Abstract
Recall that combinatorial 2sdesigns admit a classical lower bound \(b \ge\binom{v}{s}\) on their number of blocks, and that a design meeting this bound is called tight. A longstanding result of Bannai is that there exist only finitely many nontrivial tight 2sdesigns for each fixed s≥5, although no concrete understanding of ‘finitely many’ is given. Here, we use the Smith Bound on approximate polynomial zeros to quantify this asymptotic nonexistence. Then, we outline and employ a computer search over the remaining parameter sets to establish (as expected) that there are in fact no such designs for 5≤s≤9, although the same analysis could in principle be extended to larger s. Additionally, we obtain strong necessary conditions for existence in the difficult case s=4.
Keywords
Tight design Symmetric design Orthogonal polynomials Delsarte theory1 Introduction
Let v≥k≥t be positive integers and λ be a nonnegative integer. A t(v,k,λ) design, or simply a tdesign, is a pair \((V,\mathcal{B})\) where V is a vset and \(\mathcal{B}\) is a collection of ksubsets of V such that any tsubset of V is contained in exactly λ elements of \(\mathcal{B}\). The elements of V are points and the elements of \(\mathcal{B}\) are blocks. Since tdesigns are also idesigns for i≤t, the parameter t is typically called the strength of the design (when it is as large as possible). The number of blocks is usually denoted b and an easy doublecounting argument shows \(b=\lambda\binom{v}{t} /\binom{k}{t}\).
Suppose \((V,\mathcal{B})\) is a t(v,k,λ) design. Generalizing Fisher’s Inequality, RayChaudhuri and Wilson [10] showed that if t is even, say t=2s, and v≥k+s, then \(b \ge\binom{v}{s}\). If equality holds in this bound, we say \((V,\mathcal{B})\) is tight. The trivial tight 2sdesigns are those with v=k+s, where each of the \(\binom{v}{k} = \binom{v}{s}\) ksubsets of V is a block. The case of odd strength is investigated in [5]; however, it is impossible for (2s−1)designs to be tight in the sense of having \(\binom{v}{s1}\) blocks.
Returning to even strength, the full set of parameters v and k for which a tight 2sdesign exists has only been determined for s=2,3. Note that, when s=1, tight 2designs have b=v and are the ‘symmetric’ designs; see [6, 8] for surveys of this rich (yet very challenging) topic. In 1975, Ito [7] published a proof that the only nontrivial tight 4designs are the Witt 4(23,7,1) design and its complementary 4(23,16,52) design, but his proof was found to be incorrect. A few years later, Enomoto, Ito, and Noda [4] proved the weaker result that there are finitely many nontrivial tight 4designs, though still believing Ito’s initial claim to be true. Finally, in 1978, Bremner [2] successfully settled s=2 by reaffirming Ito’s result. Peterson [9] proved in 1976 that there exist no nontrivial tight 6designs.
Bannai [1] proved that there exist only finitely many nontrivial tight 2sdesigns for each s≥5. The case s=4 is quite open, and the ‘finitely many’ for s≥5 is not explicit and potentially grows with s. However, it is probably the case that there are no unknown tight 2sdesigns for s≥2.
Central to these negative results is the concept of the intersection numbers of a design. An integer 0≤μ<k is an intersection number of a t(v,k,λ) design \((V,\mathcal{B})\) if there exist different blocks \(B_{1},B_{2}\in\mathcal{B}\) such that B _{1}∩B _{2}=μ. The following strong condition was discovered first by RayChaudhuri and Wilson [10] and also implicitly by Delsarte [3].
Proposition 1.1
The polynomials Ψ _{ s } are known as the Gegenbauer polynomials.
Since a 2sdesign with v≥k+s induces at least s intersection numbers [10], it follows that the zeros of Ψ _{ s } must additionally be distinct integers for tight designs. Note also that Ψ _{ s } has no dependence on λ; indeed, for tight designs \(\lambda= \binom{v}{s}\binom{k}{2s} /\binom{v}{2s}\) and is therefore uniquely determined by v and k.
Analogously, the Lloyd polynomials L _{ e }(x) are important for the characterization of perfect eerrorcorrecting codes; see [13]. It is interesting that this characterization of perfect codes was completed long ago, while the open problems mentioned before Proposition 1.1 remain for tight designs. Our goal here is to revive the interest in tight designs and take a modest step toward the full characterization of their parameters.
