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False. The Jacobi symbol $(2/15)=(2/3)(2/5)=(-1)(-1)=1$, yet $2$ is not a quadratic residue modulo three. A solution modulo fifteen would give one modulo three, which is impossible. A positive Jacobi symbol need not indicate a square modulo a composite modulus.

Take the odd congruence class $p\equiv 3(\mathrm{mod}8)$. The congruent number elliptic curve in the standard scaled coordinates isThese are the two curves in two-isogeny descent. On $E_p$, the prime-support bound in two-isogeny descent restricts the first image to $\{1,-1,p,-p\}$. All four classes occur: the 2-torsion points give $\alpha ((0,0))=[-p^2]=[-1]$, $\alpha ((p,0))=[p]$, and $\alpha ((-p,0))=[-p]$. Thus $\#\alpha (E_p)=4$.

On $E_p^{\prime }$, the equation forces $x>0$ whenever $x\ne 0$. Its exceptional value is $[4p^2]=1$. The second image is therefore contained in $\{1,2,p,2p\}$. Use the quartic covering in a two-isogeny descent to exclude its three nontrivial candidates.

For $d=p$, a rational solution can be written with coprime integral $U,V$, and must satisfyThen $p\mid W$, so $U^4+(2V^2{)}^2\equiv 0(\mathrm{mod}p)$. Since $p\equiv 3(\mathrm{mod}4)$, $-1$ is not a quadratic residue; a sum of two squares is zero only if both are zero. This forces $p\mid U,V$, contradicting coprimality. For $d=2p$, the same reasoning applies to $W^2=2p(U^4+V^4)$ and again forces $p\mid U,V$.

For $d=2$, the equation isIt implies $W=2Z$, whence $2Z^2=U^4+p^2{V}^4$. Exactly one of $U,V$ odd is impossible by parity, while both even contradict coprimality. Thus both are odd. Since $p\equiv 3(\mathrm{mod}8)$ implies $p^2\equiv 9(\mathrm{mod}16)$, reduction modulo sixteen gives $2Z^2\equiv 1+9=10$. But the possible residues of $2Z^2$ modulo sixteen are $0,2,8$. This contradiction eliminates $d=2$. Consequently ${\alpha }^{\prime }(E_p^{\prime })=\{1\}$ and the square-class index formula for two-isogeny descent givesThe reduction argument for $y^2=x^3+kx$ used earlier bounds rational torsion by four for every nonzero integer $k$: its point-count cancellation at $q\equiv 3(\mathrm{mod}4)$ does not depend on the sign of $k$, and the same prime choices apply. Here $k=-p^2$ and all four 2-torsion points are rational. Hence $E_p(\mathbb{Q})$ consists exactly of $O,(0,0),(p,0),(-p,0)$.

To relate this calculation to congruent numbers, a rational point $(x,y)$ with $y\ne 0$ produces the right triangle with side lengthsThe identities $A^2+B^2=C^2$ and $AB/2=p$ follow from $y^2=x(x^2-p^2)$. Conversely, a positive rational right triangle of area $p$ with legs $A,B$ and hypotenuse $C$ givesDirect substitution uses $AB=2p$ and $C^2=A^2+B^2$. No such point exists here. This proves the noncongruent primes congruent to three modulo eight:

For an odd prime number $p$, the Legendre symbol is zero when $p\mid a$, one when $a$ is a nonzero quadratic residue modulo $p$, and minus one otherwise. The number-theoretic Gauss lemma says that, for $p\nmid a$, $(a/p)=(-1{)}^r$, where $r$ counts those least positive residues of $a,2a,\dots ,(p-1)a/2$ exceeding $p/2$.

