Proof. For every \(x\in\mathbb{F}_q\), we have \(\theta(x,1)=x^q-x=0\), and for the point at infinity \(\infty\), we have \(\theta(1,0)=0\). Hence \(\theta(X,Y)\) vanishes at every point of \(\mathbb{P}^1(\mathbb{F}_q)\). It follows that, for \(k\ge q+1\), the subspace \(\theta(X,Y)\cdot \mathbb{F}_q[X,Y]_{k-(q+1)}\subseteq \ker(\mathop{\mathrm{Ev}}).\) Conversely, let \(f(X,Y)=\sum_{i=0}^k a_iX^{k-i}Y^i\in \ker(\mathop{\mathrm{Ev}})\). Since \(f(1,0)=0\), we have \(a_0=0\), so \(f(X,1)=\sum_{i=1}^k a_iX^{k-i}\) is a polynomial of degree at most \(k-1\). On the other hand, \(f(x,1)=0\) for all \(x\in\mathbb{F}_q\), which implies that \(f(X,1)\) is divisible by \(X^q-X\). Thus, \(f(X,1)=(X^q-X)g(X)\) for some polynomial \(g(X)\).

If \(k<q+1\), then \(\deg f(X,1)\le k-1<q=\deg(X^q-X)\), and hence \(f(X,1)=0\). Therefore \(f=0\), and so \(\ker(\mathop{\mathrm{Ev}})=0\). If \(k\ge q+1\), then \(\deg g(X)\le k-q-1.\) Let \(h(X,Y)\in\mathbb{F}_q[X,Y]_{k-(q+1)}\) be the \((k-(q+1))\)-homogenization of \(g(X)\). Since \(a_0=0\), the \(k\)-homogenization of \(f(X,1)\) recovers \(f(X,Y)\) precisely. Therefore, \[f(X,Y) =(X^qY-XY^q)\,h(X,Y) =\theta(X,Y)\,h(X,Y).\] This shows that \(f\in \theta(X,Y)\cdot \mathbb{F}_q[X,Y]_{k-(q+1)}\), completing the proof. ◻

Proof. This is standard for projective Reed–Solomon codes; see, for example, [21], [22]. ◻

Proof. Let \(f_i(X,Y)=X^{k-i}Y^i\) with \(i\le_p k\). For any \(g=\begin{pmatrix}a&b\\ c&d\end{pmatrix}\in \mathop{\mathrm{\mathrm{GL}}}_2(\mathbb{F}_q),\) we have \[\begin{align} (g\cdot f_i)(X,Y)&=(dX-bY)^{k-i}(-cX+aY)^i\\ &=\sum_{u=0}^{k-i}\sum_{v=0}^i \binom{k-i}{u}\binom{i}{v}a^v(-b)^u(-c)^{i-v} d^{k-i-u} X^{k-(u+v)}Y^{u+v} \end{align}\] It suffices to show that whenever the coefficient of \(X^{k-j}Y^{j}\) is nonzero, one has \(j\le_p k\). So assume that the coefficient of \(X^{k-j}Y^j\) is nonzero. Then there exists a pair \((u,v)\) with \(u+v=j\) such that the corresponding summand is nonzero. In particular, \(\binom{k-i}{u}\binom{i}{v}\not\equiv 0\pmod p.\) By Lemma 2, this implies \(u\le_p k-i\) and \(v\le_p i\). Since \(i\le_p k\), the subtraction \(k-i\) involves no borrowing in base \(p\). So \(u_r\le(k-i)_r= k_r-i_r\) and \(v_r\le i_r\) for all \(r\). Hence \[u_r+v_r\le (k_r-i_r)+i_r=k_r<p\] for all \(r\). Thus the addition \(j=u+v\) involves no carrying in base \(p\), and therefore \(j_r=u_r+v_r\le k_r\) for all \(r\). Thus every monomial occurring in \(g\cdot f_i\) is of the form \(X^{k-j}Y^j\) with \(j\le_p k\). Hence \(g\cdot f_i\in W_k^{\mathrm{Luc}}\). Since the monomials \(f_i\) with \(i\le_p k\) span \(W_k^{\mathrm{Luc}}\), it follows that \(g\cdot W_k^{\mathrm{Luc}}\subseteq W_k^{\mathrm{Luc}}\) for all \(g\in \mathop{\mathrm{\mathrm{GL}}}_2(\mathbb{F}_q)\). This proves the theorem. ◻

Proof. By Lemmas 3 and 4, it suffices to show that \(\mathcal{B}_W\) is \(G\)-invariant. For any \(S\in\mathcal{B}_W\) and \(g\in \mathop{\mathrm{PGL}}_2(\mathbb{F}_q)\), choose a representative \(\widetilde{g}=\begin{pmatrix} a&b\\ c&d\end{pmatrix}\in \mathop{\mathrm{\mathrm{GL}}}_2(\mathbb{F}_q)\). If \(S\subseteq \mathbb{F}_q\), then \[\begin{align} \widetilde{g}\cdot F_S&=\prod_{t\in S}((dX-bY)-t(-cX+aY))\\ &=\prod_{t\in S}((ct+d)X-(at+b)Y)\\ &=\mu F_{g(S)}, \end{align}\] for some \(\mu\in\mathbb{F}_q^*\). If \(\infty\in S\), then \(F_S=Y\prod_{t\in S\setminus\{\infty\}}(X-tY),\) and one checks similarly that \(\widetilde{g}\cdot F_S=\mu F_{g(S)}\) for some \(\mu\in\mathbb{F}_q^*\). Since \(\widetilde{g}\cdot F_S\in W\) and \(W\) is an \(\mathbb{F}_q\)-subspace, we conclude that \(F_{g(S)}\in W\). Hence \(g(S)\in \mathcal{B}_W\), and therefore \(\mathcal{B}_W\) is \(G\)-invariant. ◻

Proof. The equivalence of \((i)\) and \((ii)\) is immediate from the definition of \(\mathcal{B}_W\). If \((ii)\) holds, then \(F_S\in W\) for some \(S\in\binom{X}{k}\), and by construction, \(F_S\) vanishes exactly at the \(k\) distinct points of \(S\). Hence \((ii)\Rightarrow(iii)\). Conversely, assume \((iii)\), and let \(f\in W\) be a nonzero homogeneous polynomial of degree \(k\) vanishing at exactly \(k\) distinct points of \(X\), say \(S\subseteq X\). Since \(f\) has degree \(k\) and all its zeros in \(X\) are simple, it follows that \(f=\mu F_S\) for some \(\mu\in\mathbb{F}_q^\times\). Therefore \(F_S\in W\), and hence \((iii)\Rightarrow(ii)\). This proves the equivalence of \((i)\), \((ii)\) and \((iii)\).

