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The Von Mangoldt function is supported on prime powers:Write $e(t)=\exp (2\pi it)$. A convenient form of the Siegel–Walfisz theorem is: for every fixed $A,B>0$, uniformly for $q\le (\log x{)}^B$ and $(a,q)=1$,The implied constant can be ineffective because of possible Siegel zeros; the modulus bound is logarithmic, not an unrestricted uniformity assertion.

Apply this to modulus five and separate the reduced residue classes:The last sum is $O(\log N)$: only powers of five can contribute in the zero residue class. By the orthogonality of roots of unity,since two is coprime to five and the sum including $a=0$ vanishes. As $\phi (5)=4$, the Siegel–Walfisz theorem yieldsThe sum is complex-valued, but the displayed error controls its absolute difference from the real main term.

Put $t=\sqrt{-30}$ and $K=\mathbb{Q}(t)$. The ring of integers is $\mathbb{Z}[t]$, and the fundamental discriminant is $D_K=-120$. For a reduced positive definite binary quadratic form $(a,b,c)$ of discriminant $-120$, reduction givesThe coefficient $b$ is even; checking $1\le a\le 6$ and $|b|\le a$, with the usual nonnegative sign choice at boundary equalities, leaves preciselyThe correspondence between proper quadratic forms and ideal classes therefore gives $h_K=4$. All four classes equal their inverses, since inversion replaces $b$ by $-b$. HenceThe three nontrivial classes may be represented by the ramified prime ideals ${\mathfrak{p}}_2=(2,t)$, ${\mathfrak{p}}_3=(3,t)$ and ${\mathfrak{p}}_5=(5,t)$. They square to principal ideals and satisfy ${\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_5=(t)$. None is principal: an integral generator of norm $2$, $3$ or $5$ would require $a^2+30b^2$ to equal that number, which is impossible. The relation then makes the three classes distinct, consistently with the four-form calculation.

ConsiderThe three rational square classes are independent, so $[H:\mathbb{Q}]=8$. Their radical product is $\sqrt{-30}$, so $K\subset H$ and $[H:K]=4$; the extension $H/K$ is abelian. To verify that it is unramified, rather than assuming that radical adjunctions are harmless at $2,3,5$, use the conductor-discriminant formula. The seven quadratic subfields have fundamental discriminantsThe formula givesThe discriminant tower identity now givesThus the relative discriminant is the unit ideal and no finite prime ramifies. The base has no real places, so there is no infinite ramification to check. This unramified abelian degree-four extension has degree $h_K$, and part (i) provesThis is the Hilbert class field of Q of square root minus thirty.

Next translate prime representation into an ideal class. For $p\nmid 30$, we claimIf a representation exists, $\alpha =2x+yt$ lies in ${\mathfrak{p}}_2$ and has $N_{K/\mathbb{Q}}\alpha =2p$. The integral ideal $(\alpha ){\mathfrak{p}}_2^{-1}$ has norm $p$, so it is a prime ideal $\mathfrak{P}$. Its class is $[{\mathfrak{p}}_2{]}^{-1}=[{\mathfrak{p}}_2]$. Conversely, if $[\mathfrak{P}]=[{\mathfrak{p}}_2]$, the product ${\mathfrak{p}}_2\mathfrak{P}$ is principal, say $(\alpha )$. Since this ideal lies in ${\mathfrak{p}}_2$, its generator has the form $\alpha =2x+yt$ with integers $x,y$. Taking norms gives $4x^2+30y^2=2p$, proving the reverse implication. This is prime representation via an ideal class, and it proves sufficiency as well as necessity.

Identify the required class in $\mathrm{Gal}(H/K)$ through the Artin map. We can use the explicit representation $17=2+15$: $2+t$ has norm $34$, so the prime above $17$ obtained from $(2+t){\mathfrak{p}}_2^{-1}$ has class $[{\mathfrak{p}}_2]$. Its Frobenius in the multiquadratic extension acts on the three radicals with signsTheir product is one, so this automorphism fixes $K$. Because $H/\mathbb{Q}$ is abelian, for any unramified rational prime $p$ that splits in $K$, the Frobenius at either prime of $K$ of norm $p$ is its rational Frobenius restricted to $H/K$. The Artin isomorphism therefore makes the required class condition equivalent toConversely, these signs already imply splitting in $K$, since their product is $(\frac{-30}{p})=1$. No extra splitting condition is missing.