For $a=2$, the residues are simply $2,4,\dots ,p-1$, so $r=(p-1)/2-\lfloor p/4\rfloor$. Examining the four odd residue classes modulo eight givesNow $2^m\equiv -1(\mathrm{mod}p)$ and $p$ is odd. The multiplicative order $d$ of $2$ divides $2m$ but not $m$. Since $m$ is a power of two, every proper divisor of $2m$ divides $m$, whence $d=2m$. Lagrange's theorem for finite groups gives $2m\mid p-1$. Because $m\ge 4$, this implies $p\equiv 1(\mathrm{mod}8)$, so the supplementary law just proved gives $(2/p)=1$. By Euler's criterion, $2^{(p-1)/2}\equiv 1(\mathrm{mod}p)$, and consequently $2m\mid (p-1)/2$. Therefore

For the real quadratic field $K=\mathbb{Q}(\sqrt{d})$ with squarefree $d>1$, Dirichlet unit theorem givesfor a fundamental unit $ϵ>1$, since the roots of unity in a real field are $\pm 1$.

If $p\mid d$ and $p\equiv 3(\mathrm{mod}4)$, a unit of norm $-1$ would have the form $(a+b\sqrt{d})/2$ with integers $a,b$ and satisfy $a^2-d{b}^2=-4$. Reducing modulo the odd prime $p$ gives $(a/2{)}^2\equiv -1(\mathrm{mod}p)$, impossible because $-1$ is not a quadratic residue for such a prime. Hence every unit has norm $+1$ in this case. In contrast, $\boxed{d=2,u=1+\sqrt{2}}$ has norm $-1$.

Fermat's little theorem states that if $p$ is prime, then $a^p\equiv a(\mathrm{mod}p)$ for every integer $a$; equivalently, $a^{p-1}\equiv 1(\mathrm{mod}p)$ when $p\nmid a$.

If $p\equiv 3(\mathrm{mod}4)$ and $x^2\equiv -1(\mathrm{mod}p)$, thenbecause $(p-1)/2$ is odd. This contradicts Fermat's theorem, so $x^2\equiv -1(\mathrm{mod}p)$ has no solution.

If the primes congruent to $1(\mathrm{mod}4)$ were $p_1,\dots ,p_k$, set $N=(2p_1\cdots p_k{)}^2+1$. Any prime divisor $q$ of $N$ is odd and has $-1$ as a quadratic residue. The preceding result forces $q\equiv 1(\mathrm{mod}4)$, but $q$ divides none of the $p_i$ because $N\equiv 1(\mathrm{mod}{p}_i)$. This contradiction proves that infinitely many primes are congruent to $1(\mathrm{mod}4)$.

True. Write the odd integer as $m=2n+1$. The identity from part (b) and $p\mid F_m$ giveConsecutive Fibonacci numbers are coprime, so $F_{n+1}$ is invertible modulo $p$. HenceThus $-1$ is a quadratic residue modulo the odd prime $p$, which occurs exactly when $\boxed{p\equiv 1(\mathrm{mod}4)}$.

An $R$-orientation of a rank-$d$ real vector bundle $E\to B$ is a locally coherent choice of generator of $H^d(E_b,E_b\setminus \{0\};R)$ in every fibre. Equivalently, it is represented by a Thom class $u_E\in H^d(D(E),S(E);R)$ restricting to the chosen generator on each fibre pair. If $e(E)\in H^d(B;R)$ is the Euler class, the Gysin sequence of its unit sphere bundle contains

The diagonal quotient defining $L(p)$ is the three-dimensional lens space as a circle bundle $S^1\to L(p)\to S^2$ with Euler class $p$ times a generator. With integral coefficients, the only nontrivial Euler-class map is multiplication by $p:H^0(S^2;\mathbb{Z})\to H^2(S^2;\mathbb{Z})$. Exactness givesModulo $p$, the Euler class vanishes, so the same Gysin sequence giveswith all other groups zero.

For the coefficient sequence $0\to \mathbb{Z}\stackrel{p}{\to }\mathbb{Z}\to {\mathbb{F}}_p\to 0$, the long exact sequence from a coefficient sequence containsThe last map is zero and both middle groups have order $p$, so $\delta$ is an isomorphism. Reduction $H^2(L(p);\mathbb{Z})\to H^2(L(p);{\mathbb{F}}_p)$ is likewise an isomorphism. Their composite is the Bockstein isomorphism for a three-dimensional lens space

If $a^{\prime }=na$ is another generator, linearity of the Bockstein homomorphism and bilinearity of the cup product giveMoreover $t(a)\ne 0$: $\beta (a)$ is nonzero, and the Poincare duality pairing $H^1\times H^2\to H^3\cong {\mathbb{F}}_p$ is nondegenerate. If $h:L(p)\to L(p)$ is an orientation-reversing homotopy equivalence and $h^{\ast }a=na$, naturality givesCancelling the nonzero $t(a)$ yields $n^2\equiv -1(\mathrm{mod}p)$. Thus $-1$ must be a quadratic residue modulo $p$.

