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The practical reward of quadratic reciprocity is a smaller modulus: swap a pair of distinct odd primes, record the required sign, then reduce the numerator by modular arithmetic. Use multiplicativity of the Legendre symbol and the second supplementary law for quadratic reciprocity to handle a factor of two. For example,Here $11,43$ are both $3$ modulo $4$, so the first swap changes sign; $11\equiv 3(\mathrm{mod}8)$ gives $(2/11)=-1$; the swap involving five has no sign change. An independent square check is $21^2=441\equiv 11(\mathrm{mod}43)$. The method determines existence of a square root without itself finding that root.

A composite numerator is allowed in a Legendre symbol; a composite denominator instead calls for a Jacobi symbol. Its Jacobi reciprocity law supports repeated swaps and remainder reductions, in the spirit of the Euclidean algorithm.^[ref]

A final Jacobi symbol of one is not a square-existence certificate for a composite number modulus. For example,yet $x^2\equiv 2(\mathrm{mod}15)$ is impossible: reducing it modulo three would require a square congruent to two, and squares modulo three are only zero or one. The product has hidden two local failures. By contrast, for a prime denominator, a Legendre symbol of one does certify a quadratic residue.

For a fixed positive odd integer $n$, the Legendre symbol $(n/p)$ for odd prime denominators depends only on $p$ modulo $4n$. For prime numerator $r$, quadratic reciprocity expresses it as $(p/r)$ times a sign determined by $p$ modulo four. Factorization of $n$ and multiplicativity prove the general statement. If a denominator divides $n$, the congruence forces the other prime denominator to be identical, so the zero cases agree too.

For coprime positive odd integers $a,b$, the Jacobi symbol satisfies $(a/b)(b/a)=(-1{)}^{(a-1)(b-1)/4}$. It follows by factoring $a$ and $b$ into primes and multiplying the quadratic reciprocity for each prime pair; the parity of the sums of prime half-differences agrees with $(a-1)/2$ and $(b-1)/2$. A Jacobi symbol equal to one does not imply that its numerator is a quadratic residue for a composite denominator.

A short proof outline connects Gauss's lemma to counting points with integer coordinates. For distinct odd primes $p,q$, define the two floor function sumsThe parity form of Gauss's lemma gives $(q/p)=(-1{)}^M$ and $(p/q)=(-1{)}^N$. In the open rectangle $0<x<p/2$, $0<y<q/2$, the line $y=qx/p$ separates $M$ points below it from $N$ points above it. None lies on the line, because $py=qx$ and coprimality would force $p\mid x$. The total is therefore $(p-1)(q-1)/4$. Combining that count with the parity identities gives quadratic reciprocity.^[ref]

For prescribing a factor by elliptic-curve tripling, quadratic reciprocity gives $(3/p)=1$ when $p\equiv 1(\mathrm{mod}12)$. Choose $r\in {\mathbb{F}}_p$ with $r^2=3$ and choose an integer $a$ by the Chinese remainder theorem so thatThen $a^2+6a-3\equiv 0(\mathrm{mod}p)$ and it is four modulo $q$.

The curve $E:y^2=x^3+ax-a$ has discriminant $-16a^2(4a+27)$. At $p$, $a\ne 0$, since otherwise $r=3/2$ would force $p=3$. Also $4a+27=15+8r$ cannot vanish: squaring $8r=-15$ would give $192=225(\mathrm{mod}p)$, forcing $p=3$ or 11, neither in the required class. At $q$, $a=1$ makes $a(4a+27)=31$, and $q\equiv 5(\mathrm{mod}12)$ excludes $q=31$. Thus both reductions are nonsingular.

The point $P=(1,1)$ is on every curve in this family. Doubling it uses the invertible denominator two and givesModulo $p$ this difference vanishes, and the doubling ordinate is $-1$, so $2P=-P$ and $3P$ is the identity. Modulo $q$, taking $a=1$ gives $x(2P)=2$, so $2P$ and $P$ have distinct abscissas and their sum is a finite point.

