in this case, which completes the proof.

Proposition 14. If 3 | a and 3 | b, then 4 cannot be written as the sum of 3 cubes in LQ_{a,b}.

Proof. Suppose x, y, z ∈ LQ_{a,b} are such that 4 = x^{3} +y^{3} +z^{3}. Examining the real coefficients of Eqn. (7), we get the following (similar to Eqn. (8)):

x_{0}^{3} − 3x_{0}P_{x} + y_{0}^{3} − 3y_{0}P_{y} + z_{0}^{3} − 3z_{0}P_{z} = 4.  (14)

Note that since 3 | a and 3 | b, we have P_{x} ≡ P_{y} ≡ P_{z} ≡ 0 (mod 3); therefore Eqn. (14) becomes

x_{0}^{3} + y_{0}^{3} + z_{0}^{3} ≡ 4 (mod 9),

which has no integer solutions.

Propositions 6, 12, 13, and 14 then complete the proof of Theorem 4.

4 Acknowledgments

The authors would like to thank the McDaniel College Student-Faculty Summer Research Fund and Research and Creativity Fund for supporting their research.

References

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[4] A. Cooke, S. Hamblen, and S. Whitfield, Sums of squares in quaternion rings, Involve 10 (2017), 651–664.

[5] D. J. Newman and M. Slater, Waring’s problem for the ring of polynomials, J. Number Theory 11 (1979), 477–487.

[6] P. Pollack, Waring’s problem for integral quaternions, Indag. Math. 29 (2018), 1259– 1269.

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2010 Mathematics Subject Classification: Primary 11P05; Secondary 11R52. Keywords: Waring’s problem, quaternion integer, sum of cubes.

Received October 6 2019; revised versions received February 22 2020; February 23 2020. Published in Journal of Integer Sequences, March 20 2020.

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