Now assume that \(k\le q\). We show that \((iii)\) and \((iv)\) are equivalent. Let \(0\neq f\in W\). Then the zero coordinates of \(\mathop{\mathrm{Ev}}(f)\in C_W\) are precisely the points of \(X\) at which \(f\) vanishes. Hence \[\mathop{\mathrm{wt}}(\mathop{\mathrm{Ev}}(f))=(q+1)-\#\{x\in X:f(x)=0\}.\] Therefore, \(f\) vanishes at exactly \(k\) distinct points of \(X\) if and only if \(\mathop{\mathrm{Ev}}(f)\) has weight \(q+1-k\). By Corollary 1, every nonzero codeword of \(C_W\) has weight at least \(q+1-k\). Consequently, \(C_W\) contains a codeword of weight \(q+1-k\) if and only if \(d(C_W)=q+1-k.\) Thus \((iii)\) and \((iv)\) are equivalent, completing the proof. ◻

Proof. By Proposition 3, the hypothesis implies that \((X,\mathcal{B}_W)\) is a \(3\)-\((q+1,k,\mu)\) design for some \(\mu\). For each minimum-weight codeword \(c=\mathop{\mathrm{Ev}}(f)\in C_W\), its zero set \(Z(c):=\{\,x\in X:c_x=0\,\}\) is a block of \(\mathcal{B}_W\). Since \(\mathop{\mathrm{Supp}}(c)=X\setminus Z(c)\), the supports of the minimum-weight codewords are precisely the complements of the blocks in \(\mathcal{B}_W\). Since the complements of the blocks in a \(t\)-design again form a \(t\)-design, the conclusion follows. ◻

Proof. Since \(C_W\subseteq \mathcal{P}_k\), taking duals yields \(\mathcal{P}_k^\perp\subseteq C_W^\perp\). By Proposition 1, the code \(\mathcal{P}_k\) is MDS with parameters \([q+1,k+1,q+1-k]\), and hence its dual \(\mathcal{P}_k^\perp\) is also MDS, with parameters \([q+1,q-k,k+2]\). In particular, \(d(\mathcal{P}_k^\perp)=k+2\). Since \(\mathcal{P}_k^\perp\subseteq C_W^\perp\), we obtain \(d(C_W^\perp)\le d(\mathcal{P}_k^\perp)=k+2\), proving \((i)\).

For a fixed integer \(w\), let \(\mathcal{B}_w(C_W^\perp):=\{\mathop{\mathrm{Supp}}(c):c\in C_W^\perp,\;\mathop{\mathrm{wt}}(c)=w\}.\) To prove \((ii)\), it suffices to show that \(\mathcal{B}_w(C_W^\perp)\) is \(\mathop{\mathrm{PGL}}_2(\mathbb{F}_q)\)-invariant. Let \(S=\mathop{\mathrm{Supp}}(y)\in \mathcal{B}_w(C_W^\perp),\)where \(y\in C_W^\perp\) and \(\mathop{\mathrm{wt}}(y)=w\). Take any \(g\in \mathop{\mathrm{PGL}}_2(\mathbb{F}_q)\), and choose a representative \(\widetilde{g}\in \mathop{\mathrm{\mathrm{GL}}}_2(\mathbb{F}_q)\) of \(g\). Since \(\mathop{\mathrm{\mathrm{GL}}}_2(\mathbb{F}_q)\) acts on \(\mathcal{P}_k\) by monomial transformations, there exists a monomial matrix \(M=DP\) such that \(\widetilde{g}\cdot c=Mc\) for all \(c\in \mathcal{P}_k,\) where \(D\) is diagonal with nonzero diagonal entries and \(P\) is a permutation matrix. Now define \(z:=M^{-T}y.\) We claim that \(z\in C_W^\perp\). Indeed, for any \(c\in C_W\), the \(\mathop{\mathrm{\mathrm{GL}}}_2(\mathbb{F}_q)\)-invariance of \(C_W\) implies that \(M^{-1}c\in C_W\). Thus, we have \[c^Tz = c^TM^{-T}y = (M^{-1}c)^Ty = 0,\] since \(y\in C_W^\perp\). This confirms that \(z\in C_W^\perp\). Since \(M^{-T}=D^{-1}P,\) the matrix \(M^{-T}\) is again monomial. Thus it preserves Hamming weight, and \(\mathop{\mathrm{wt}}(z)=\mathop{\mathrm{wt}}(y)=w.\) Moreover, multiplication by \(D^{-1}\) does not change the support, while the permutation matrix \(P\) acts on coordinates as \(g\). Therefore \[\mathop{\mathrm{Supp}}(z)=g(\mathop{\mathrm{Supp}}(y))=g(S).\] Hence \(g(S)\in \mathcal{B}_w(C_W^\perp).\) This shows that \(\mathcal{B}_w(C_W^\perp)\) is \(G\)-invariant. If \(C_W^\perp\) contains a codeword of weight \(w\), then \(\mathcal{B}_w(C_W^\perp)\neq\varnothing\). Therefore, by Lemmas 3 and 4, \((X,\mathcal{B}_w(C_W^\perp))\) is a \(3\)-\((q+1,w,\lambda_w)\) design for some \(\lambda_w\). This proves \((ii)\). The final assertion follows by taking \(w=d(C_W^\perp),\) provided that \(d(C_W^\perp)\ge 3.\) ◻