The supplementary law for $2$, quadratic reciprocity for $-3$ and $5$, and then the Chinese remainder theorem turn these sign conditions intoTheir four simultaneous residue classes modulo $120$ are $17,23,47,113$. Finally, $2$ is represented by $(x,y)=(\pm 1,0)$, whereas $3$ and $5$ cannot be represented: $y\ne 0$ gives a value at least $15$, and $y=0$ gives twice a square. We have proved the complete criterion

Fix integers $q\ge 1$ and $a$ with $(a,q)=1$. We prove the Dirichlet theorem on primes in arithmetic progressions by showing that the prime Dirichlet sum in this residue class diverges as its real exponent decreases to one.

First justify the required Dirichlet L-function facts. If $\chi$ is a nonprincipal Dirichlet character modulo $q$, its sum over a full period is zero: choose a unit $b$ with $\chi (b)\ne 1$, and multiplication by $b$ permutes the residue classes, forcing the sum to equal $\chi (b)$ times itself. Therefore the partial sums $C_{\chi }(u)={\sum }_{n\le u}\chi (n)$ are bounded by $q$. Partial summation giveswhich supplies holomorphic continuation near one, with locally uniform convergence. For the principal Dirichlet character ${\chi }_0$,has a simple pole at one, with residue $\phi (q)/q$.

The assumption in the question gives nonvanishing at one for real nonprincipal characters. We must also prove it for nonreal characters. For a nonreal $\chi$, its square ${\chi }^2$ is nonprincipal. For real $\sigma >1$, logarithms of the Euler products giveFor a prime not dividing $q$, its logarithmic coefficient is $3+4\mathrm{Re}\chi (p{)}^k+\mathrm{Re}\chi (p{)}^{2k}=2(1+\cos \theta {)}^2\ge 0$, where $\chi (p{)}^k=e^{i\theta }$. Primes dividing $q$ contribute only the positive zeta term. If $L(1,\chi )$ vanished to order $m\ge 1$, the product would be $O((\sigma -1{)}^{4m-3})$ and tend to zero: the squared-character factor is bounded because ${\chi }^2$ is nonprincipal. This is a contradiction. HenceThis step uses the given real-character hypothesis and the Euler product positivity for L-function nonvanishing for all remaining characters.

In $\sigma >1$ choose the logarithm furnished by the absolutely convergent Euler product. Its prime-power expansion givesThe error from powers $k\ge 2$ is uniformly bounded as $\sigma \downarrow 1$, by ${\sum }_p{\sum }_{k\ge 2}1/(k{p}^{k\sigma })<\infty$. For nonprincipal $\chi$, nonvanishing and holomorphic continuation at one provide a bounded logarithm in a neighborhood of one. The Euler-product branch differs from that branch by a constant integral multiple of $2\pi i$ on the final real interval, so it too is bounded. ThusFor the principal character, its simple pole instead gives

The Orthogonality of Dirichlet characters gives, for primes not dividing $q$,Multiply by $p^{-\sigma }$ and sum. Absolute convergence allows the interchange for $\sigma >1$, soThe right side tends to infinity. If there were only finitely many primes in the progression, the left side would have a finite limit at one. This contradiction proves infinitely many primes occur in every reduced residue class. For $q=1$ or $2$ the character argument simply has no nonprincipal terms and gives the same conclusion.

For the final deduction the modulus $q$ is fixed. Let $A(t)=\pi (t;q,a)$. The given consequence of the Siegel–Walfisz theorem is $A(t)\sim t/(\phi (q)\log t)$. Partial summation, including the prime $2$ through the lower endpoint $2^{-}$ when appropriate, givesThe boundary is $O_q(1/\log x)$. Write $A(t)=t(1+\epsilon (t))/(\phi (q)\log t)$ with $\epsilon (t)\to 0$. The main integral is $(\log \log x-\log \log 2)/\phi (q)$. The error is $o(\log \log x)$: given $\eta >0$, choose $T$ such that $|\epsilon (t)|\le \eta$ for $t\ge T$. Its integral from $2$ to $T$ is a fixed constant, and its absolute value from $T$ to $x$ is at most $\eta \log \log x/\phi (q)$ plus a fixed constant. Divide by $\log \log x$ and then let $\eta \downarrow 0$. We obtain the reciprocal primes in a fixed arithmetic progression asymptoticThe argument uses the supplied prime-counting asymptotic in this last part; the preceding infinitude proof did not assume the Prime number theorem in arithmetic progressions.