The Dedekind factorization theorem says that, when a rational prime $p$ does not divide the index of an order $\mathbb{Z}[\theta ]\subseteq {\mathcal{O}}_K$, a factorizationof the minimal polynomial modulo $p$ gives

Let $K=\mathbb{Q}(\sqrt{d})$ with square-free $d$, and let $p$ be odd. The resulting splitting of rational primes in a quadratic field is:

Finally suppose $d>0$ and
$\alpha =A/C+(B/C)\sqrt{d}$ has norm $-1$, with integers
$A,B,C$ chosen coprime. ThenIf an odd $p\equiv 3(\mathrm{mod}4)$ ramified, then $p\mid d$. Reduction modulo $p$ gives
$A^2\equiv -C^2(\mathrm{mod}p)$. Since $-1$ is not a square modulo such a prime, $p\mid A,C$. Dividing the equation's divisibility shows $p\mid B$ as well (use that square-free $d$ has $p$-adic valuation one), contradicting coprimality. Hence no prime $p\equiv 3(\mathrm{mod}4)$ ramifies.

Let $N=\lfloor \sqrt{x}\rfloor$ and formFor sufficiently large $x$, every factor is positive and at most $x$. Every odd prime divisor $p$ of $P$ makes $2$ a quadratic residue modulo $p$, so part (a) gives $p\equiv 1$ or $7(\mathrm{mod}8)$.

For an odd prime power $p^a$, the congruence $n^2\equiv 2(\mathrm{mod}{p}^a)$ has at most two residue classes, because each root modulo $p$ is simple and lifts uniquely. ThereforeSince $\log p/(p-1)\le \log 3/2$ for odd $p$,The prime $2$ contributes at most $N\log 2$, because $v_2(n^2-2)\le 1$. Ifthen, for all sufficiently large $x$, $2\log x\le N\log 3/2$ and hence

On the other hand, the product contains at least $N/3$ factors for large $N$, and every one satisfiesonce $x$ is sufficiently large. ThusComparing the two estimates givesfor every sufficiently large $x$. This is the elementary logarithmic lower bound for primes in the quadratic-residue classes of two.

Write a matrix as $(r,b)$, where $r=a^2$ is a nonzero quadratic residue modulo $11$. The square subgroup isMatrix multiplication becomesThus $H$ is the affine semidirect product of cyclic groups of orders eleven and fiveThere are five choices for $r$ and eleven for $b$, so $|H|=55$. It is nonabelian, since

LetConjugation by the complement givesHence the ten nonidentity translations split into the two orbitseach of size five. For $r\ne 1$,Since $1-r\ne 0$, varying $t$ gives every $b\in {\mathbb{F}}_{11}$. Conjugation cannot change $r$ in the abelian quotient $H/N$. Therefore, withthe seven conjugacy classes areof sizes $1,5,5,11,11,11,11$, agreeing with the conjugacy classes in the affine semidirect product of orders eleven and five.

The commutators with $(4,0)$ generate every translation because multiplication by $4-1=3$ is invertible in ${\mathbb{F}}_{11}$. Hence the commutator subgroup is $N$, and the abelianization isPut $\zeta =e^{2\pi i/5}$. The one-dimensional characters factor through the abelianization, giving five characters

For the remaining characters, put $\omega =e^{2\pi i/11}$ and define a character of $N$ byThe complement has two free orbits on the nontrivial ${\theta }_m$, indexed by $m\in Q$ and $m\in 2Q$. By induction from an abelian normal subgroup with a free character orbit, the charactersare irreducible of degree five and vanish outside $N$.