Proof. For \(t\in\mathbb{F}_q\), one has \[X-tY=(\xi^qU-\xi V)-t(U-V)=(\xi^q-t)U-(\xi-t)V=(\xi^q-t)(U-\kappa(t)V).\] For \(t=\infty\), the corresponding factor is \(Y=U-V=U-\kappa(\infty)V.\) Therefore, for \(S\subseteq \mathbb{P}^1(\mathbb{F}_q)\) and \(|S|=k\), we have \[F_S(\xi^qU-\xi V,U-V)=c_S\,G_{\kappa(S)}(U,V)\] for some nonzero scalar \(c_S\in\mathbb{F}_{q^2}^\times\). Then, \(F_S\in W\) if and only if \(c_S\,G_{\kappa(S)}(U,V)\in \widetilde{W}\). Since \(\widetilde{W}\) is an \(\mathbb{F}_{q^2}\)-vector subspace, it is equivalent to \(G_{\kappa(S)}(U,V)\in \widetilde{W}\). By definitions of \(\mathcal{B}_W\) and \(\widetilde{\mathcal{B}}_W\), we have \(\kappa(S)\in \widetilde{\mathcal{B}}_W\) if and only if \(G_{\kappa(S)}\in \widetilde{W}\). It follows that \[\widetilde{\mathcal{B}}_W =\{\kappa(S):S\in\mathcal{B}_W\} =\left\{\,T\in\binom{U_{q+1}}{k}:G_T(U,V)\in\widetilde{W}\,\right\}.\] Since \(\kappa\) is a bijection from \(\mathbb{P}^1(\mathbb{F}_q)\) onto \(U_{q+1}\), the incidence structure \((U_{q+1},\widetilde{\mathcal{B}}_W)\) is isomorphic to \((\mathbb{P}^1(\mathbb{F}_q),\mathcal{B}_W).\) Therefore, if \(\mathcal{B}_W\neq\varnothing\), Proposition 2 implies that \((U_{q+1},\widetilde{\mathcal{B}}_W)\) is a \(3\)-\((q+1,k,\lambda)\) design for some \(\lambda\). ◻

Proof. By Proposition 5, a \(k\)-subset \(T\subseteq U_{q+1}\) lies in \(\widetilde{\mathcal{B}}_W\) if and only if \(G_T(U,V)\in\widetilde{W}\). By (4 ), this is equivalent to the coefficient conditions \(\sum_{a=0}^k(-1)^a\lambda_{r,a}e_a(T)=0,\) for \(1\le r\le m\). This completes the proof. ◻

Proof. Let \(H= \begin{pmatrix} -1 & \xi\\ -1 & \xi^q \end{pmatrix}\in \mathop{\mathrm{\mathrm{GL}}}_2(\mathbb{F}_{q^2}).\) Then \((H\cdot f)(U,V)=f(\xi^qU-\xi V,U-V)\) for every \(f\in \mathbb{F}_{q^2}[X,Y]_k.\) Extend scalars from \(\mathbb{F}_q\) to \(\mathbb{F}_{q^2}\) and set \[W_{k,\mathbb{F}_{q^2}}^{\mathrm{Luc}} := \mathop{\mathrm{span}}_{\mathbb{F}_{q^2}}\{U^{k-a}V^a:a\le_p k\} \subseteq \mathbb{F}_{q^2}[U,V]_k.\] Since the proof of Theorem 1 uses only Lemma 2 and the binomial expansion, it remains valid over any extension field of characteristic \(p\). Hence \(W_{k,\mathbb{F}_{q^2}}^{\mathrm{Luc}}\) is invariant under \(\mathop{\mathrm{\mathrm{GL}}}_2(\mathbb{F}_{q^2})\), and in particular \(H\cdot W_{k,\mathbb{F}_{q^2}}^{\mathrm{Luc}}=W_{k,\mathbb{F}_{q^2}}^{\mathrm{Luc}}.\) Thus, in the notation of Section 3.3, the transformed subspace \(\widetilde{W}\) attached to \(W_k^{\mathrm{Luc}}\) is precisely \(W_{k,\mathbb{F}_{q^2}}^{\mathrm{Luc}}\). By Proposition 5, a \(k\)-subset \(T\subseteq U_{q+1}\) lies in \(\widetilde{\mathcal{B}}_{W_k^{\mathrm{Luc}}}\) if and only if \(G_T(U,V)\) belongs to \(W_{k,\mathbb{F}_{q^2}}^{\mathrm{Luc}}\). Writing \(G_T(U,V)=\sum_{a=0}^k(-1)^ae_a(T)U^{k-a}V^a,\) we see that \(G_T(U,V)\in W_{k,\mathbb{F}_{q^2}}^{\mathrm{Luc}}\) if and only if the coefficients of \(U^{k-a}V^a\) vanish for all \(a\not\le_p k\), that is, if and only if \(e_a(T)=0\) for all \(a\not\le_p k\). ◻

Proof. Write \(T=\{u_1,\dots,u_k\}\subseteq U_{q+1}\). Since \(u_i^q=u_i^{-1}\) for every \(i\in\{1,\cdots,k\}\), we have \(e_a(T)^q=e_a(u_1^q,\dots,u_k^q)=e_a(u_1^{-1},\dots,u_k^{-1}),\) for each \(0\le a\le k\). On the other hand, for nonzero \(u_1,\dots,u_k\), \[e_a(u_1^{-1},\dots,u_k^{-1})=\frac{e_{k-a}(u_1,\dots,u_k)}{e_k(u_1,\dots,u_k)}.\] Hence, \(e_{k-a}(T)=e_k(T)e_a(T)^q.\) Since \(e_k(T)=\prod_{u\in T}u\neq 0\), it follows that \(e_a(T)=0\) is equivalent to \(e_{k-a}(T)=0.\) Moreover, by Lemma 2, we have \(a\le_p k\) if and only if \(\binom{k}{a}\not\equiv 0 \pmod p\), and similarly \(k-a\le_p k\) if and only if \(\binom{k}{k-a}\not\equiv 0 \pmod p\). Since \(\binom{k}{a}=\binom{k}{k-a}\), it follows that \(a\le_p k\) exactly when \(k-a\le_p k,\) and hence \(a\not\le_p k\) exactly when \(k-a\not\le_p k.\) Thus the vanishing conditions occur in pairs \((a,k-a)\), and it suffices to impose one condition from each pair, namely those with \(1\le a\le \lfloor k/2\rfloor\). ◻

Proof. By Theorem 2, the block set \(\widetilde{\mathcal{B}}_{W_k^{\mathrm{Luc}}}\) is defined by equations \(e_a(T)=0,\) for \(a\in F_{k,p}.\) By Lemma 6, the conditions \(e_a(T)=0\) and \(e_{k-a}(T)=0\) are equivalent. Hence two indices in the same orbit under \(a\mapsto k-a\) give equivalent equations, and thus the block set can be defined by a single independent equation. ◻

Proof. Since the involution \(a\mapsto k-a\) has orbits of size at most \(2\), the set \(F_{k,p}\) consists of a single orbit if and only if \(|F_{k,p}|\in\{1,2\}\). Writing \(k=\sum_{r\ge 0}k_rp^r,\) we have \(|F_{k,p}|=(k+1)-\prod_{r\ge 0}(k_r+1).\)