The outline is as follows. In Sect. 2, we review the work of Bannai in [1] on the asymptotic structure of the zeros of Ψ _{ s }. Extending this, we obtain some exact bounds relevant to this analysis. Section 3 summarizes the techniques for exhausting small cases s≥5, and Sect. 4 is devoted to a partial analysis of the case s=4. An appendix of tables following the main text will prove useful to the interested reader.
2 Bannai’s analysis and the Smith bound
2.1 Notation
Assuming a tight design, let x _{ i }, for \(i =  \lfloor\frac {s}{2} \rfloor,\ldots,(0),\ldots, \lfloor\frac {s}{2} \rfloor\), denote the zeros of Ψ _{ s } listed in increasing order. For example, the zeros of Ψ _{4} and Ψ _{5} are denoted x _{−2}<x _{−1}<x _{1}<x _{2} and x _{−2}<x _{−1}<x _{0}<x _{1}<x _{2}, respectively.
Note τ=2 implies v=2k+1. Moreover, if v<2k, we may complement blocks, replacing k with v−k and obtain v>2k. This is discussed further in Sect. 2.2.

for any β _{0}, there are only finitely many tight 2sdesigns with β≤β _{0}; and

there exists β _{0} (depending only on s), such that there are no nontrivial tight 2sdesigns with β>β _{0}.
Here, our main goal is to compute such a β _{0} explicitly for 5≤s≤9 and, by searching across all pairs (v,k) for which β≤β _{0}, show that there are in fact zero nontrivial tight 2sdesigns for these s.
2.2 Symmetry with respect to the parameter τ
In the analytic work which follows, it is helpful to obtain a lower bound on τ. As discussed above, we may complement blocks to assume v≥2k. The following was mentioned but not fully proven in [1].
Lemma 2.1
Let s≥1. There are no tight 2sdesigns with v=2k.
Proof
Now, we are able to justify assuming that τ≥2 for nonexistence of tight designs.
Proposition 2.2
Let s≥1. If there exists a nontrivial tight 2sdesign with τ<2, then there also exists a nontrivial tight 2sdesign with τ≥2.
Proof
Suppose \(\mathcal{D}\) is a nontrivial tight 2s(v,k,λ) design with τ<2. This means k≤v≤2k−1 because v≠2k by Lemma 2.1, and so the complementary 2s(v,v−k,λ′) design of \(\mathcal{D}\) is a nontrivial tight 2sdesign with τ≥2. □
Incidentally, Bannai and Peterson ruled out the case τ=2, observing that it yields symmetric zeros of Ψ _{ s } about their mean \(\overline{\alpha}\). This is a key observation.
2.3 Hermite polynomials
Proposition 2.4
 (i)
If s is odd and s≥5, then \(\xi_{1}^{2} < \sqrt{3}\).
 (ii)
If s is even and s≥8, then \(\xi_{2}^{2}\xi_{1}^{2} < \sqrt{3}\).
 (iii)
If s=6, then \(1.0 < \frac{\xi_{2}^{2}\xi_{1}^{2}}{3} < 1.1\), \(3.5 < \frac{\xi_{3}^{2}\xi_{1}^{2}}{3} < 3.6\), and \(3.34634 < \frac {\xi_{3}^{2}\xi_{1}^{2}}{\xi_{2}^{2}\xi_{1}^{2}} < 3.34635\).
Proof
Items (i) and (ii) are referenced in Bannai’s Proposition 13 and proven on page 126 of [12]. Item (iii) can be verified numerically. See Appendix A. (Note that Bannai’s Proposition 13 (iii) actually contains an error.) □
2.4 The Smith bound
We now state a useful result for explicitly finding β _{0}. Sometimes known as the Smith bound, it is a consequence of the Gershgorin circle theorem.
Theorem 2.5
([11])
We will see from Propositions 2.6 and 2.7 that the z _{ i } are wellapproximated by the ξ _{ i } as β→∞, independently of τ.
Proposition 2.6
Proof
2.5 Bounding G _{ s } in terms of β
In the next proposition, it is helpful to think of the \(G_{s}^{(i)}(\xi_{i})\) as functions of β and τ.
Proposition 2.7
The necessary ingredients for this result were proved in [1], although the bound was not directly stated in this form. Therefore, we omit the proof and instead focus on how to (carefully) obtain B _{ i } and C _{ i } for small s using some basic computer algebra.
