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Use the usual positive-definite convention for binary quadratic forms with negative discriminant. Reduction by integral changes of variables of determinant one produces a form $a{x}^2+bxy+c{y}^2$ with $|b|\le a\le c$, and with $b\ge 0$ on the boundary. For completeness, replacing $x$ by $x+ny$ makes $|b|\le a$; if then $c<a$, interchange the variables, reversing one sign to keep determinant one. The new positive leading coefficient is smaller. Repeating terminates and gives a reduced form.

For discriminant $-7$, $4ac-b^2=7$ and the reduction inequalities give $3a^2\le 7$. Thus $a=1$, $b$ is odd and $|b|\le 1$, and $c=2$. The boundary convention chooses $b=1$. Hence there is exactly one positive-definite equivalence class, represented by $x^2+xy+2y^2$.

For an odd prime $p\ne 7$ represented by this form, the identity $4p=(2x+y{)}^2+7y^2$ shows that $-7$ is a square modulo $p$. Here $p\nmid y$, because otherwise $p\mid x$ and $p^2$ would divide the represented number. Conversely, if $-7$ is a square modulo $p$, choose odd $b$ with $b^2\equiv -7(\mathrm{mod}p)$. Then $(p,b,(b^2+7)/(4p))$ is an integral positive-definite form of discriminant $-7$. By the preceding reduction it is equivalent to the given form, so its value $p$ at $(1,0)$ is represented by that form. Quadratic reciprocity gives $(-7/p)=(p/7)$, whose value is one precisely for $p\equiv 1,2,4(\mathrm{mod}7)$. The exceptional primes are represented: $2=Q(0,1)$ and $7=Q(1,-2)$. ThereforeWithout the positive-definite convention, the first printed assertion is false: $-x^2-xy-2y^2$ also has discriminant $-7$ but cannot be equivalent to a positive-definite form, because an invertible change of variables preserves its sign.

For the odd prime $p$, the Fermat-Euler theorem gives $x^{p-1}\equiv 1(\mathrm{mod}p)$. Therefore the residue class $z=x^{(p-1)/2}$ obeys $(z-1)(z+1)\equiv 0(\mathrm{mod}p)$. Since a prime number dividing a product divides one of its factors,If $x$ is a quadratic residue, choose $y$ with $y^2\equiv x(\mathrm{mod}p)$. Then $y$ is nonzero modulo $p$, and Fermat's little theorem givesThis proves the quadratic residue direction of Euler's criterion.

For coefficients $b_n$ supported on an interval of length $H$, write $B=\sum b_n$ and $B_p(a)={\sum }_{n\equiv a(\mathrm{mod}p)}{b}_n$. The variance form of the large sieve statesThe constant is absolute. One may take the explicit right side $(Q^2+2\pi H)\sum |b_n{|}^2$, by orthogonality of roots of unity and the exponential-sum large sieve proved in Question 2.

Take $H=\lfloor N^2\rfloor$, $Q=N$, and $b_n$ the indicator function of the $N^{\epsilon }$-smooth numbers up to $N^2$. Applying the given smooth-number density with parameter $\epsilon /2$ giveswith a harmless adjustment of the constant for integer endpoints. If an odd prime $p\le N$ has least quadratic nonresidue $n(p)>N^{\epsilon }$, then $p>N^{\epsilon }$: a quadratic nonresidue always occurs among $1,\dots ,p-1$. Every prime factor of every selected smooth number is thus a nonzero quadratic residue modulo $p$. By the multiplicativity of the Legendre symbol, every selected number is a nonzero quadratic residue modulo $p$.

There are $(p-1)/2$ nonzero quadratic nonresidue classes, and $B_p(a)=0$ on all of them. Their contribution to the variance is at leastIf $R$ denotes the number of exceptional primes, the variance form of the large sieve, with $\sum |b_n{|}^2=B$, yields $R{B}^2\ll N^{2}B$. ConsequentlyThis is the bounded exceptional primes for least quadratic nonresidues argument. Using an interval of length $N^2$ is what matches the $Q^2$ term; an interval of length $N$ would not give a bounded exceptional set.

Take sieve level $D=x^{1/3}$ and coefficient cutoff $L=\sqrt{D}=x^{1/6}$. Let $\mathcal{P}$ be the finite set of odd primes at most $L$. Give $n=p+2$ one unit of weight for each prime $p\le x$. For an odd squarefree integer $d$ on these primes, the sieve distribution isThe residue $-2$ is coprime to every such $d$, and $g(p)=1/(p-1)\in (0,1)$ for all permitted primes. Values of $g$ outside the permitted set can be chosen arbitrarily in $(0,1)$ at primes and extended multiplicatively on squarefree integers.