Consequently the attempted addition computing $3P$ modulo $N$ asks to invert a denominator divisible by $p$ and not by $q$. Its numerator is nonzero modulo $p$, since the two ordinates there are minus one and one, so this is not a removable coincidence. The Euclidean gcd used for inversion returnsa proper factor. Thus the specified elliptic-curve computation successfully factors $N$.

A nonzero residue class $a(\mathrm{mod}p)$ is a quadratic residue if $a\equiv x^2(\mathrm{mod}p)$ for some nonzero $x$. If it is a square, Fermat's little theorem gives $a^{(p-1)/2}=x^{p-1}\equiv 1$. Conversely, the map $x\mapsto x^2$ on the nonzero classes identifies exactly the pairs $\{x,-x\}$, giving $(p-1)/2$ distinct squares. They are all roots of $X^{(p-1)/2}-1$ over the field ${\mathbb{F}}_p$; a polynomial of this degree has at most that many roots. Its roots are therefore precisely the squares. This proves Euler's criterionApplying it to $-1$ gives $\boxed{(-1/p)=1⟺p\equiv 1(\mathrm{mod}4)}$. The exclusion of $a=0$ is the usual nonzero-residue convention needed for this equivalence.

For distinct odd primes $p,q$, quadratic reciprocity statesUsing the multiplicativity of the Legendre symbol, and the supplementary law $(2/p)=(-1{)}^{(p^2-1)/8}$,The first factor is $1$ because $73\equiv 1(\mathrm{mod}8)$; the second is $(73/3)=(1/3)=1$ because the reciprocity sign is positive. Thus $\boxed{73\text{ is a quadratic residue modulo }127}$.

The discriminant of a binary quadratic form is $d=b^2-4ac$; it is unchanged by a unimodular matrix change of integral variables and satisfies $d\equiv 0$ or $1(\mathrm{mod}4)$. For such a $d$, the discriminant criterion for prime representation by a binary quadratic form says that some integral binary quadratic form of discriminant of a binary quadratic form $d$ represents an odd prime $p$ exactly when $d$ is a square modulo $p$, including zero. Equivalently,These versions agree: a square root modulo $p$ can be chosen even when $d\equiv 0(\mathrm{mod}4)$ and odd when $d\equiv 1(\mathrm{mod}4)$, by adding $p$ if needed, and then the Chinese remainder theorem applies.

For completeness, if $f(x,y)=p$, then $(x,y)$ is primitive, since any common divisor would have its square divide $p$. Complete $(x,y)$ to a basis of ${\mathbb{Z}}^2$ with determinant one using Bézout's identity. The transformed binary quadratic form has leading coefficient $p$ and discriminant of a binary quadratic form $B^2-4pC=d$. Conversely, a suitable $B$ makes $[p,B,(B^2-d)/(4p)]$ an integral binary quadratic form representing $p$ at $(1,0)$.

For $x^2+3y^2$, the discriminant of a binary quadratic form is $-12$. The prime $3$ is represented. If $p\ne 3$ is represented, reducing modulo $3$ gives $p\equiv x^2\equiv 1(\mathrm{mod}3)$, which also excludes $p=2$. To prove sufficiency for $p\equiv 1(\mathrm{mod}3)$, quadratic reciprocity givesThe criterion constructs a positive definite binary quadratic form $[p,B,C]$ of discriminant of a binary quadratic form $-12$. It is primitive: a common divisor of its coefficients would divide $p$, and if that divisor were $p$, then $p^2$ would divide $12$, impossible for $p\ne 2,3$.

We must still establish that this form is equivalent to $x^2+3y^2$, rather than merely some form of the same discriminant of a binary quadratic form. In a positive definite integral binary quadratic form, replace $x$ by $x+ky$ to arrange $|b|\le a$. If $c<a$, apply $(x,y)\mapsto (-y,x)$ and repeat. Each such exchange strictly decreases the positive integer leading coefficient, so this reduction algorithm for a positive definite binary quadratic form terminates with $|b|\le a\le c$. Then $12=4ac-b^2\ge 3a^2$, so $a\le 2$. With $a=1$, parity forces $b=0$, giving $c=3$. With $a=2$, the only integral choices are $b=\pm 2,c=2$, and these are not primitive. Thus every primitive positive definite binary quadratic form of discriminant of a binary quadratic form $-12$ is equivalent to $[1,0,3]$, and preserves its represented integers. Therefore

Put $\omega =(1+\sqrt{229})/2$. Since $229\equiv 1(\mathrm{mod}4)$ is squarefree,We first compute the ideal class group, rather than assume the proposed extension has the full Hilbert-class-field degree.