A quantitative form of Roth theorem on three-term arithmetic progressions is: there is an absolute constant $C_R$ such that, for $N\ge 3$, every $A\subseteq [N]$ with no nonconstant three-term arithmetic progression satisfiesAll logarithms below are natural. We prove the classical Roth bound for three-term progressions with an unoptimized $C_R=100$. The empty set is immediate; write $\delta =|A|/N>0$.

First establish a quantitative density increment. Suppose $N\ge 4\cdot 10^6{\delta }^{-4}$. Put $M=2N+1$, embed $I=[N]$ in $G={\mathbb{Z}}_M$, and extend all interval functions by zero. Let $a=1_A$, $u=\delta 1_I$ and $f=a-u$, the balanced subset indicator. DefineThe relation in the sum is modulo $M$. For interval elements its integer discrepancy has absolute value less than $M$, so it is precisely the integer relation. Thus $\Lambda (a,a,a)=|A|=\delta N$. The full interval count is the number of endpoint pairs of the same parity:Consequently, since $N\ge 4{\delta }^{-2}$,

Orthogonality of the additive characters givesBoth identities follow by expanding the sums and using ${\sum }_r{e}^{2\pi irt/M}=M$ for $t=0$ in $G$, and zero otherwise; the latter follows from a finite geometric series. Since $M$ is odd, $r\mapsto -2r$ permutes the frequencies. Telescope the difference asThe Cauchy-Schwarz inequality and the displayed Parseval identity on a finite group bound each term by $\delta N{\max }_r|\hat{f}(r)|$. Indeed, $\Vert a{\Vert }_2^2=\delta N$ and $\Vert u{\Vert }_2^2={\delta }^{2}N$, so every product of the two remaining L2 norms is at most $\delta N$. HenceThe zero coefficient vanishes because $\sum f=0$, so choose a nonzero frequency $\theta =r/M$ giving this bound.

Let $Q=\lfloor \sqrt{N}\rfloor$. The pigeonhole principle applied to the $Q+1$ fractional parts $0,\theta ,\dots ,Q\theta$ in $Q$ intervals gives an integer $1\le q\le Q$ with $\Vert q\theta {\Vert }_{\mathbb{R}/\mathbb{Z}}\le 1/Q$. Partition each residue class modulo $q$ in $[N]$ into consecutive arithmetic progressions of exactlyelements, leaving fewer than $L$ remainder points per residue. The total remainder $R_0$ has size at most $qL\le {\delta }^{2}N/1000$. The size hypothesis gives $L\ge {\delta }^2\sqrt{N}/2000$. On a full arithmetic progression $P$, the phase $e^{-2\pi i\theta x}$ varies from its value at the first point by at most $2\pi L/Q\le 4\pi {\delta }^2/1000$.

Let $F_P={\sum }_{x\in P}f(x)$. Replacing the phase by its first value on each full arithmetic progression and estimating the remainder using $|f|\le 1$ givesAlso $|{\sum }_P{F}_P|=|{\sum }_{R_0}f|\le {\delta }^{2}N/1000$. Therefore the sum of the positive $F_P$ isSince the total length of these arithmetic progressions is at most $N$, at least one has $F_P/|P|\ge {\delta }^2/64$. Thus the step just proved isAn affine map taking this arithmetic progression to $[|P|]$ preserves the absence of nonconstant three-term arithmetic progressions.

Now iterate. Write $N_j,{\delta }_j$ for successive lengths and subset densities, with $N_0=N$ and ${\delta }_0=\delta$. Since every subset density is at most one,Consequently $T=\lceil 65/\delta \rceil$ successful steps are impossible. On the other hand, subset densities never decrease, so with $B=\log (2000/{\delta }^2)$,Suppose $\delta \log \log N>100$. Since $T\le 66/\delta$,Let $D=\log (4\cdot 10^6{\delta }^{-4})$. We have $2B+D<31+8\log (1/\delta )\le 39/\delta <e^{54/\delta }$. Therefore every length through step $T$ satisfies $\log N_j>D$, and hence $N_j\ge 4\cdot 10^6{\delta }^{-4}\ge 4\cdot 10^6{\delta }_j^{-4}$. The density increment lemma remains applicable for all $T$ steps, contradicting the reciprocal-density bound.