For the Selberg sieve normalizing sum, $k(p)=g(p)/(1-g(p))=1/(p-2)$ andHere is a direct proof of this lower bound. Restrict the Euler product to $3\le p\le y=L^{1/4}$. The Mertens first theorem gives ${\sum }_{3\le p\le y}\log p/(p-1)=\log y+O(1)$, which is at most $\frac{1}{2}\log L$ for large $L$. The Mertens second theorem givesThe truncated Euler-product lower bound, with $\theta =1/2$, now gives $J\ge W/2\gg \log L$.

Use the minimizing Selberg sieve weights from Question 3. Their absolute values are at most one, so the Selberg least-common-multiple weights satisfy $|{\lambda }_d|\le 3^{\omega (d)}$. Parts (a) and (b) control the remainder:The first factor uses the Bombieri–Vinogradov theorem; $D=x^{1/3}$ is below $x^{1/2}/{\log }^{B}x$ for all sufficiently large $x$. The second uses the permitted ${\sum }_{q\le D}{9}^{\omega (q)}/\varphi (q)\ll {\log }^{9}D$. The PDF has the exponent nine, which the TeX transcribes incorrectly. Choose $A=14$ to obtain $O(x/{\log }^{3}x)$.

Every twin prime pair with $p+2>L$ survives the finite sieve, while the pairs with $p+2\le L$ number at most $L$. The Selberg upper-bound sieve thus givesTherefore the twin-prime count is $O(x/{\log }^{2}x)$. The finite cutoff on the forbidden primes is essential: sieving by all odd primes would also discard the large prime values $p+2$ that we are trying to count.

A pointed category has a zero object, both initial and terminal. Factoring through that object gives a zero morphism $0_{A,B}$ between any two objects. In this paper a semi-additive structure means commutative-monoid enrichment: every hom-set is a commutative monoid with additive zero, and composition distributes over addition in both variables. No existence of finite products in a category is included in this last definition. This convention matters for the final one-object example; under the stronger convention requiring finite biproducts, that example would not be a semi-additive category.

Let $c_{A,B}:A+B\to A\times B$ be the given canonical isomorphism. Transfer the coproduct injections across $c_{A,B}$, so $A\times B$ is also a coproduct in a category, with injectionsWrite ${\nabla }_B:B\times B\to B$ for the morphism whose composites with $i_1,i_2$ are both $1_B$. For $f,g:A\to B$, define the biproduct-induced addition of morphisms byThe zero is the existing zero morphism. We verify the laws from the universal properties, without assuming addition in advance.

All finite canonical maps from coproducts to products are isomorphisms, by induction from the binary ones and the zero object. Thus both $(B\times B)\times B$ and $B\times (B\times B)$ are ternary coproducts in a category. The two iterated fold maps to $B$ agree on each of the three injections, hence are equal. Applying this to $⟨f,g,h⟩$ gives associativity. The interchange $B\times B\to B\times B$ swaps the two injections, so its composite with ${\nabla }_B$ is ${\nabla }_B$, giving commutativity. Finally $⟨f,0⟩=i_{1}f$, so $f+0=f$, and similarly $0+f=f$.

Precomposition is additive because $⟨f,g⟩u=⟨fu,gu⟩$. For $v:B\to C$, the morphisms $v{\nabla }_B$ and ${\nabla }_C(v\times v)$ agree on both injections, so they agree; consequently $v(f+g)=vf+vg$. Composition with a zero morphism is zero. This proves the commutative-monoid enrichment.

It is unique. In any such enrichment, the additive zero morphisms agree with the pointed ones, since each map to or from the zero object belongs to a singleton hom-set. The projections of $i_1{p}_1+i_2{p}_2$ from $A\times B$ are respectively $p_1,p_2$ by bilinearity, soFor $A=B$, composing on the left by ${\nabla }_B$ and on the right by $⟨f,g⟩$ forces the displayed formula for $f+g$. Thus the biproducts determine exactly one such enrichment.