Assume first that \(p\) is odd. Let \(t\) be the largest index such that \(k_t\ne 0\), so that \(k=ap^t+b,\) for \(1\le a\le p-1\) and \(0\le b<p^t\). Write \[b=\sum_{r=0}^{t-1}b_rp^r,\qquad A(b):=\prod_{r=0}^{t-1}(b_r+1).\] Then \(|F_{k,p}|=a(p^t-A(b))+|F_{b,p}|.\) If \(t\ge 2\) and \(F_{k,p}\ne\varnothing\), then \(b\) is not of the form \(p^t-1\), hence \(A(b)\le p^{t-1}(p-1),\) so \(p^t-A(b)\ge p^{t-1}.\) Therefore \[|F_{k,p}|\ge ap^{t-1}\ge p^{t-1}\ge p\ge 3,\] contradicting \(|F_{k,p}|\in\{1,2\}\). Thus \(t=1\), so \(k=ap+b\), for \(1\le a\le p-1\) and \(0\le b\le p-1.\) In this case \(|F_{k,p}|=ap+b+1-(a+1)(b+1)=a(p-b-1).\) Requiring \(a(p-b-1)\in\{1,2\}\) yields exactly the three possibilities \[(a,b)=(1,p-3),\quad (1,p-2),\quad (2,p-2),\] that is, \(k\in\{2p-3,\,2p-2,\,3p-2\}.\)

Now assume that \(p=2\). Write \(k=2^t+b\), where \(t\) is the highest nonzero binary digit of \(k\) and \(0\le b<2^t\). If \(t\ge 3\) and \(F_{k,2}\ne\varnothing\), then the same argument gives \(|F_{k,2}|\ge 2^{t-1}\ge 4,\) again impossible. Hence \(t\le 2\). Since \(k\ge 3\), the case \(t=1\) gives only \(k=3\), for which \(F_{3,2}=\varnothing\). Thus \(t=2\), so \(k=4+b,\) where \(0\le b\le 3.\) A direct check gives \[|F_{4,2}|=3,\qquad |F_{5,2}|=2,\qquad |F_{6,2}|=3,\qquad |F_{7,2}|=0.\] Therefore the only possibility is \(k=5\). ◻

Proof. It suffices to determine the forbidden set \(F_{k,p}\) in each case.

1. If \(p=2\) and \(k=5\), then \(5=(101)_2\), so \(F_{5,2}=\{2,3\}.\) By Lemma 6, the conditions \(e_2(T)=0\) and \(e_3(T)=0\) are equivalent, so the single independent equation is \(e_2(T)=0\).

2. If \(p\) is odd and \(k=2p-3\), then \(k=(1,p-3)_p\), and one checks that \(F_{2p-3,p}=\{p-2,p-1\}.\) These two indices form one orbit under \(a\mapsto k-a\), so the single independent equation is \(e_{p-2}(T)=0\).

3. If \(p\) is odd and \(k=2p-2\), then \(k=(1,p-2)_p\), and \(F_{2p-2,p}=\{p-1\}.\) Thus the single equation is \(e_{p-1}(T)=0\).

4. If \(p\) is odd and \(k=3p-2\), then \(k=(2,p-2)_p\), and \(F_{3p-2,p}=\{p-1,2p-1\}.\) These two indices again form one orbit under \(a\mapsto k-a\), so the single independent equation is \(e_{p-1}(T)=0\).

 ◻

Proof. Since \(p\mid k\), the least base-\(p\) digit of \(k\) is zero. Hence, if \(i\le_p k\), then the least base-\(p\) digit of \(i\) is also zero, so \(p\mid i\). It follows that every monomial \(X^{k-i}Y^i\) occurring in \(W_k^{\mathrm{Luc}}\) has both exponents divisible by \(p\). Therefore every polynomial \(f\in W_k^{\mathrm{Luc}}\) can be written in the form \[f(X,Y)=\sum_j c_j X^{pa_j}Y^{pb_j}.\] Since \(\mathbb{F}_q\) is perfect, each \(c_j\) has a \(p\)-th root in \(\mathbb{F}_q\), and hence \(f(X,Y)=g(X,Y)^p\) for some homogeneous polynomial \(g(X,Y)\in \mathbb{F}_q[X,Y]\). Consequently every zero of \(f\) in \(\mathbb{P}^1(\mathbb{F}_q)\) occurs with multiplicity divisible by \(p\). In particular, \(f\) cannot vanish at \(k\) distinct points of \(\mathbb{P}^1(\mathbb{F}_q)\). By Proposition 3, this implies \(\mathcal{B}_{W_k^{\mathrm{Luc}}}=\varnothing\). ◻

Proof. For each \(j\), write \[f_j(X,Y)=\sum_{i_j\le_p \ell_j} a_{j,i_j}X^{\ell_j-i_j}Y^{i_j}.\] Consider a monomial occurring in the product \(f_1\cdots f_s\). Its exponent of \(Y\) is of the form \(i=i_1+\cdots+i_s\) with \(i_j\le_p \ell_j\) for all \(j\). Write \[\ell_j=\sum_{t\ge 0}(\ell_j)_tp^t,\qquad i_j=\sum_{t\ge 0}(i_j)_tp^t,\qquad k=\sum_{t\ge 0}k_tp^t.\] Since \(i_j\le_p \ell_j\), we have \((i_j)_t\le (\ell_j)_t\) for all \(j,t\). As the sum \(\ell_1+\cdots+\ell_s\) is carry-free, for each \(t\) one has \((\ell_1)_t+\cdots+(\ell_s)_t=k_t<p.\) Hence \((i_1)_t+\cdots+(i_s)_t\le (\ell_1)_t+\cdots+(\ell_s)_t=k_t<p\) for all \(t\). Therefore the addition \(i=i_1+\cdots+i_s\) is also carry-free in base \(p\), and so \(i_t=(i_1)_t+\cdots+(i_s)_t\le k_t\) for all \(t\). Thus \(i\le_p k\). It follows that every monomial occurring in \(f_1\cdots f_s\) is of the form \(X^{k-i}Y^i\) with \(i\le_p k\), and hence \(f_1(X,Y)\cdots f_s(X,Y)\in W_k^{\mathrm{Luc}}.\) ◻