Thus $\delta \log \log N\le 100$, proving the boxed theorem. The proof includes the Fourier analysis on a finite abelian group detection, conversion to an integer arithmetic progression, and the length bookkeeping that produces the double logarithm.

Let $d=\gcd (a,m)$. If the linear congruence $ar\equiv b(\mathrm{mod}m)$ holds, then $b=ar-ms$ for some integer $s$, so the greatest common divisor $d$ divides $b$. Conversely, Bézout's identity gives integers $u,v$ with $au+mv=d$. If $b=dj$, multiply this identity by $j$ to obtain $a(uj)\equiv b(\mathrm{mod}m)$. HenceFor the number of solutions, write $a=d{a}^{\prime }$, $m=d{m}^{\prime }$ and $b=d{b}^{\prime }$. The linear congruence becomes $a^{\prime }r\equiv b^{\prime }(\mathrm{mod}{m}^{\prime })$, where $a^{\prime },m^{\prime }$ are coprime integers. Multiplication by $a^{\prime }$ is invertible modulo $m^{\prime }$ by Bézout's identity, so the solutions form one residue class $r\equiv r_0(\mathrm{mod}{m}^{\prime })$. Modulo $m=d{m}^{\prime }$, these are exactlyThey are distinct: if two have the same residue class modulo $m$, then $d{m}^{\prime }$ divides $(j-\ell )m^{\prime }$, forcing $d\mid j-\ell$; with $0\le j,\ell <d$, this forces $j=\ell$. Every other solution differs from $r_0$ by a multiple of $m^{\prime }$, so these exhaust the $d$ solutions.

Finally,Combining this with the existence criterion gives the requested equivalent modular congruence.

Take a generator $u$ of the multiplicative group of a finite field ${\mathbb{F}}_p^{\times }$, which has order $p-1$. Write $x=u^a$. Since any possible nonzero base is $y=u^r$, the condition of being a $k$th power isBy part (a), this linear congruence is soluble exactly when $d=\gcd (k,p-1)$ divides $a$. MeanwhileThus the required test isThe exponent residue classes divisible by $d$ are $0,d,\dots ,p-1-d$, so there are exactly $(p-1)/d$ distinct nonzero $k$th powers. Equivalently, part (a) shows each value has exactly $d$ preimages under the power map.

The greatest common divisor issince $19$ and $44$ are coprime integers. Part (b) therefore gives $\boxed{1012/23=44}$ distinct $437$th powers in the nonzero multiplicative group of a finite field ${\mathbb{F}}_{1013}^{\times }$. If “powers mod $1013$” includes all residue classes, the additional value $0=0^{437}$ gives $\boxed{45}$ values including zero. The distinction is relevant because part (b) explicitly works in the nonzero group.

A Hausdorff, noncompact topology. The condition says precisely that each point of $U$ lies in a residue class $n+k\mathbb{Z}$ contained in $U$. These classes form a basis: the intersection of two classes containing a common point contains the class through that point with modulus equal to the least common multiple of their moduli. Thus arbitrary unions obey the condition, and so do finite intersections. Also $\varnothing$ and $\mathbb{Z}$ obey it. This is the profinite topology on $\mathbb{Z}$.

If $a\ne b$, choose $k>|a-b|$. Then $a+k\mathbb{Z}$ and $b+k\mathbb{Z}$ are disjoint open neighbourhoods, proving the Hausdorff property. To test compactness, defineEvery $F_n$ is nonempty and both open and closed: its complement is a finite union of other residue classes. Since $a_{n+1}-a_n=4^n$, $F_{n+1}\subseteq F_n$. If $x$ belonged to their intersection, then $3x+1$ would be divisible by $4^n$ for every $n$, forcing $3x+1=0$, impossible for an integer. Therefore $U_n=\mathbb{Z}\setminus F_n$ is an increasing open cover of $\mathbb{Z}$. Any finite subcollection has union $U_N$ for its largest index and misses the nonempty $F_N$. This proves that the profinite topology on the integers is not compact.