The real quadratic Minkowski bound for ideal classes is $\sqrt{229}/2<8$. Thus every ideal class has an integral representative of norm at most seven. The prime two is inert because $229\equiv 5(\mathrm{mod}8)$, while seven is inert because $229\equiv 5(\mathrm{mod}7)$ is a nonsquare. The inert ideal $(2)$ has norm four and is principal; a prime above seven has norm forty-nine. The only possible nonprincipal prime factors of these small representatives are therefore primes above three and five.

Reduction of $X^2-X-57$ gives roots $0,1$ modulo three and $2,4$ modulo five. WriteTheir conjugate pairs multiply to $(3)$ and $(5)$. Direct norm calculations giveThe element $6+\omega$ vanishes at the residue roots $0$ modulo three and $4$ modulo five, and not at the conjugate roots. Its absolute norm forces each exponent to be one. Likewise $15+2\omega$ vanishes only at the root zero modulo three, and its norm forces exponent three. ThusThese also give the factorizations suggested by the hint: $(13+\sqrt{229})=(2)PQ$ and $(16+\sqrt{229})=P^3$. The classes of $P,Q$ generate the ideal class group; the first relation makes $[Q]=[P{]}^{-1}$ and the second makes $[P{]}^3=1$. Hence the class number divides three.

It is not one. The algebraic integerso it is a unit. If $P$ were principal, its generator $a=x+y\omega$ would have norm $\pm 3$. Multiply $a$ by a suitable integer power of $ϵ$ so thatIts conjugate then also has absolute value at most $\sqrt{3ϵ}$ because $|a{a}^{\prime }|=3$. ThereforeThe final inequality is $12ϵ<229$, immediate from the displayed value of $ϵ$. Thus the integer $y$ is zero, which makes $|N(a)|=x^2=3$ impossible. ConsequentlyThis proves the ideal class group of Q of square root 229. No claim that $ϵ$ is a fundamental unit is needed. Its norm minus one also realizes all real sign patterns after multiplication by $-1$, so the ordinary and narrow class numbers agree.

Now let $f(X)=X^3-4X-1$ and let $H$ be its splitting field. The possible rational roots $\pm 1$ are not roots, so $f$ is irreducible. Its discriminant isThus its Galois group is $S_3$, since an irreducible cubic has group $A_3$ precisely when its discriminant is a square. Its quadratic subfield is $\mathbb{Q}(\sqrt{229})=K$, and $H/K$ is cyclic of degree three. If $\alpha$ is one root, then $H=K(\alpha )$.

For every rational prime $\ell \ne 229$, the reduction of $f$ is separable. Its roots lift in an unramified extension containing its residue splitting field, by Hensel's lemma. Hence its splitting field is unramified over $\mathbb{Q}$ at all these primes, and consequently over $K$ there.

At 229 there is exactly one double residue root and one simple root:The simple root lifts to $r\in {\mathbb{Q}}_{229}$, and division gives $f=(X-r)q$ with $q$ monic quadratic. The discriminant-product identity isHere $q(r)=f^{\prime }(r)$ is a unit, since $r$ has the simple residue root. Therefore the quadratic discriminant has valuation one, and its root field is the ramified quadratic field ${\mathbb{Q}}_{229}(\sqrt{229})$. This is already the completion of $K$. The local splitting field has no further extension over it: the simple root lies in ${\mathbb{Q}}_{229}$ and the other two lie in that quadratic field. Thus $H/K$ is unramified even at 229. Equivalently its rational inertia is a transposition and has trivial intersection with $A_3$.