Proof. Since the sets \(S_1,\dots,S_s\) are pairwise disjoint, we have \[F_S(X,Y)=\prod_{j=1}^s F_{S_j}(X,Y).\] As \(F_{S_j}\in W_{\ell_j}^{\mathrm{Luc}}\) for each \(j\), Lemma 7 gives \(F_S\in W_k^{\mathrm{Luc}}.\) Hence \(S\in \mathcal{B}_{W_k^{\mathrm{Luc}}}\). ◻

Proof. Since \(H\) is the subgroup of \(d\)-th roots of unity in \(\mathbb{F}_q^\times\), we have \[F_H(X,Y)=\prod_{a\in H}(X-aY)=X^d-Y^d.\] Only the coefficients in positions \(0\) and \(d\) occur, and these are always allowed. Hence \(H\in \mathcal{B}_{W_d^{\mathrm{Luc}}}\). Now set \(S':=H\cup\{\infty\}\). Then \[F_{S'}(X,Y)=Y(X^d-Y^d).\] The only exponents of \(Y\) occurring here are \(1\) and \(d+1\). The exponent \(d+1\) is always allowed, while \(1\le_p d+1\) holds exactly when the least base-\(p\) digit of \(d+1\) is nonzero, equivalently when \(p\nmid (d+1)\). Therefore, under this additional assumption, \(H\cup\{\infty\}\in \mathcal{B}_{W_{d+1}^{\mathrm{Luc}}}\). ◻

Proof. The polynomial vanishing exactly on \(\mathbb{F}_{p^m}\) is \(\prod_{a\in\mathbb{F}_{p^m}}(X-aY) = X^{p^m}-XY^{p^m-1}\). Therefore, the associated polynomial for \(S = \mathbb{F}_{p^m}\cup\{\infty\}\) is \[F_S(X,Y) = Y(X^{p^m}-XY^{p^m-1}) = X^{p^m}Y - XY^{p^m}.\] The exponents of \(Y\) occurring in \(F_S(X,Y)\) are exactly \(1\) and \(p^m\). For \(k = p^m+1\), its base-\(p\) expansion simply consists of \(1\)’s at the \(m\)-th and \(0\)-th positions. It is then immediate that \(1 \le_p k\) and \(p^m \le_p k\). Thus, \(F_S \in W_{p^m+1}^{\mathrm{Luc}}\), which yields \(S\in \mathcal{B}_{W_{p^m+1}^{\mathrm{Luc}}}\). ◻

Proof. If \(m\mid e\), then Proposition 10 shows that \(\mathcal{B}_{W_{q_0+1}^{\mathrm{Luc}}}\neq \varnothing.\) Conversely, suppose that \(\mathcal{B}_{W_{q_0+1}^{\mathrm{Luc}}}\neq \varnothing.\) Since \(\mathop{\mathrm{PGL}}_2(\mathbb{F}_q)\) acts sharply \(3\)-transitively on \(\mathbb{P}^1(\mathbb{F}_q)\) and preserves \(\mathcal{B}_{W_{q_0+1}^{\mathrm{Luc}}}\), it suffices to study the blocks containing \(\{\infty,0,1\}\). Let \(S\in \mathcal{B}_{W_{q_0+1}^{\mathrm{Luc}}}\) contain \(\{\infty,0,1\}\), and write \(S=\{\infty\}\cup T,\) where \(T\subseteq \mathbb{F}_q\) has cardinality \(q_0\) and contains \(0\) and \(1\). Define \(P_T(Z):=\prod_{t\in T}(Z-t)\in \mathbb{F}_q[Z].\) Then \[F_S(X,Y)=Y\prod_{t\in T}(X-tY)=Y^{q_0+1}P_T(X/Y).\] Since \(W_{q_0+1}^{\mathrm{Luc}}=\mathop{\mathrm{span}}_{\mathbb{F}_q}\{X^{q_0+1},X^{q_0}Y,XY^{q_0},Y^{q_0+1}\},\) and every term of \(F_S(X,Y)\) contains a factor \(Y\), it follows that \(P_T(Z)=Z^{q_0}+cZ+d\) for some \(c,d\in \mathbb{F}_q\). As \(0\in T\), we have \(d=0\). As \(1\in T\), we have \(0=P_T(1)=1+c,\) so \(c=-1\). Therefore \(P_T(Z)=Z^{q_0}-Z.\) Thus \(T\) is precisely the set of roots in \(\mathbb{F}_q\) of the polynomial \(Z^{q_0}-Z\). The roots of \(Z^{p^m}-Z\) in \(\mathbb{F}_q\) form the unique subfield of \(\mathbb{F}_q\) of order \(p^{\gcd(m,e)}\), namely \(\mathbb{F}_{p^{\gcd(m,e)}}\). Since \(|T|=q_0=p^m\), we must have \(\gcd(m,e)=m\), that is, \(m\mid e.\) We have shown that \(\mathcal{B}_{W_{q_0+1}^{\mathrm{Luc}}}\neq \varnothing\) if and only if \(m\mid e.\)

Assume now that \(m\mid e\). Then the argument above shows that the unique block containing \(\{\infty,0,1\}\) is \(\mathbb{F}_{q_0}\cup\{\infty\}.\) By the sharp \(3\)-transitivity of \(\mathop{\mathrm{PGL}}_2(\mathbb{F}_q)\), every \(3\)-subset of \(\mathbb{P}^1(\mathbb{F}_q)\) is therefore contained in exactly one block of \(\mathcal{B}_{W_{q_0+1}^{\mathrm{Luc}}}\). Hence \((\mathbb{P}^1(\mathbb{F}_q),\mathcal{B}_{W_{q_0+1}^{\mathrm{Luc}}})\) is a Steiner system \(S(3,q_0+1,q+1)\), equivalently a \(3\)-\((q+1,q_0+1,1)\) design. Finally, \[|\mathcal{B}_{W_{q_0+1}^{\mathrm{Luc}}}|=\frac{\binom{q+1}{3}}{\binom{q_0+1}{3}} =\frac{q(q^2-1)}{q_0(q_0^2-1)}.\] This completes the proof. ◻

Proof. This gives the weight-\(3\) and weight-\(5\) terms in the known weight distribution of the ternary Melas code; see [28]. ◻