Algorithm 2.8
 1.
Using (1.1), substitute (2.1), (2.2), (2.3), and (2.5) into (2.6). To defer floatingpoint precision issues, we first replace ξ _{ i } in (2.5) by a symbolic parameter r.
 2.This results in an expression for \(G_{s}^{(i)}(r)\) as a rational function of β, sayHere, the denominator is$$G_s^{(i)}(r) (\beta,\tau) = \frac{p(r,\beta,\tau)}{q(\beta,\tau)}. $$$$ q(\beta,\tau) = \beta^s \binom{vs}{s} = \beta^s \binom{\tau^4(\tau 1)^{2} \beta^2 +\tau+s1}{s}. $$(2.7)
 3.Observe that q is positive for β>0 and τ≥2, and that a lower bound on q isThis is obtained by replacing each factor in the falling factorial of (2.7) by τ ^{4}(τ−1)^{−2} β ^{2}.$$\tilde{q}(\beta,\tau) = \frac{1}{s!} \beta^{3s} \tau^{4s} (\tau1)^{2s}. $$
 4.
The numerator p(r,β,τ) is, for general r, a polynomial of degree 3s in β. However, for r=ξ _{ i }, Proposition 2.7 shows the two top coefficients, namely of β ^{3s } and β ^{3s−1}, vanish. Again, to maintain symbolic algebra, we artificially replace these coefficients by zero and call this polynomial \(\tilde {p}(r,\beta,\tau)\).
 5.We haveNote that for r=ξ _{ i }, the right hand side is a polynomial in β ^{−1}.$$\beta^2 G_s^{(i)}(r) (\beta,\tau) \le \frac{\beta^2 \tilde {p}(r,\beta ,\tau)}{\tilde{q}(\beta,\tau)}. $$
 6.Consider the coefficient κ _{ j }(r,τ) of β ^{3s−j } in \(\beta^{2} \tilde{p}(r,\beta,\tau)\). With r=ξ _{ i }, compute (or upperbound) the maximaThen, estimating termbyterm,$$M_{j}= \sup_{\tau\ge2} \frac{\kappa_j(\xi_i,\tau)}{\frac {1}{s!}(\tau1)^{2s} \tau^{4s}}. $$for all τ≥2.$$\bigl\beta^2 G_s^{(i)}(\xi_i) (\beta, \tau)\bigr \le M_0+M_1 \beta^{1} + M_2 \beta^{2} + \cdots $$
 7.
Construct B _{ i },C _{ i } so that β>B _{ i } implies M _{0}+M _{1} β ^{−1}+M _{2} β ^{−2}+⋯≤C _{ i }. Note that with sufficiently large B _{ i } and a safe choice of C _{ i }, it suffices to estimate the first few coefficients M _{ j }.
We should remark that for small s, Algorithm 2.8—even the calculation of all 3s−1 coefficient maxima M _{ j }—is essentially instantaneous on today’s personal computers. Moreover, deferring the use of floatingpoint arithmetic to step 5—when τ is eliminated—makes our subsequent use of floatingpoint numbers M _{ j } quite mild. Indeed, there is virtually no loss in taking M _{ j } as (integer) ceilings of the suprema, so that estimating for C _{ i } can be performed in ℚ.
See Appendix B for the results of this calculation for each 5≤s≤9 and all relevant indices i.
2.6 Bounding the zeros
We are now ready for our main result of this section. This is in Bannai’s paper [1], but with no attempt to control β.
Proposition 2.9
Proof
3 The case s≥5
3.1 Estimates for large β
The goal here is to provide formulas for the smallest β _{0} possible (see the end of Sect. 2.1) using the B _{ i } and C _{ i } constructed in Algorithm 2.8. This task is simplified under the conditions that B _{ i } is independent of i and C _{ i }=C _{−i }. There is no loss of generality in assuming this because we can simply take \(\overline{B}\) to be the maximum of the B _{ i } and \(\overline {C_{i}}=\max\{ C_{i},C_{i}\}\), and then redefine each \(B_{i} = \overline{B}\) and \(C_{i} = C_{i} = \overline{C_{i}}\). In fact, this is not necessary for our explicit constructions because the constants in Appendix B satisfy the above conditions.
Again, for convenience, we denote H _{ s−1}(ξ _{ i }) by D _{ i } in the following proofs.