The positive cubic discriminant gives three distinct real roots, so $H$ is totally real and there is no infinite ramification over $K$. We have exhibited an everywhere-unramified cyclic extension of degree $h_K=3$. The Hilbert-class-field characterization in part (i) now proves

Finally let $p\ne 229$ be rational prime. A prime ideal of norm $p$ is principal if and only if its Artin symbol in $H_K/K$ is trivial. If it is principal, a generator has norm $\pm p$; multiplying by the norm-minus-one unit if necessary gives norm $+p$. Conversely an element of norm $p$ generates a prime ideal of norm $p$. Hence representation by $x^2+xy-57y^2$ is equivalent to $p$ splitting in $K$ with trivial Artin symbol in $H_K/K$, or to complete splitting of $p$ in $H_K/\mathbb{Q}$.

Let ${\tau }_p\in S_3$ be its rational Frobenius, defined up to conjugacy. Its restriction to the discriminant quadratic field is its sign. The splitting criterion for that quadratic field isFor odd $p$ this follows by quadratic reciprocity from the character of discriminant 229; for $p=2$ both criteria are negative since $229\equiv 5(\mathrm{mod}8)$. The polynomial is separable modulo $p$, and the degrees of its factors are the cycle lengths of ${\tau }_p$. Thus a root modulo $p$ means that ${\tau }_p$ fixes a root. An even permutation of three letters with a fixed point is the identity: a nonidentity three-cycle has none. Therefore the two stated conditions together are exactly complete splitting in $H_K$, andThis is prime representation by the norm form of Q of square root 229, with the norm sign and the prime two explicitly covered.

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

For an odd prime $p$, the Legendre symbol $(a/p)$ is zero if $p\mid a$, one if $a$ is a nonzero square modulo $p$, and minus one otherwise. By Gauss lemma, $(2/p)=(-1{)}^N$, where $N$ counts those least positive residues of $2,4,\dots ,p-1$ which exceed $p/2$. They correspond to integers $j>p/4$ among $1\le j\le (p-1)/2$, henceFor $p\equiv 1,3,5,7(\mathrm{mod}8)$, its parity is respectively $0,1,1,0$, the same as that of $(p^2-1)/8$. ThereforeUse multiplicativity and quadratic reciprocity for the requested composite numerator. Since $91=7\cdot 13$,The first reciprocity sign is negative because both primes are $3(\mathrm{mod}4)$; the subsequent signs are positive because one prime is $1(\mathrm{mod}4)$. Thus $(91/167)=-1$.

Let $g$ generate the cyclic group of nonzero residues, whose order $p-1=2^{2^n}$ is a power of two. Write $a=g^j$. Its multiplicative order is $(p-1)/\gcd (j,p-1)$, so it is a primitive root precisely when $j$ is odd. The quadratic residues are exactly the even powers of $g$. Thus the primitive roots are precisely the quadratic nonresidues, with nonzero residue classes understood.

For $n\ge 1$, $p\equiv 1(\mathrm{mod}4)$, so quadratic reciprocity gives $(7/p)=(p/7)$. Since $2^3\equiv 1(\mathrm{mod}7)$ and $2^n$ alternates between $1$ and $2$ modulo three, $p$ is respectively $3$ or $5$ modulo seven. The nonzero squares modulo seven are $1,2,4$. Consequently $(7/p)=-1$, andThere is a genuine endpoint exception if the printed form permits $n=0$: then $p=3$ and $7\equiv 1(\mathrm{mod}3)$, which has order one and is not primitive. The first equivalence remains true for $p=3$, but the final assertion requires excluding that prime.

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.

The Chinese remainder theorem says that for pairwise coprime positive moduli $m_i$, the map $\mathbb{Z}/(\prod m_i)\to {\prod }_i\mathbb{Z}/m_i$ is an isomorphism; in particular prescribed residues have a unique solution modulo the product.

Write $n=p^{e}m$, with $e\ge 2$ and $(p,m)=1$. Choose $z\equiv 1+p^{e-1}(\mathrm{mod}{p}^e)$ and $z\equiv 1(\mathrm{mod}m)$. The Chinese remainder theorem supplies $z$. The binomial theorem gives $z^p\equiv 1(\mathrm{mod}{p}^e)$, since every nonconstant term is divisible by $p^e$; also $z^p\equiv 1(\mathrm{mod}m)$. Thus $z^p\equiv 1(\mathrm{mod}n)$, whereas $z\not\equiv 1(\mathrm{mod}n)$. Its multiplicative order is exactly $p$.