Proof. Fix the three points \(T_0:=\{\infty,0,-1\}\subseteq \mathbb{P}^1(\mathbb{F}_q).\) By definition, \(\lambda_2=\#\{\,B\in \mathcal{B}_{W_7^{\mathrm{Luc}}}:T_0\subseteq B\,\}.\) Every such block has the form \[B=\{\infty,0,-1,x_1,x_2,x_3,x_4\},\] where \(x_1,x_2,x_3,x_4\in\mathbb{F}_q^\times\setminus\{-1\}\) are pairwise distinct. For \(1\le i\le 4\), set \(e_i=\sigma_i(x_1,x_2,x_3,x_4).\) Then \[\prod_{j=1}^4(X-x_jY)=X^4-e_1X^3Y+e_2X^2Y^2-e_3XY^3+e_4Y^4,\] and hence \[\begin{align} F_B(X,Y) &=Y\cdot X\cdot (X+Y)\prod_{j=1}^4(X-x_jY) \\ &=X^6Y+(1-e_1)X^5Y^2+(e_2-e_1)X^4Y^3 +(e_2-e_3)X^3Y^4+(e_4-e_3)X^2Y^5+e_4XY^6. \end{align}\] Since \[W_7^{\mathrm{Luc}}=\mathop{\mathrm{span}}_{\mathbb{F}_q}\{X^7,X^6Y,X^4Y^3,X^3Y^4,XY^6,Y^7\},\] we obtain \(B\in \mathcal{B}_{W_7^{\mathrm{Luc}}}\) if and only if \(1-e_1=0,\;e_4-e_3=0,\) that is, \(e_1=1,\;e_3=e_4.\) Since \[x_1+x_2+x_3+x_4=e_1,\qquad x_1^{-1}+x_2^{-1}+x_3^{-1}+x_4^{-1}=\frac{e_3}{e_4},\] it follows that \(B\in \mathcal{B}_{W_7^{\mathrm{Luc}}}\) if and only if \[x_1+x_2+x_3+x_4=1,\; x_1^{-1}+x_2^{-1}+x_3^{-1}+x_4^{-1}=1.\] Let \(\Lambda:= \Bigl\{ \{x_1,x_2,x_3,x_4\}\subseteq \mathbb{F}_q^\times\setminus\{-1\}: x_i\;\text{pairwise distinct}, \;\sum_{i=1}^4x_i=1,\;\sum_{i=1}^4x_i^{-1}=1 \Bigr\}.\) Then \(\lambda_2=|\Lambda|.\) Define \(P:=\{(c,t):c\in M(q),\;\mathrm{wt}(c)=5,\;t\in \mathop{\mathrm{Supp}}(c)\}.\) Since every weight-\(5\) codeword has exactly five support positions, \(|P|=5A_5.\) For \((c,t)\in P\), write \[c=\sum_{x\in\mathop{\mathrm{Supp}}(c)}\varepsilon_xe_x,\qquad \varepsilon_x\in\{\pm 1\}\subseteq\mathbb{F}_3.\] For each \(x\in \mathop{\mathrm{Supp}}(c)\setminus\{t\}\), define \(u_x:=-\varepsilon_x\varepsilon_t\,\frac{x}{t}.\) Set \(\Phi(c,t):=\{u_x:x\in \mathop{\mathrm{Supp}}(c)\setminus\{t\}\}.\) We claim that \(\Phi\) defines a map \(\Phi:P\longrightarrow \Lambda.\) Since \(c\in M(q)\), we have \(\sum_{x\in\mathop{\mathrm{Supp}}(c)}\varepsilon_xx=0,\) and \(\sum_{x\in\mathop{\mathrm{Supp}}(c)}\varepsilon_xx^{-1}=0.\) Separating the term indexed by \(t\) and dividing by \(-\varepsilon_tt\) and \(-\varepsilon_tt^{-1}\), respectively, we obtain \[\sum_{x\in\mathop{\mathrm{Supp}}(c)\setminus\{t\}}u_x=1, \qquad \sum_{x\in\mathop{\mathrm{Supp}}(c)\setminus\{t\}}u_x^{-1}=1.\] Thus only distinctness and the exclusion of the value \(-1\) remain to be checked.

First, no \(u_x\) can be equal to \(1\). Indeed, if \(u_x=1\), then \(-\varepsilon_x\varepsilon_t\,\frac{x}{t}=1,\) so \(\varepsilon_xx=-\varepsilon_tt,\) and \(\varepsilon_xx^{-1}=-\varepsilon_tt^{-1}.\) Deleting the coordinates \(x\) and \(t\) would then produce a codeword of weight \(3\) in \(M(q)\), contradicting \(A_3=0\).

Next, no \(u_x\) can be equal to \(-1\). Suppose \(u_x=-1\) for some \(x\ne t\), and let the remaining normalized values be \(v_1,v_2,v_3\). Then \[v_1+v_2+v_3=-1,\qquad v_1^{-1}+v_2^{-1}+v_3^{-1}=-1.\] If \[e_1''=v_1+v_2+v_3,\qquad e_2''=v_1v_2+v_1v_3+v_2v_3,\qquad e_3''=v_1v_2v_3,\] then \(e_1''=-1\) and \(e_2''/e_3''=-1\), so \(e_2''=-e_3''\). Hence \(v_1,v_2,v_3\) are the roots of \[T^3-e_1''T^2+e_2''T-e_3''=T^3+T^2+e_2''T+e_2''=(T+1)(T^2+e_2'').\] Thus one of the \(v_i\) equals \(-1\). Repeating the same argument gives at least two normalized values equal to \(-1\), which is impossible, since \(u_x=-1\) is equivalent to \(x=\varepsilon_x\varepsilon_tt\), and for fixed \(t\) there is at most one such \(x\neq t\). Finally, the values \(u_x\) are pairwise distinct. Suppose \(u_x=u_y\) for distinct \(x,y\in \mathop{\mathrm{Supp}}(c)\setminus\{t\}\). Excluding the case \(u_x=-1\) already treated, write \(u_x=u_y=u\neq -1\), and let the remaining two normalized values be \(v,w\). Then \[u+u+v+w=1,\qquad u^{-1}+u^{-1}+v^{-1}+w^{-1}=1.\] Since the characteristic is \(3\), this becomes \[v+w=1+u,\qquad v^{-1}+w^{-1}=1+u^{-1}=\frac{u+1}{u}.\] Hence \[\frac{v+w}{vw}=\frac{u+1}{u},\] so, because \(u\neq -1\), we obtain \(vw=u.\) Therefore \(v\) and \(w\) are the roots of \[T^2-(1+u)T+u=(T-1)(T-u).\] Thus one of \(v,w\) equals \(1\), contradicting the fact already proved that no normalized value equals \(1\). We have shown that \(\Phi(c,t)\in \Lambda\), so \(\Phi\) is well defined.