Proposition 3.1
 (i)There exists β _{1} such that, whenever β>β _{1},$$y_1+y_{1}2y_0 < 1. $$
 (ii)There exists β _{2} such that, whenever β>β _{2},for \(2\le i\le\lfloor\frac{s}{2}\rfloor\).$$y_i+y_{i}y_{i1}y_{(i1)} < 1+ \frac{\xi_i^2\xi_{i1}^2}{\xi_{i1}^2\xi_{i2}^2}y_{i1}+y_{(i1)}y_{i2}y_{(i2)} $$
 (iii)
There exists β _{0}(s) such that, whenever β>β _{0}(s) and y _{ i }+y _{−i }−y _{ i−1}−y _{−(i−1)} is an integer for \(1\le i\le\lfloor\frac{s}{2}\rfloor\), it is necessarily the case that y _{ i }+y _{−i }−y _{ i−1}−y _{−(i−1)}=0 for each i.
Proof
 (i)Observe that since τ≥2,Define$$ 0 \le2(\lambda_1\lambda_0) = \biggl(1\frac{2}{\tau} \biggr)^2\frac{\xi_1^2}{3} < \frac {\xi_1^2}{3}. $$(3.1)If \(\epsilon_{1}=\epsilon_{0}\frac{C_{1}D_{0}}{C_{0}D_{1}}\), then$$\epsilon_0 = \frac{1}{2} \biggl(1\frac{\xi_1^2}{3} \biggr) \biggl(1+\frac {C_1D_0}{C_0D_1} \biggr)^{1}\quad \text{and} \quad \beta_1 = \widehat{\beta}(0,\epsilon_0). $$Hence for β>β _{1}, By (3.1),$$\widehat{\beta}(1,\epsilon_1)=\beta_1\quad \text {and}\quad 2\epsilon_0+2\epsilon_1=1 \frac{\xi_1^2}{3}. $$and the claim follows.$$ \biggl(1\frac{\xi_1^2}{3} \biggr) < y_1+y_{1}2y_0 < 1 $$
 (ii)For \(2\le i\le\lfloor\frac{s}{2}\rfloor\), letNote if \(2\le i\le\lfloor\frac{s}{2}\rfloor\), then$$a_i=\frac{\xi_i^2\xi_{i1}^2}{\xi_{i1}^2\xi_{i2}^2}\quad \text{and}\quad \epsilon_i = \frac{1}{2} \biggl(1+(1+a_i) \frac {C_{i1}D_i}{C_iD_{i1}}+a_i\frac{C_{i2}D_i}{C_iD_{i2}} \biggr)^{1}. $$Define \(\beta_{2} = \max\{\widehat{\beta}(i,\epsilon_{i}):2\le i\le\lfloor\frac{s}{2}\rfloor \}\). For β>β _{2} and working as in (i),$$ (\lambda_i\lambda_{i1}) = ( \lambda_{i1}\lambda_{i2})a_i. $$(3.2)Using (3.2) and (3.3) again with i−1 replacing i, as required.$$ \bigly_i+y_{i}y_{i1}y_{(i1)}2( \lambda_i\lambda_{i1})\bigr < 2\epsilon_i \biggl( 1+\frac{C_{i1}D_i}{C_iD_{i1}} \biggr). $$(3.3)
 (iii)
Set β _{0}(s)=max{β _{1},β _{2}} and assume that β>β _{0}(s) and y _{ i }+y _{−i }−y _{ i−1}−y _{−(i−1)} is an integer for \(1\le i\le\lfloor\frac{s}{2}\rfloor\). By (i), y _{1}+y _{−1}−2y _{0}=0 since it is an integer whose absolute value is less than 1. Assume that y _{ i−1}+y _{−(i−1)}−y _{ i−2}−y _{−(i−2)}=0 for some \(2\le i\le\lfloor \frac{s}{2}\rfloor\). Then (ii) gives the result that y _{ i }+y _{−i }−y _{ i−1}−y _{−(i−1)} is also less than one and hence equal to 0 since it is an integer, so by induction y _{ i }+y _{−i }−y _{ i−1}−y _{−(i−1)}=0 for \(1\le i\le\lfloor \frac{s}{2}\rfloor\), and so the proof is complete. □
Proposition 3.2
 (i)There exists β _{1} such that, whenever β>β _{1},$$y_2+y_{2}y_1y_{1} < 1. $$
 (ii)There exists β _{2} such that, whenever β>β _{2},for \(3\le i\le\lfloor\frac{s}{2}\rfloor\).$$y_i+y_{i}y_{i1}y_{(i1)} < 1+ \frac{\xi_i^2\xi_{i1}^2}{\xi_{i1}^2\xi_{i2}^2}y_{i1}+y_{(i1)}y_{i2}y_{(i2)} $$
 (iii)
There exists β _{0}(s) such that, whenever β>β _{0}(s) and y _{ i }+y _{−i }−y _{ i−1}−y _{−(i−1)} is an integer for \(2\le i\le\lfloor\frac{s}{2}\rfloor\), it is necessarily the case that y _{ i }+y _{−i }−y _{ i−1}−y _{−(i−1)}=0 for each i.