For odd $n=\prod p_i^{e_i}$ define the Jacobi symbol $(a/n)={\prod }_i(a/p_i{)}^{e_i}$, using the Legendre symbol, with $(a/1)=1$. It is zero if $a$ shares a prime factor with $n$. For example $(2/9)=(2/3{)}^2=1$, but the square residues modulo nine are $0,1,4,7$, so two is not a square.

For positive odd coprime $u,v$, Jacobi reciprocity isFactor $u=\prod p_i^{a_i}$ and $v=\prod q_j^{b_j}$ and multiply quadratic reciprocity for each prime pair. Its sign exponent is ${\sum }_{i,j}{a}_i{b}_j(p_i-1)(q_j-1)/4$. For odd numbers, $(rs-1)/2\equiv (r-1)/2+(s-1)/2(\mathrm{mod}2)$, so this exponent has the same parity as $(u-1)(v-1)/4$. Multiplicativity of the Legendre symbol gives the two Jacobi products, proving the formula. If $u,v$ are not coprime, both sides of the displayed reciprocity equality are zero.

For the previously constructed $z$, every prime divisor of $n$ sees residue one, so $(z/n)=1$. But $p\nmid (n-1)/2$, because $p\mid n$ and $p$ is odd. The order-$p$ unit therefore has $z^{(n-1)/2}\not\equiv 1(\mathrm{mod}n)$. This supplies a unit failing the proposed test.

Finally the passing units form the kernel of the homomorphismMultiplicativity makes this a homomorphism, and the failing unit shows its kernel is proper. A proper subgroup has index at least two. Hence the test fails on at least half the units modulo $n$.

For an integral binary quadratic form $[a,b,c]=a{x}^2+bxy+c{y}^2$ with $a>0$ and $D=b^2-4ac<0$, the reduced positive definite binary quadratic form convention is $|b|\le a\le c$, with $b\ge 0$ on either boundary $|b|=a$ or $a=c$. Since $|D|=4ac-b^2\ge 3a^2$, the enumeration of reduced binary quadratic forms for $D=-23$ has $a\le \sqrt{23/3}<3$. The integer $b$ is odd. For $a=1$, only $b=1$ is retained and $c=6$; for $a=2$, both $b=\pm 1$ give $c=3$. Thus the complete list isA proper representation of a prime uses a primitive representing vector. Completing that vector to a matrix in ${\mathrm{SL}}_2(\mathbb{Z})$ transforms a representing form to $[p,b,c]$, so its discriminant of a binary quadratic form gives $b^2\equiv -23(\mathrm{mod}p)$. Conversely, choose an odd $b$ satisfying this congruence and put $c=(b^2+23)/(4p)$. This is integral, and $[p,b,c]$ is positive definite and represents $p$ at $(1,0)$. This proves the discriminant criterion for prime representation by a binary quadratic form here.

For $p\ne 23$, quadratic reciprocity gives $(-23/p)=(p/23)$. Including the ramified prime, the answer is thereforeFor example $[23,23,6]$ supplies a proper representation of $23$. Reduction transfers every such representation to one of the three displayed forms.

For odd positive coprime integers, quadratic reciprocity for the Jacobi symbol isIf the integers are not coprime, both symbols vanish; equivalently the usual signed reciprocity identity still holds. To deduce it from quadratic reciprocity for the Legendre symbol, factor $m={\prod }_i{p}_i^{a_i}$ and $n={\prod }_j{q}_j^{b_j}$. Multiplication of the prime reciprocity identities gives the sign exponentModulo two, each bracket is respectively $(m-1)/2$ and $(n-1)/2$: for odd $u,v$, $(uv-1)/2\equiv (u-1)/2+(v-1)/2(\mathrm{mod}2)$. This proves the composite-denominator law, including repeated prime factors.

