Proof. For 1\le n\le m-2, neither na/m nor (n+1)a/m is an integer. The floor identity \left\lfloor\frac{k(m-a)}m\right\rfloor = k-1-\left\lfloor\frac{ka}m\right\rfloor therefore gives d_n(a)+d_n(m-a)=1. The endpoint terms satisfy d_0(a)=d_0(m-a)=0, \qquad d_{m-1}(a)=d_{m-1}(m-a)=1. There are b-2 interior elements of G. Since a is not divisible by b, \left\lfloor\frac ab\right\rfloor + \left\lfloor\frac{m-a}b\right\rfloor =b-1. Substitution in (1) gives the first identity. The map a\mapsto m-a carries U_s bijectively onto U_{b-s}. Averaging the first identity over a fiber gives the second, and subtracting the paired means gives the third. ◻

Proof. The subgroup H=\{1+jb:0\le j\le b-1\} is the kernel of reduction from U_m to (\mathbb Z/b\mathbb Z)^\times. A primitive character modulo b^2 is nontrivial on H. Character orthogonality gives \sum_{u\in H}\overline\chi(u)=0. Every fiber U_s is a multiplicative coset of H, so its character sum also vanishes. The fiber mean is constant on that coset, which proves the second identity. ◻

Proof. Put P_\chi=\sum_{k=1}^{b-1}\overline\chi(k), \qquad \alpha=\overline\chi(b+1). The terms indexed by n=r(b+1) give S_G(\chi)=Q_\chi-\alpha P_\chi, where Q_\chi = \sum_{r=0}^{b-2} \overline\chi\bigl(r(b+1)+1\bigr). The reflection b^2-\bigl(r(b+1)+1\bigr) =(b-1-r)(b+1) maps the arguments in Q_\chi bijectively onto \{j(b+1):1\le j\le b-1\}. Oddness gives -Q_\chi = \alpha\sum_{j=1}^{b-1}\overline\chi(j) = \alpha P_\chi. Thus Q_\chi=-\alpha P_\chi, which proves the result. ◻

Proof. Multiplication by n permutes U_m. Substitution by n^{-1}a gives \sum_{a\in U_m} \overline\chi(a) \left\{\frac{na}{m}\right\} = \chi(n)\sum_{a\in U_m} \overline\chi(a)\frac am. This is the first identity. Subtracting it from \frac nm\sum_{a\in U_m}a\,\overline\chi(a) gives the second. ◻

Proof. Write \varphi=\varphi(m). By Lemma 2, the centering term contributes zero. Hence \varphi\,\widehat S^\circ(\chi) = \sum_{a\in U_m}S(a)\overline\chi(a).

The constant term in (1) contributes zero. Also, \left\{\frac ab\right\} is constant on each fiber U_s. Primitive fiber cancellation gives \sum_{a\in U_m} \left\{\frac ab\right\}\overline\chi(a)=0. It follows that -\sum_{a\in U_m} \left\lfloor\frac ab\right\rfloor\overline\chi(a) = -\frac1b\sum_{a\in U_m}a\,\overline\chi(a) = -bB_{1,\overline\chi}.

At one endpoint, d_0(a)=0. At the other, d_{m-1}(a)=1. Every nontrivial character sums to zero, so both endpoint slices contribute zero. Every remaining diagonal index has the form n=r(b+1) with 1\le r\le b-2. Both n and n+1 are units modulo m, so Lemma 4 gives \sum_{a\in U_m}d_n(a)\overline\chi(a) = \bigl[1+\chi(n)-\chi(n+1)\bigr]B_{1,\overline\chi}.