Proof
 (i)Since τ≥2,Define$$ 0 \le2(\lambda_2\lambda_1) = \biggl(1\frac{2}{\tau} \biggr)^2\frac{\xi_2^2\xi_1^2}{3} < \frac{\xi_2^2\xi_1^2}{3}. $$(3.4)If \(\epsilon_{2}=\epsilon_{1}\frac{C_{2}D_{1}}{C_{1}D_{2}}\), then$$\epsilon_1 = \frac{1}{2} \biggl(1\frac{\xi_2^2\xi_1^2}{3} \biggr) \biggl(1+\frac{C_2D_1}{C_1D_2} \biggr)^{1}\quad \text{and}\quad \beta_1 = \widehat{\beta}(1,\epsilon_1). $$Hence for β>β _{1},$$\widehat{\beta}(2,\epsilon_2)=\beta_1\quad \text {and}\quad 2\epsilon_1+2\epsilon_2=1 \frac{\xi_2^2\xi_1^2}{3}. $$By (3.4),$$\bigly_2+y_{2}y_1y_{1}2( \lambda_2\lambda_1)\bigr < 2\epsilon_1+2 \epsilon_2=1\frac{\xi_2^2\xi_1^2}{3}. $$and the claim follows.$$ \biggl(1\frac{\xi_2^2\xi_1^2}{3} \biggr) < y_2+y_{2}y_1y_{1} < 1 $$
 (ii)
Define ϵ _{ i } and β _{2} in as in the proof of Proposition 3.1(ii), but omit i=2.
 (iii)
Imitate the proof of Proposition 3.1(iii). □
In the case s=6, \(\frac{\xi_{2}^{2}\xi_{1}^{2}}{3} > 1\). Hence it is impossible to choose a β _{1} to guarantee that y _{2}+y _{−2}−y _{1}−y _{−1}=0 whenever it is an integer and β>β _{1}. However, combining the integrality requirements of y _{2}+y _{−2}−y _{1}−y _{−1} and y _{3}+y _{−3}−y _{1}−y _{−1} allows us to handle this case.
Proposition 3.3
Let s=6. There exists β _{0}(6) such that, whenever β>β _{0}(6) and (y _{2}+y _{−2}−y _{1}−y _{−1}), (y _{3}+y _{−3}−y _{1}−y _{−1}) are both integers, it is necessarily the case that y _{2}+y _{−2}−y _{1}−y _{−1}=y _{3}+y _{−3}−y _{1}−y _{−1}=0.
Proof
On the other hand, suppose y _{2}+y _{−2}−y _{1}−y _{−1}=1. Then 2(λ _{2}−λ _{1})>1−ϵ _{2} and so 2(λ _{3}−λ _{1})>(1−ϵ _{2})a=a−ϵ _{2} a. Hence y _{3}+y _{−3}−y _{1}−y _{−1}>a−ϵ _{2} a−ϵ _{3}=3, a contradiction to (3.5).
It follows that y _{2}+y _{−2}−y _{1}−y _{−1}=y _{3}+y _{−3}−y _{1}−y _{−1}=0. □
To summarize, we have the following reworking of Proposition 17 in [1], but with explicit β _{0}.
Theorem 3.4
For each s≥5, there are no tight 2sdesigns with β>β _{0}(s).
Proof
 Case 1:

s odd and ≥5 implies y _{ i }+y _{−i }−y _{ i−1}−y _{−(i−1)}=0 for \(1\le i\le\lfloor\frac {s}{2}\rfloor\).
 Case 2:

s even and ≥8 implies y _{ i }+y _{−i }−y _{ i−1}−y _{−(i−1)}=0 for \(2\le i\le\lfloor\frac {s}{2}\rfloor\).