We now compute the cardinality of each fiber. Fix \(U=\{u_1,u_2,u_3,u_4\}\in \Lambda.\) We first show that \(1\notin U\). Suppose, to the contrary, that \(U=\{1,v_1,v_2,v_3\}\). Since \(U\in\Lambda\), we have \[v_1+v_2+v_3=0, \qquad v_1^{-1}+v_2^{-1}+v_3^{-1}=0.\] Let \[e_1''=v_1+v_2+v_3,\qquad e_2''=v_1v_2+v_1v_3+v_2v_3,\qquad e_3''=v_1v_2v_3.\] Then \(e_1''=0\) and \(e_2''/e_3''=0\), so \(e_2''=0\). Hence \(v_1,v_2,v_3\) are the three roots of \(T^3-e_3''\). Since \(q=3^m\), the Frobenius map \(x\mapsto x^3\) is a bijection on \(\mathbb{F}_q\), so the polynomial \(T^3-e_3''\) has a unique root in \(\mathbb{F}_q\), counted with multiplicity. This contradicts the fact that \(v_1,v_2,v_3\) are pairwise distinct. Therefore \(1\notin U\). Next we show that \(U\) contains no pair \(\{u,-u\}\). Suppose, to the contrary, that \(U=\{u,-u,v,w\}\). Since \(U\in\Lambda\), we have \(v+w=1\) and \(v^{-1}+w^{-1}=1\). Thus \(\frac{v+w}{vw}=1\). Because \(v+w=1\), it follows that \(vw=1\). Hence \(v\) and \(w\) are the two roots of \[T^2-(v+w)T+vw=T^2-T+1=(T+1)^2\] in characteristic \(3\). Therefore \(v=w=-1\), contradicting both the distinctness of the elements of \(U\) and the condition \(U\subseteq \mathbb{F}_q^\times\setminus\{-1\}\). So \(U\) contains no pair \(\{u,-u\}\).

Choose \(t\in\mathbb{F}_q^\times,\) which gives \(q-1\) choices, choose \(\varepsilon_t\in\{\pm 1\},\) which gives \(2\) choices, and choose arbitrary sign \(\delta_1,\delta_2,\delta_3,\delta_4\in\{\pm 1\},\) which gives \(2^4=16\) choices. For \(1\le i\le 4\), define \(x_i:=\delta_i u_it\) and \(\varepsilon_{x_i}:=-\delta_i\varepsilon_t\). Set \(c:=\varepsilon_te_t+\sum_{i=1}^4\varepsilon_{x_i}e_{x_i}.\) We claim that the five positions \(t,x_1,x_2,x_3,x_4\) are pairwise distinct. First, if \(x_i=t\), then \(\delta_i u_i=1\), so \(u_i=\delta_i\in\{\pm1\}\). Since \(U\subseteq \mathbb{F}_q^\times\setminus\{-1\}\) and we have already proved that \(1\notin U\), this is impossible. Next, if \(x_i=x_j\) for some \(i\ne j\), then \(\delta_i u_i=\delta_j u_j\), hence \(u_i=\delta_i\delta_j\,u_j\). If \(\delta_i\delta_j=1\), then \(u_i=u_j\), contradicting the distinctness of the elements of \(U\). If \(\delta_i\delta_j=-1\), then \(u_i=-u_j\), contradicting the fact just proved that \(U\) contains no pair \(\{u,-u\}\). Therefore \(t,x_1,x_2,x_3,x_4\) are pairwise distinct, and so \(\mathop{\mathrm{wt}}(c)=5\). Exactly as in the construction above, one checks that \(c\in M(q)\), \(\mathrm{wt}(c)=5\), and \(\Phi(c,t)=U.\) Conversely, every preimage of \(U\) arises in this way: if \((c,t)\in\Phi^{-1}(U)\) and \[c=\varepsilon_t e_t+\sum_{i=1}^4\varepsilon_{x_i}e_{x_i},\] then, after indexing the four elements of \(\mathop{\mathrm{Supp}}(c)\setminus\{t\}\) so that \(u_i=-\varepsilon_{x_i}\varepsilon_t\frac{x_i}{t}\) for \(1\le i\le 4\), we recover \(x_i=\delta_i u_i t\) and \(\varepsilon_{x_i}=-\delta_i\varepsilon_t\), with \(\delta_i:=-\varepsilon_{x_i}\varepsilon_t\in\{\pm1\}\). Thus every element of \(\Phi^{-1}(U)\) is obtained uniquely from a choice of \[t\in\mathbb{F}_q^\times,\qquad \varepsilon_t\in\{\pm1\},\qquad (\delta_1,\delta_2,\delta_3,\delta_4)\in\{\pm1\}^4.\] Therefore \[|\Phi^{-1}(U)|=(q-1)\cdot 2\cdot 16=32(q-1)\] for every \(U\in \Lambda\). Counting \(P\) in two ways gives \[5A_5=|P|=\sum_{U\in\Lambda}|\Phi^{-1}(U)|=32(q-1)|\Lambda|=32(q-1)\lambda_2,\] and hence \(\lambda_2=\frac{5A_5}{32(q-1)}.\) Applying Lemma 8, we obtain \[\lambda_2=\frac{q^2+(( -1)^m-14)q+36}{24}.\] The formula for \(|\mathcal{B}_{W_7^{\mathrm{Luc}}}|\) follows from \[|\mathcal{B}_{W_7^{\mathrm{Luc}}}|=\lambda_2\frac{\binom{q+1}{3}}{\binom{7}{3}}.\] The final assertion follows from Proposition 2. ◻

Proof. Suppose, to the contrary, that \(e_2(y_1,\dots,y_5,1,1)=0.\) Let \[g(X)=\prod_{i=1}^5(X-y_i)=X^5+a_4X^4+a_3X^3+a_2X^2+a_1X+a_0.\] Since \(g(X)(X-1)^2=\prod_{i=1}^5(X-y_i)(X-1)^2,\) the roots of \(g(X)(X-1)^2\) are precisely \(y_1,\dots,y_5,1,1\). Hence the coefficient of \(X^5\) in this polynomial is \(e_2(y_1,\dots,y_5,1,1),\) and therefore vanishes by assumption. Since the characteristic is \(3\), we have \((X-1)^2=X^2+X+1.\) Thus \[g(X)(X-1)^2=g(X)(X^2+X+1),\] and the coefficient of \(X^5\) in this product is \(1+a_4+a_3\), so \(1+a_4+a_3=0.\)