SetThe quadratic periods modulo eleven give the nonsquare sum $\overline{\eta }=(-1-i\sqrt{11})/2$. Multiplication by a square preserves $Q$ and multiplication by a nonsquare exchanges the two square classes, so the complete character table isFinally,and there are seven rows for the seven conjugacy classes. Thus these are all irreducible characters, as described by the irreducible characters of the affine semidirect product of orders eleven and five.

Write an integral binary quadratic form asA positive definite form is a reduced positive definite binary quadratic form whenwith $b\ge 0$ on the boundary $|b|=a$ or $a=c$. The class number of a negative discriminant $h(d)$ is the number of proper equivalence classes of primitive positive definite integral forms of discriminant $d$.

Every such class has a reduced representative. For a reduced form,soThere are only finitely many possible integers $a$ and $b$, and thenis determined. Hence $h(d)<\infty$. There is at least one class: the principal form isandThus $1\le h(d)<\infty$, in agreement with the enumeration of reduced binary quadratic forms.

Now let $q\equiv 3(\mathrm{mod}8)$ be prime. The two formsare primitive, positive definite, reduced, and have discriminant $-8q$. They are not properly equivalent: $f_0$ represents one, while the least positive value of $f_1$ is two. Therefore

Assume henceforth that $h(-8q)=2$, so these are the only two classes. Iffor a prime $p>q$, then $x$ is odd andMoreover $p\equiv x^2(\mathrm{mod}q)$, so $p$ is a quadratic residue modulo $q$. This proves the necessity.

Conversely, suppose $p\equiv \pm 1(\mathrm{mod}8)$ and $p$ is a quadratic residue modulo $q$. By quadratic reciprocity, these two conditions implyIndeed, for $p\equiv 1(\mathrm{mod}8)$ both relevant signs are positive, while for $p\equiv 7(\mathrm{mod}8)$ the signs from $(-8/p)$ and reciprocity with $q\equiv 3(\mathrm{mod}4)$ cancel. Choose an even integer $b$ such thatThenis a primitive positive definite integral form of discriminant $-8q$ and represents $p$. It belongs to one of the two classes $f_0,f_1$.

The second form cannot represent such a prime. Ifthen $y$ is odd and, since $q\equiv 3(\mathrm{mod}8)$,This contradicts $p\equiv \pm 1(\mathrm{mod}8)$. Hence $[p,b,c]$ lies in the principal class $f_0$, and proper equivalence of binary quadratic forms preserves represented integers. ThereforeThis is the prime representation when h of minus eight q equals two.

A primitive root modulo $N$ is a unit of order $\phi (N)$; a nonzero quadratic residue modulo an odd prime $p$ is a square.

If a primitive root $g$ modulo $p$ were $x^2$, then its order would divide $(p-1)/2$, the order of the subgroup of squares, contradicting ${\mathrm{ord}}_p(g)=p-1$.

Modulo $11$, $2$ has order $10$, so the primitive roots are $2^k$ for $\gcd (k,10)=1$:Finally $2^{10}=1024\equiv 56\not\equiv 1(\mathrm{mod}121)$, so the lifting criterion makes $2$ primitive modulo $121$. It is plainly the smallest possible positive primitive root.

If $q\equiv 3(\mathrm{mod}8)$ is prime and $h(-8q)=2$, then for every prime $p>q$,for some integers $x,y$ exactly when $p\equiv \pm 1(\mathrm{mod}8)$ and $p$ is a quadratic residue modulo $q$.

The two-isogeny descent quartics associated with square classes $2$ and $-2$ on $Y^2=X^3+24X^2+64X$ have no rational points. A rational solution can be scaled to integral $W,U,V$ with coprime integers $U,V$. If $U$ is odd, reduction modulo sixteen gives $2$ or $10$ for the positive class, and $14$ or $6$ for the negative class; none is a quadratic residue. If $U=2u$ is even, then $V$ is odd and the right side is $32(\pm u^4+3u^2{V}^2\pm V^4)$, with the same sign on the two signed terms. The bracket is odd, so its 2-adic valuation is exactly five, impossible for a nonzero square. These two parity cases exhaust primitive solutions.