The two endpoint terms in \overline{S_G(\chi)} both equal 1. Therefore the sum over the b-2 interior slices is B_{1,\overline\chi} \bigl[b-\overline{S_G(\chi)}\bigr]. Combining this with the floor contribution gives \varphi\,\widehat S^\circ(\chi) = -bB_{1,\overline\chi} + B_{1,\overline\chi} \bigl[b-\overline{S_G(\chi)}\bigr] = -B_{1,\overline\chi}\overline{S_G(\chi)}. ◻

Proof. The functional equation for a primitive odd character gives B_{1,\overline\chi} = -\frac b\pi\, \varepsilon(\overline\chi)L(1,\chi) with the stated Gauss-sum convention [1]. Substitute this identity and (3) into Theorem 5. Taking absolute values and applying Lemma 3 gives the last line. ◻

Proof. For either odd modulus q=b or q=b^2, character orthogonality followed by the odd projector gives \sum_{\substack{\chi\bmod q\\\chi(-1)=-1}}\chi(y) = \frac{\varphi(q)}2 \left( \mathbf 1_{y\equiv1\;(\bmod q)} - \mathbf 1_{y\equiv-1\;(\bmod q)} \right). The imprimitive odd characters modulo b^2 are precisely the lifts of the odd characters modulo b. Subtraction proves the kernel formula. Setting y=1 gives the population. ◻

Proof. Write x_r=r(b+1)+1, \qquad 0\le r\le b-2. These are distinct units modulo b^2, and their reductions modulo b run once through the nonzero residue classes. Equation (3) reduces the theorem to \sum_{\chi\in\mathcal P_b}|A(\chi)|^2 = b|\mathcal P_b|.

Expand the left side and apply Lemma 7 with y=x_kx_j^{-1}. The positive congruence modulo b^2 holds only when j=k. Its contribution is \frac{\varphi(b^2)}2(b-1) = \frac b2(b-1)^2. The negative congruence modulo b^2 would require (j+k)(b+1)\equiv-2\pmod{b^2}. Since (b-1)(b+1)\equiv-1\pmod{b^2}, this forces j+k\equiv2b-2\pmod{b^2}. The range 0\le j+k\le2b-4 makes that impossible.

Modulo b, the positive congruence again holds on the b-1 diagonal pairs. The negative congruence is j+k=b-2, which also has b-1 ordered solutions. Their two kernel contributions have equal magnitude and opposite sign. They cancel. The surviving term is \sum_{\chi\in\mathcal P_b}|A(\chi)|^2 = \frac b2(b-1)^2 = b|\mathcal P_b|. Multiplication by four through (3) proves the theorem. ◻

Proof. Parseval with the normalization (2) gives \sum_{\chi\bmod m}|\widehat S^\circ(\chi)|^2 = \frac1{\varphi(m)} \sum_{a\in U_m}|S^\circ(a)|^2. Lemma 1 makes every even coefficient vanish. Every imprimitive character modulo b^2 is induced from modulus b and is constant on each fiber U_s. Fiber centering makes every imprimitive odd coefficient vanish as well. Only primitive odd characters remain.

Substitution of Corollary 6 gives \frac{b^2}{\pi^2\varphi(m)^2} \sum_{\chi\ {\rm primitive\ odd}} |L(1,\chi)|^2|S_G(\chi)|^2 = \frac{E_b}{\varphi(m)}. Rearranging proves the identity. ◻

Proof. The first identity is Theorem 8. Divide Theorem 9 by 2b(b-1)^2 and use \varphi(b^2)=b(b-1). ◻

Proof. Put \psi=\overline\chi and z=\psi(2). The residue 2 generates U_{25} and has order 20. Write z=e^{2\pi i j/20}. Oddness makes j odd, while primitivity modulo 25 gives 5\nmid j. Thus \gcd(j,20)=1, and z is a primitive twentieth root of unity.

Since 3\equiv2^7\pmod{25}, \qquad 4\equiv2^2\pmod{25}, the short character sum is P_5(\chi) = \sum_{k=1}^{4}\psi(k) = p(z), \qquad p(x)=1+x+x^2+x^7. Lemma 3 gives |S_G(\chi)|=2|p(z)|.