 Case 3:

s=6 implies y _{2}+y _{−2}−y _{1}−y _{−1}=y _{3}+y _{−3}−y _{1}−y _{−1}=0.
3.2 Searching over small β
We now turn to small values of β, for which the problem becomes finite.
Algorithm 3.5
 1.
Compute β _{0} from the B _{ i },C _{ i } as in the previous section.
 2.
By Propositions 2.2 and 2.3, we may restrict attention to τ>2. Since \(\alpha= \beta^{2}/(1\frac{1}{\tau})^{2}\), it follows that \(\alpha< 4\beta_{0}^{2}\). Now, since \(\alpha= \bigl(s\overline {\alpha }+\binom{s}{2} \bigr)/s\) and \(s\overline{\alpha}\) is an integer, we have \(\alpha\in\frac{1}{s} \mathbb{Z}\). This gives an explicit finite number of admissible α, as Bannai observed in [1].
 3.Note that, under the assumption of a tight design, the expressionis an integer. This is because Proposition 5 in [1] asserts that the coefficient of x ^{ s−2} in the monic polynomial \(s!\varPsi_{s}(x)/\binom{vs}{s}\) is$$ \binom{s}{2}\alpha \biggl(\alpha+\frac {2\alpha\tau\alpha +2}{\alpha\tau^2+\tau+1} \biggr) $$(3.6)and the latter term is always an integer.$$\binom{s}{2}\alpha \biggl(\alpha+\frac{2\alpha\tau\alpha +2}{\alpha\tau^2+\tau+1} \biggr)+ \binom{s}{3} \biggl(3\alpha+\frac{3s1}{4} \biggr), $$
 4.Fix α as in Step 2. Put n=k−s=ατ and defineas in (3.6). As τ>2, we may take n _{min}(α)=max{s,⌊2α⌋+1} as a lower bound for n.$$g_{\alpha}(n) := \binom{s}{2}\alpha^2 \biggl(1+ \frac{2n\alpha +2}{n^2+n+\alpha} \biggr) $$
 5.
Since \(g'_{\alpha}(n) < 0\) for all n≥n _{min}(α), it suffices to loop on integers n from n _{min}(α) until n _{max}(α), where \(g_{\alpha}(n_{\max}(\alpha)) \le\lfloor\binom{s}{2}\alpha^{2} \rfloor+1\). Any pairs (k,v) which give integral g _{ α }(n) are obtained by k=n+s and \(v = \frac{n^{2}+n}{\alpha}+2s1\).
 6.
In principle, at this point the zeros of Ψ _{ s } for these pairs (k,v) can be analyzed. However, in practice we found it sufficient in all cases to merely see that \(\lambda= \binom{v}{s}\binom {k}{2s}/\binom {v}{2s}\) was never even an integer.
We wrote a C program that implements Algorithm 3.5 for a given s and β _{0}, but with an important optimization. For n near n _{max}(α), \(g'_{\alpha}(n)\) is very small so it would be inefficient to loop over n in this region. Therefore, the program loops over integer values of g _{ α }(n) from ⌈g _{ α }(n _{max}(α))⌉ and checks the integrality of the corresponding n until the derivative becomes larger than a certain threshold (in absolute value), at which point it begins looping over n from n _{min} to a much smaller n _{max}. The program is available by contacting the authors.
Our calculations of β _{0} in step 1 are displayed in Appendix B. We can report that the method succeeds for 5≤s≤9, and probably much higher s. We have chosen to avoid continued searches for s>9 until new ideas are obtained. In particular, it would be interesting if s≥s _{0} could be excluded for nontrivial tight 2sdesigns.
Theorem 3.6
For each 5≤s≤9, there are no nontrivial tight 2sdesigns.
4 The case s=4
We are able to obtain more precise conditions in the following result.
Proposition 4.1
If there exists a nontrivial tight 8design with parameters v and k, then k>25,000 and f(k,v)=0, where f(k,v) is as in Appendix C.
Proof
It is possible to apply the MordellWeil Theorem to the Diophantine equation f(k,v)=0 and obtain a crude finiteness result. We omit the details. In any case, some very recent work of Ziqing Xiang has used careful estimates on the curve f(k,v)=0 to completely rule out the existence of nontrivial tight 8designs.
Notes
Acknowledgements
The authors would like to thank Jane Wodlinger for helpful discussions. Research of the authors is supported by NSERC.
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