Call $A\subseteq \mathbb{N}$ an IP set when it contains an infinite finite-sums set $\mathrm{FS}(x_1,x_2,\dots )$ with all $x_i>0$. We first need the relative consequence of the Hindman theorem: a finite union can be an IP set only if one of its constituents is an IP set. This does not follow by simply applying the theorem to a coloring of all integers, because its monochromatic finite-sums set might lie outside the given set.

Here is a reduction that proves the partition regularity of IP sets. Suppose $\mathrm{FS}(x_i)\subseteq A$ and $A$ is finitely colored. Index the generators by $i\ge 0$. For $n>0$, let $S(n)$ be the set of positions of the nonzero digits in its binary expansion, and color $n$ by the color of ${\sum }_{i\in S(n)}{x}_i$. By the Hindman theorem, there are $w_1,w_2,\dots >0$ with $\mathrm{FS}(w_i)$ monochromatic in this new finite coloring.

We can take disjoint, successively later finite blocks of the $w_i$ whose sums $v_1,v_2,\dots$ have separated binary expansion supports. Start with $v_1=w_1$. Having chosen finitely many blocks, take $K$ larger than every binary position used in their sums. On a fresh tail of the $w_i$, consider $2^K+1$ partial sums, including the zero partial sum. Two have the same residue class modulo $2^K$, by the pigeonhole principle. Their positive difference is a consecutive block sum $v_{j+1}$ divisible by $2^K$. Discard all indices up to that block's end and repeat. Thus every bit of $v_{j+1}$ is above every bit in the preceding block sums.

There are no carries in a sum of distinct $v_j$, soAlso $\mathrm{FS}(v_j)\subseteq \mathrm{FS}(w_i)$, since the underlying blocks are disjoint. Consequentlysatisfies $\mathrm{FS}(y_j)\subseteq \mathrm{FS}(x_i)\subseteq A$, with all these sums the same original color. This proves the relative assertion. In particular, if $A_1\cup \cdots \cup A_r$ contains an infinite finite-sums set, color its points by the first constituent containing them and apply this argument; one $A_j$ is an IP set.

Let $\mathcal{I}$ be the family of sets that are not IP sets. It is closed under subsets and, by the preceding argument, under finite unions. It contains every finite set, since positive successive partial sums of an infinite sequence are unbounded. It does not contain $\mathbb{N}$. Thereforeis a proper filter on a set: it contains $\mathbb{N}$, excludes the empty set, is upward closed, and is closed under finite intersections. Its members are exactly the IP-star sets, the sets meeting every IP set. Indeed, failure to meet an IP set is equivalent to having an IP complement. This identifies the sets that must belong to the required ultrafilter.

Apply the ultrafilter lemma to extend the IP-star filter $\mathcal{F}$ to an ultrafilter $\mathcal{U}$. For completeness, Zorn lemma applies to its proper filter extensions ordered by inclusion, because the union of a chain is still a proper filter on a set. A maximal extension decides every set $A$: if $A$ cannot be adjoined while keeping the filter proper, some filter member is disjoint from $A$, forcing $A^c$ into the filter. This is the defining ultrafilter property.

If $A\in \mathcal{U}$ were not an IP set, then $A^c\in \mathcal{F}\subseteq \mathcal{U}$, contradicting properness. Hence every member of $\mathcal{U}$ contains an infinite finite-sums set. Conversely, any ultrafilter with finite-sums members must contain every IP-star set, since it cannot contain such a set's non-IP complement. Thus this construction captures precisely the forced filter, without making an unnecessary claim about idempotence.

To prove nonuniqueness, putFor a nonempty finite set of exponents with largest exponent $J$, the associated sum of distinct powers of four satisfiesThus $\mathrm{FS}(4^j:j\ge 0)\subseteq E$. Doubling the inequality givesso $\mathrm{FS}(2\cdot 4^j:j\ge 0)\subseteq E^c$. These alternating dyadic intervals contain disjoint IP sets.

Every member of $\mathcal{F}$ meets $E$ and also $E^c$, because it meets every IP set. A finite intersection of members of $\mathcal{F}$ is again a member, so both $\mathcal{F}\cup \{E\}$ and $\mathcal{F}\cup \{E^c\}$ have the finite intersection property. Extend the proper filters they generate to ultrafilters ${\mathcal{U}}_0$ and ${\mathcal{U}}_1$. Each extends $\mathcal{F}$, so the previous argument makes each an ultrafilter with finite-sums members. But $E\in {\mathcal{U}}_0$ and $E^c\in {\mathcal{U}}_1$, so

Again use integer coefficients after clearing denominators and put $S={\sum }_i{a}_i$. If $S=0$, choose any odd positive integer $u$ and take all $x_i=u$. This gives a monochromatic solution in any colouring.