Now $261=9\cdot 29$, and 9 contributes a square factor. Since $317\equiv 1(\mathrm{mod}4)$, reciprocity and reduction of the numerator giveA second reciprocity step gives $(3/29)=(29/3)(-1{)}^{14}=(2/3)=-1$. Thus the requested Jacobi symbol is $-1$.

The polynomial is Eisenstein at every prime number $p$, so a polynomial root $\alpha$ generates a totally ramified cubic field $F/{\mathbb{Q}}_p$. The splitting-field Galois group is a transitive subgroup of $S_3$, hence $C_3$ or $S_3$. Its polynomial discriminant isThe product of the three polynomial root differences is a square root of $\Delta$, transformed by the sign of the polynomial root permutation. Thus the group is $C_3$ exactly when $\Delta$ is a square, and otherwise it is $S_3$. In the latter case the splitting field isIndeed adjoining one polynomial root leaves a quadratic factor, whose polynomial discriminant is $\Delta /f^{\prime }(\alpha {)}^2$, so the square root of $\Delta$ supplies its remaining polynomial roots.

Suppose first that $p>3$. The unit $-(4p+27)$ reduces to $-27$, and Hensel's lemma makes it a square exactly when $-3$ is a square modulo $p$. Quadratic reciprocity givesIf $p\equiv 1(\mathrm{mod}3)$, the field generated by $\alpha$ is already the cyclic splitting field, totally ramified of degree three. If $p\equiv 2(\mathrm{mod}3)$, the nonsquare unit gives an unramified quadratic field $E={\mathbb{Q}}_p(\sqrt{\Delta })$. Since $F$ and $E$ have coprime degrees, $[FE:{\mathbb{Q}}_p]=6$. Its residue degree is at least two and its ramification index at least three; their product is six, so they are exactly two and three. Thus its inertia group is the normal $C_3$ inside $S_3$.

In both cases the ramification index is three and is not divisible by $p$. The wild inertia group is trivial, so all positive lower groups are trivial. As a different check, $f^{\prime }(\alpha )=3{\alpha }^2+p$ has valuation two in $F$ when $p>3$, agreeing with tame different exponent $3-1$.

For $p=3$ the polynomial discriminant is $-351=-3^3\cdot 13$, which is nonsquare. Hence $G=S_3$. The quadratic subfield can be written $E={\mathbb{Q}}_3(\beta )$ with ${\beta }^2=-39$. It is ramified of degree two, by the Eisenstein polynomial $T^2+39$. Together with the totally ramified cubic field this forces $e(L/{\mathbb{Q}}_3)$ to be divisible by both two and three. Since $[L:{\mathbb{Q}}_3]=6$, the splitting field is totally ramified. In its normalized valuation,

All differences between distinct polynomial roots have the same valuation, since $S_3$ permutes the polynomial root pairs and preserves the absolute value on a field. Alsohas valuation six, so each polynomial root difference has valuation three. A nonidentity three-cycle fixes $\beta$, and thereforeThe transposition fixing $\alpha$ negates $\beta$, so sends $\pi$ to $-\pi$ and has displacement valuation $v_L(2\pi )=1$. The other transpositions have the same value by conjugacy. The uniformizer criterion now determines every group:For the wild case the different exponent is $5+2=7$. Independently, the cubic subfield has different exponent $v_F(3({\alpha }^2+1))=3$, and the quadratic extension $L/F$ is tame with exponent one. The tower formula gives $1+2\cdot 3=7$, confirming the filtration. Its positive upper break is $1/2$, because $[G_0:G_1]=2$. This completes ramification of the splitting field of X3 plus pX plus p for every odd prime number.

Write $Q(x,y)=a{x}^2+bxy+c{y}^2$ and associate the symmetric matrix $M=\left(\begin{array}{cc}a& b/2\\ b/2& c\end{array}\right)$. Define the left group action of $S{L}_2(\mathbb{Z})$ by $(g\cdot Q)(v)=Q(g^{-1}v)$; using $Q(gv)$ instead describes the same equivalence classes. The transformed matrix is $g^{-T}M{g}^{-1}$, so its determinant is unchanged. Hence the discriminant of a binary quadratic formis invariant under proper equivalence of binary quadratic forms.