Let a_r be the least positive residue of 2^r modulo 25. For 0\le r\le9, (a_0,\ldots,a_9) = (1,2,4,8,16,7,14,3,6,12). The relations a_{r+10}=25-a_r, \qquad z^{r+10}=-z^r pair the Bernoulli sum into 25B_{1,\psi} = \sum_{r=0}^{9}(2a_r-25)z^r. Reduction by \Phi_{20}(z)=z^8-z^6+z^4-z^2+1=0 gives 25B_{1,\psi}=-10q(z), where q(x) = 1+2x+3x^2+x^3-2x^4+x^5+x^6+2x^7.

The Laurent polynomial identity \begin{aligned} x^7\bigl( q(x)q(x^{-1})-5p(x)p(x^{-1}) \bigr) ={}& \Phi_{20}(x)\\ &\cdot \bigl(-3+x^2+5x^3+x^4-3x^6\bigr) \end{aligned} is exact. Setting x=z gives |q(z)|^2=5|p(z)|^2. Since B_{1,\psi}=-(2/5)q(z), |B_{1,\psi}|^2 = \frac45|p(z)|^2. The two conclusions follow. ◻

Proof. At modulus 25, |B_{1,\overline\chi}|=\frac5\pi|L(1,\chi)|. Theorem 11 gives |S_G(\chi)| = \frac{5\sqrt5}{\pi}|L(1,\chi)|. Substitution in Theorem 5 proves the result. ◻

Proof. Let \zeta=e^{2\pi i/20}. The primitive odd characters modulo 25 correspond to the eight exponents j\in(\mathbb Z/20\mathbb Z)^\times. For p(x)=1+x+x^2+x^7, put d(x)=p(x)p(x^{-1}). Its exact Laurent expansion is \begin{aligned} d(x) ={}& 4+2(x+x^{-1})+(x^2+x^{-2})\\ &+(x^5+x^{-5})+(x^6+x^{-6})+(x^7+x^{-7}). \end{aligned}

Write c_{20}(n) = \sum_{j\in(\mathbb Z/20\mathbb Z)^\times}\zeta^{jn} for the Ramanujan sum. The even-power coefficients of d(x)^2, reduced modulo x^{20}-1, are \begin{aligned} 32 &+14(x^2+x^{-2})+7(x^4+x^{-4})\\ &+17(x^6+x^{-6})+9(x^8+x^{-8})+2x^{10}. \end{aligned} Terms with odd exponent do not contribute because c_{20}(n)=0 for odd n. The standard Ramanujan-sum formula [2] gives \begin{array}{c|rrrrrr} n&0&2&4&6&8&10\\ c_{20}(n)&8&2&-2&2&-2&-8. \end{array} Therefore \begin{aligned} \sum_{j\in(\mathbb Z/20\mathbb Z)^\times} |p(\zeta^j)|^4 ={}& 32(8)+28(2)+14(-2)\\ &+34(2)+18(-2)+2(-8)\\ ={}&300. \end{aligned} Since |S_G(\chi)|=2|p(\zeta^j)|, \sum_{\chi\ {\rm primitive\ odd}}|S_G(\chi)|^4=4800.

The cyclotomic identity gives |B_{1,\overline\chi}|^2 = \frac15|S_G(\chi)|^2. Theorem 5 now yields \sum_{\chi\ {\rm primitive\ odd}} |\widehat S^\circ(\chi)|^2 = \frac1{2000} \sum_{\chi\ {\rm primitive\ odd}}|S_G(\chi)|^4 = \frac{12}{5}. Parseval gives \frac{E_5}{20}=\frac{12}{5}, so E_5=48.

Finally, Corollary 12 and Parseval give \sum_{\chi\ {\rm primitive\ odd}}|L(1,\chi)|^4 = \frac{4\pi^4}{625}E_5. Substitution of E_5=48 finishes the proof. ◻