For the last part, the underlying one-object category has endomorphism monoid $(\mathbb{N},\cdot )$, including $0$, and identity $1$. Its usual addition gives a commutative-monoid enrichment. Any permutation $\sigma$ of the prime numbers extends, by unique prime factorization, to a multiplicative monoid automorphism ${\varphi }_{\sigma }:\mathbb{N}\to \mathbb{N}$, fixing $0,1$. Transport addition byThis is a commutative monoid operation with zero $0$, and multiplication distributes over it: applying ${\varphi }_{\sigma }$ reduces each distributive law to the ordinary one in $\mathbb{N}$. Thus each operation supplies a semi-additive structure on the same fixed composition law.

For each odd prime $p$, take $\sigma$ to interchange $2$ and $p$, fixing every other prime. ThenDifferent choices of $p$ give different additions, and there are infinitely many prime numbers. Hence there are infinitely many distinct semi-additive structures, even though these transported structures are isomorphic as enriched categories. The underlying one-object category has no terminal object, because its endomorphism set is not a singleton, so it indeed has no finite products in a category.

Euler's criterion givesIf either factor is divisible by $p$, both asserted values are zero. Otherwise the two values lie in $\{1,-1\}$, whose elements are distinct modulo an odd prime. The modular congruence is therefore equality:Finally, $a\equiv b(\mathrm{mod}p)$ means they give the same residue class and thus the same set of square-congruence solutions. Divisibility by $p$ is also identical. Hence $\left(\frac{a}{p}\right)=\left(\frac{b}{p}\right)$.

For an integer $a$ and an odd prime $p$, the Legendre symbol isThe value one refers to a nonzero quadratic residue; divisibility by $p$ is recorded separately by zero.

The quadratic reciprocity states that for distinct odd primes $p,q$,With $q=3$, it gives $\left(\frac{3}{p}\right)=(-1{)}^{(p-1)/2}\left(\frac{p}{3}\right)$. The nonzero square modulo three is one, so $\left(\frac{p}{3}\right)=1$ when $p\equiv 1(\mathrm{mod}3)$, and $-1$ when $p\equiv -1(\mathrm{mod}3)$. An odd prime other than three is congruent to $1,5,7$, or $11$ modulo twelve. Evaluating the two signs givesThus

For an odd prime $p$, the Legendre symbol is zero if $p\mid a$, one if $a$ is a nonzero quadratic residue modulo $p$, and minus one if it is a quadratic nonresidue. ThusIt depends only on the congruence class of $a$ modulo $p$.

For distinct odd primes $p,q$, the quadratic reciprocity law isThus the two symbols have opposite signs exactly when both primes are three modulo four; otherwise they agree. Distinctness excludes zero symbols.

For an odd prime $p$, the Legendre symbol is zero if $p\mid a$, one if $a$ is a nonzero quadratic residue modulo $p$, and minus one if it is a quadratic nonresidue. ThusIt depends only on the congruence class of $a$ modulo $p$.

For distinct odd primes $p,q$, the quadratic reciprocity law isThus the two symbols have opposite signs exactly when both primes are three modulo four; otherwise they agree. Distinctness excludes zero symbols.

Euler's criterion states, for an odd prime $p$,Apply it to $a=-1$. The integer $(-1{)}^{(p-1)/2}$ is $1$ when $p\equiv 1(\mathrm{mod}4)$ and $-1$ when $p\equiv 3(\mathrm{mod}4)$. Since those signs are distinct modulo an odd prime, the modular congruence determines the actual Legendre symbol:This is the first supplementary law for quadratic reciprocity.

For an odd prime $p$, the Legendre symbol isIt depends only on the congruence class of the integer $a$. The values $1$ and $-1$ distinguish a nonzero quadratic residue from a quadratic nonresidue; zero is a separate case.

For distinct odd primes $p,q$, quadratic reciprocity statesHere the Legendre symbol records whether a nonzero congruence class is a quadratic residue. The first supplementary law for quadratic reciprocity and second supplementary law for quadratic reciprocity giveThus minus one is a square exactly when $p\equiv 1(\mathrm{mod}4)$, and two is a square exactly when $p\equiv 1$ or $7(\mathrm{mod}8)$.

To test the specified congruence class, factor $111=3\cdot 37$. Since $2017\equiv 1(\mathrm{mod}4)$, multiplicativity of the Legendre symbol and quadratic reciprocity yieldThere is no sign change on swapping $19$ and $37$ because $37\equiv 1(\mathrm{mod}4)$. Finally $18=2\cdot 3^2$, so multiplicativity of the Legendre symbol gives $(18/19)=(2/19)=-1$, since $19\equiv 3(\mathrm{mod}8)$. Therefore

For an odd prime $p$, Euler's criterion statesFor $p\nmid a$, the Legendre symbol is $1$ for a quadratic residue and $-1$ for a quadratic nonresidue; if $p\mid a$, both sides are zero.