Write \[g(X)=X^5-e_1X^4+e_2X^3-e_3X^2+e_4X-e_5,\] where \(e_j=e_j(y_1,\dots,y_5)\). Since each \(y_i\in U_{q+1}\), it satisfies \(y_i^q=y_i^{-1}\). We have \[e_j(y_1,\dots,y_5)^q=e_j(y_1^{-1},\dots,y_5^{-1})=\frac{e_{5-j}(y_1,\dots,y_5)}{e_5(y_1,\dots,y_5)}\] for \(0\le j\le 5\). Therefore \(e_4=e_5e_1^q,\) and \(e_3=e_5e_2^q.\)

Since \[a_4=-e_1,\quad a_3=e_2,\quad a_2=-e_3,\quad a_1=e_4,\quad a_0=-e_5,\] it follows that \(a_1=a_0a_4^q\) and \(a_2=a_0a_3^q.\) Taking \(q\)-th powers in \(1+a_4+a_3=0\), we obtain \(1+a_4^q+a_3^q=0.\) Hence \[\begin{align} g(1) &=1+a_4+a_3+a_2+a_1+a_0 \\ &=(1+a_4+a_3)+a_0(1+a_4^q+a_3^q)=0. \end{align}\] Thus \(1\) is a root of \(g(X)\), which means that \(y_i=1\) for some \(i\). This contradicts the assumption that \(y_1,\dots,y_5\in U_{q+1}\setminus\{1\}\). Therefore \(e_2(y_1,\dots,y_5,1,1)\neq 0.\) ◻

Proof. Set \(\mathcal{B}^*:=\{(B,x):B\in \widetilde{\mathcal{B}}_{W_7^{\mathrm{Luc}}},\;x\in B\}.\) It follows that \(|\mathcal{B}^*|=7\,|\widetilde{\mathcal{B}}_{W_7^{\mathrm{Luc}}}|.\) We claim that the map \[\phi:\mathcal{B}^*\longrightarrow \mathcal{A},\qquad (B,x)\longmapsto B\setminus\{x\},\] is a bijection. Let \((B,x)\in \mathcal{B}^*\), and set \(A:=B\setminus\{x\}.\) Since \(e_2(B)=0\), we have \(e_2(A)+xe_1(A)=0.\) Hence \(e_2(A)^{q+1}=(-1)^{q+1}x^{q+1}e_1(A)^{q+1}=e_1(A)^{q+1},\) because \(q+1\) is even and \(x\in U_{q+1}\) implies \(x^{q+1}=1\). Using the symmetry relation \(e_{6-i}(A)=e_6(A)e_i(A)^q,\) it is equivalent to \(e_4(A)e_2(A)=e_5(A)e_1(A).\) Thus \(A\in \mathcal{A}\), so \(\phi\) is well defined. Conversely, let \(A=\{x_1,\dots,x_6\}\in \mathcal{A}.\) We first show that \(e_1(A)\neq 0\). Otherwise the defining equation for \(\mathcal{A}\) gives \(e_2(A)^{q+1}=e_1(A)^{q+1}=0,\) hence \(e_2(A)=0\). Dividing by \(x_6\), we obtain \[e_1\!\left(\frac{x_1}{x_6},\dots,\frac{x_5}{x_6},1\right)=0, \qquad e_2\!\left(\frac{x_1}{x_6},\dots,\frac{x_5}{x_6},1\right)=0.\] Therefore \[e_2\!\left(\frac{x_1}{x_6},\dots,\frac{x_5}{x_6},1,1\right)=e_2\!\left(\frac{x_1}{x_6},\dots,\frac{x_5}{x_6},1\right)+1\cdot e_1\!\left(\frac{x_1}{x_6},\dots,\frac{x_5}{x_6},1\right)=0\] contradicting Lemma 9. Thus \(e_1(A)\neq 0\).

Now define \(x:=-\frac{e_2(A)}{e_1(A)}.\) Since \(A\in \mathcal{A}\), we have \(x^{q+1}=1\), so \(x\in U_{q+1}\). We next show that \(x\notin A\). Suppose, for instance, that \(x=x_1\). Then \(e_2(x_1,x_1,x_2,\dots,x_6)=0.\) Dividing by \(x_1\) gives \(e_2\!\left(1,1,\frac{x_2}{x_1},\dots,\frac{x_6}{x_1}\right)=0,\) again contradicting Lemma 9. Hence \(x\notin A\). Set \(B:=A\cup\{x\}.\) Then \(e_2(B)=e_2(A)+xe_1(A)=0,\) so \(B\in \widetilde{\mathcal{B}}_{W_7^{\mathrm{Luc}}}\) and clearly \(\phi(B,x)=A.\) Thus \(\phi\) is surjective. Finally, we show that \(\phi\) is injective. If \(\phi(B_1,x_1)=\phi(B_2,x_2)=A\), then we have \(e_2(B_1)= e_2(A)+x_1e_1(A)=0\) and \(e_2(B_2)= e_2(A)+x_2e_1(A)=0\). Since \(e_1(A)\neq0\), we have \(x_1=x_2\) and then \(B_1=B_2\).

Therefore \(\phi\) is a bijection, and hence \[|\mathcal{A}|=|\mathcal{B}^*|=7\,|\widetilde{\mathcal{B}}_{W_7^{\mathrm{Luc}}}|.\] This completes the proof. ◻

Proof. By (1 ), we have \[|\mathcal{A}|=\lambda_1\frac{\binom{q+1}{3}}{\binom{6}{3}}, \qquad |\widetilde{\mathcal{B}}_{W_7^{\mathrm{Luc}}}|=\lambda_2\frac{\binom{q+1}{3}}{\binom{7}{3}}.\] Using \(|\mathcal{A}|=7|\widetilde{\mathcal{B}}_{W_7^{\mathrm{Luc}}}|\) in Proposition 12, we obtain \[\lambda_1\frac{\binom{q+1}{3}}{20} = 7\lambda_2\frac{\binom{q+1}{3}}{35},\] and therefore \(\lambda_1=4\lambda_2.\) Substituting the value of \(\lambda_2\) from Theorem 4, we get \(\lambda_1=\frac{q^2+(( -1)^m-14)q+36}{6}.\) ◻