For necessity, suppose $S\ne 0$ and choose $K$ with $2^K>|S|$. Colour the odd positive integers by their residue classes modulo $2^K$. In any monochromatic candidate, $x_i\equiv u(\mathrm{mod}{2}^K)$ for a common odd $u$. The equation would imply $uS\equiv 0(\mathrm{mod}{2}^K)$. Since $u$ is coprime to $2^K$, this forces $2^K\mid S$, impossible for nonzero $S$ with $|S|<2^K$. Thus $\boxed{\text{regular over odd positive integers}⟺\sum _i{a}_i=0}$. This is one-equation partition regularity over odd integers; its obstruction concerns the full coefficient sum rather than just a zero-sum subset.

No. The ten residue classes modulo $10$ form a finite partition, so an ultrafilter $p$ contains exactly one class $C_r$. The addition formula shows that $p+p$ contains $C_{2r}$: for $n\in C_r$, the translate $C_{2r}-n$ is $C_r$ and belongs to $p$. If $p$ is an idempotent ultrafilter, uniqueness of its selected class gives $2r\equiv r(\mathrm{mod}10)$, hence $r=0$.

Thus every idempotent ultrafilter contains the multiples of $10$, and cannot also contain their complement. This is the zero-residue constraint for idempotent ultrafilters, valid for every finite modulus.

For a primitive Dirichlet character the previous formula gives $|\hat{\chi }(a)|=|\chi (a)||\hat{\chi }(1)|$. The Plancherel theorem for the unitary finite transform yieldsThus $|\hat{\chi }(1)|=1$, the normalized Gauss sum of a Dirichlet character magnitude.

For an imprimitive character modulo $p^k$ with $k\ge 2$, its values on residues are periodic modulo $p^{k-1}$: descent preserves the values on units, and divisibility by $p$ is unchanged by that shift. Splitting the sum into these residue classes gives a factor ${\sum }_{j=0}^{p-1}e(-j/p)=0$. Hence $\hat{\chi }(1)=0$. If $k=1$, the only imprimitive character is principal, and its Gauss sum is ${\sum }_{n=1}^{p-1}e(-n/p)=-1$, giving $|\hat{\chi }(1)|=p^{-1/2}$. All cases satisfy the required bound.

Let $\chi$ be the quadratic character, equivalently the Legendre symbol. Zero is included among the square residue classes. For every integer $n$ its square-class indicator isFor a nonzero quadratic residue the right side is one, for a nonresidue it is zero, and for a multiple of $q$ it is one. Summing over the interval, the character contribution is $O(\sqrt{q}\log q)$ by the previous part, while the number of multiples of $q$ is $N/q+O(1)$. Thus the required count isThis counts the integers in the interval whose residue classes are squares. When the interval exceeds one period, repeated appearances are counted; the displayed main term could not describe a count of distinct residue classes for arbitrary $N$.

An even Dirichlet character satisfies $\chi (-1)=1$, while an odd Dirichlet character satisfies $\chi (-1)=-1$. Write $a=0$ or $1$ for its character parity and, for $x>0$, define the Dirichlet character theta functionFor a primitive Dirichlet character whose conductor of a Dirichlet character is $q>1$, the term at zero is zero. Put $\tau (\chi )={\sum }_{r\mathrm{mod}q}\chi (r)e(r/q)$, using the positive exponential, and ${\epsilon }_{\chi }=\tau (\chi )/(i^a\sqrt{q})$. The primitive Gauss sum of a Dirichlet character has magnitude $\sqrt{q}$, so $|{\epsilon }_{\chi }|=1$. The theta transformation isThus the powers are $x^{-1/2}$ in the even case and $x^{-3/2}$ in the odd case; the odd root number contains $1/i$. The conjugate character is necessary for a nonreal character. These formulas also follow by applying Poisson summation to the Gaussian function on each residue class, and to its derivative for odd parity. For the primitive principal Dirichlet character whose conductor of a Dirichlet character is one, use the ordinary Jacobi theta function with constant term one; its transformation has root number one.