The product is a single power because the arithmetic progression of exponents has sum $2017\cdot 2018/2$:Since $2017\equiv 1(\mathrm{mod}8)$, the second supplementary law for quadratic reciprocity gives $(2/2017)=1$. Hence Euler's criterion gives $2^{1008}\equiv 1(\mathrm{mod}2017)$. Reduce the exponent using $1009=1008+1$ and $2017=2\cdot 1008+1$:The least nonnegative remainder is therefore $2$.

If $g$ is a primitive root modulo an odd prime $p$, thenis the exact condition for $g$ to remain a primitive root modulo every power $p^m$. Its multiplicative order modulo $p^2$ must be a multiple of $p-1$ and a divisor of $p(p-1)$, so the condition makes that order $p(p-1)$. The result then follows from lifting a primitive root to odd prime powers.

For example, two is primitive modulo three and $2^2=4\not\equiv 1(\mathrm{mod}9)$. It is therefore primitive modulo every $3^m$. The assumption that $p$ is odd is essential; the unit groups of higher powers of two are not cyclic groups.

With the one-element unit group convention at $n=1$, a positive integer modulus admits a primitive root exactly whenwhere $p$ is an odd prime and $a\ge 1$.

For existence, the finite subgroup of a field multiplicative group is cyclic theorem supplies a primitive root $g$ modulo $p$. At least one of $g$ and $g+p$ satisfies the primitive-root lifting criterion: if $g^{p-1}\equiv 1(\mathrm{mod}{p}^2)$, thenThus lifting a primitive root to odd prime powers gives every $p^a$. The Chinese remainder theorem for unit groups identifies $U(2p^a)$ with $U(p^a)$ because $U(2)$ is trivial. The moduli two and four are checked directly.

For exclusion, a cyclic group has at most two solutions of $x^2=1$. If a modulus has two distinct odd prime factors, or is divisible by both four and an odd prime number, independent sign choices under the Chinese remainder theorem give at least four such units. A pure power $2^a$ with $a\ge 3$ also has the four distinct roots $\pm 1$ and $\pm (1+2^{a-1})$. These cover every omitted case.

When a primitive root exists, their number is $\phi (\phi (n))$. If $g$ generates the unit group of size $m=\phi (n)$, its power $g^j$ has multiplicative order $m/\gcd (j,m)$, so exactly the coprime exponent classes generate. Here $\phi$ is Euler's totient function.

For an odd prime $p$, put $\omega =e^{2\pi i/p}$ and let $\chi$ be the Legendre symbol. The quadratic Gauss sum isWith this choice of $\omega$ and the positive real square root of $p$, its exact value isThe modulus can be established without determining the sign. Expanding $G(\chi )\overline{G(\chi )}$ and putting $a=tb$ givesThe inner sum is $p-1$ for $t=1$ and $-1$ otherwise, while ${\sum }_{t=1}^{p-1}\chi (t)=0$ because the nonzero quadratic residues and quadratic nonresidues are equally numerous. Also, complex conjugation followed by $a\mapsto -a$ gives $\overline{G(\chi )}=\chi (-1)G(\chi )$. Hence the sum is real for $p\equiv 1(\mathrm{mod}4)$ and purely imaginary for $p\equiv 3(\mathrm{mod}4)$. These two observations determine its magnitude and its real or imaginary axis; choosing the sign in the exact evaluation is an additional step.

For every odd prime $p$, the Legendre symbol correlation satisfiesThe small negative value measures an exact dependence between neighboring square tests; it is not obtained by treating them as independent.

Count the pairs $(x,v)\in {\mathbb{F}}_p^2$ satisfying $v^2=x(x+1)$. For each $x$, their number is $1+(x(x+1)/p)$: it is one for zero, two for a nonzero quadratic residue and zero for a quadratic nonresidue. The total is therefore $p$ plus the desired sum.

Set $u=2x+1$ and $w=2v$. Since $2$ is invertible in this finite field, this is a bijection onto the solutions ofChoose any $r=u-w\ne 0$. Then $u+w=r^{-1}$, uniquely determining $u=(r+r^{-1})/2$ and $w=(r^{-1}-r)/2$. There are exactly $p-1$ choices of $r$. Hence $p+{\sum }_x(x(x+1)/p)=p-1$, proving the correlation identity.