A proper representation here means $Q(x,y)=n$ with $\gcd (x,y)=1$. Such a vector extends to the first column of an integer determinant-one matrix by Bezout identity. After that change of variables the form has leading coefficient $n$, and discriminant $-35$, so its middle coefficient $b^{\prime }$ satisfies $(b^{\prime }{)}^2\equiv -35(\mathrm{mod}4n)$. Conversely, such a congruence gives the positive definite integral formwhich represents $n$ at $(1,0)$. It is primitive: a common divisor of all three coefficients would have its square dividing the squarefree discriminant 35.

For clarity, enumerate all the proper equivalence classes at this discriminant. Substitutions $(x,y)\mapsto (x+my,y)$ first arrange $|b|\le a$; if $c<a$, the determinant-one substitution $(x,y)\mapsto (-y,x)$ makes the leading coefficient smaller. Repeating terminates because this coefficient is a positive integer. Thus every class has a reduced representative with $|b|\le a\le c$. From $4ac-b^2=35$ it follows that $3a^2\le 35$, giving $a\le 3$. Checking $a=1,2,3$, with $b$ odd, yieldsThe two signs in the first pair are equivalent by replacing $x$ with $x-y$, and in the second pair by the determinant-one swap. Thus the two forms in the question exhaust the classes.

The primitive representation criterion at discriminant minus thirty-five now follows. For odd $n$ prime to 35, the congruence modulo $4n$ is solvable exactly when $-35$ is a square modulo every prime divisor of $n$. Necessity is immediate. For sufficiency, a nonzero root modulo each odd prime $q\mid n$ lifts to all required prime powers because its derivative $2b^{\prime }$ is invertible; combine the lifts by the Chinese remainder theorem, also imposing $b^{\prime }\equiv 1(\mathrm{mod}2)$. This gives the modulus-four condition since $-35\equiv 1(\mathrm{mod}4)$. We concludeBy quadratic reciprocity, the symbol equals $(q/5)(q/7)$. In particular each prime divisor must be a quadratic residue at both 5 and 7, or a nonresidue at both. The condition is on each prime divisor, not merely on the product Jacobi symbol of $n$.

For odd coprime positive integers $a={\prod }_i{p}_i^{e_i}$ and $b={\prod }_j{q}_j^{f_j}$, define the Jacobi symbol by multiplying Legendre symbols over the prime factorization of the denominator. Multiplicativity in the numerator givesBy quadratic reciprocity for the Legendre symbol, this is $(-1{)}^{UV}$, where $U={\sum }_i{e}_i(p_i-1)/2$ and $V={\sum }_j{f}_j(q_j-1)/2$. For odd $x,y$, $(xy-1)/2\equiv (x-1)/2+(y-1)/2(\mathrm{mod}2)$, since $(x-1)(y-1)/2$ is even. Thus $U\equiv (a-1)/2$ and $V\equiv (b-1)/2(\mathrm{mod}2)$. We obtainIf $a,b$ are not coprime, both symbols are zero and the equivalent identity $(a/b)=(-1{)}^{(a-1)(b-1)/4}(b/a)$ still holds. The convention $(a/1)=1$ handles denominator one.

Herbrand quotient and local norm indices. Let $G=⟨\sigma ⟩$ have order $n$. For an additive $G$-module $M$, put $D=\sigma -1$ and $N=1+\sigma +\cdots +{\sigma }^{n-1}$. The relevant Tate cohomology of a cyclic group groups areIf both groups are finite, their size ratio is the Herbrand quotientMultiplicative modules use products for $N$ and $D(m)=\sigma (m)/m$. A finite module has quotient one: $\#\ker N=\#M/\#NM$, while $\#DM=\#M/\#M^G$, and the two cohomology orders are equal. The six-term periodic cohomology sequence shows $h(B)=h(A)h(C)$ for a short exact sequence $0\to A\to B\to C\to 0$ when these groups are finite. Thus the quotient is unchanged by finite-index changes of lattices. For the trivial $G$-module $\mathbb{Z}$, the zeroth group is $\mathbb{Z}/n\mathbb{Z}$ and the negative first group is zero, so $h_G(\mathbb{Z})=n$.