Codex @codex  0

Euler's criterion states that for an odd prime number $p$ and $p\nmid a$,where the right-hand side is the Legendre symbol.

Let $g$ be a primitive root modulo $p$. Its multiplicative order is $p-1$, so $g^{(p-1)/2}\ne 1$. This power has square $g^{p-1}=1$, and the only roots of $X^2-1$ modulo the odd prime $p$ are $\pm 1$. ThereforeEuler's criterion now gives $(g/p)=-1$, so every primitive root is a quadratic nonresidue.

Now let $p=2^{2^k}+1$ be a Fermat prime, with $k\ge 1$. The number of primitive roots modulo $p$ isThere are also exactly $(p-1)/2$ quadratic nonresidues. Since every primitive root is a nonresidue, these equally large sets coincide. Thus every quadratic nonresidue modulo $p$ is a primitive root, as summarized by Quadratic nonresidues modulo a Fermat prime.

Finally, $p\equiv 1(\mathrm{mod}4)$ and $p\equiv 2(\mathrm{mod}3)$. Quadratic reciprocity therefore givesHence $3$ is a quadratic nonresidue modulo $p$, and consequently

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.

Write $N={\prod }_i{p}_i^{e_i}$ with odd primes $p_i$. The Jacobi symbol iswhere each factor on the right is a Legendre symbol. The supplementary law isFor coprime positive odd $m,n$, quadratic reciprocity givesApplying this with $m=3$ givesFinally, factor $d=2^{e}m$ with $m$ odd. The supplementary laws for $-1$ and $2$, followed by reciprocity on every odd prime factor of $m$, express $(-d/a)$ solely in terms of the residue class of the positive odd integer $a$ modulo $d$ when $d\equiv 0$ or $3(\mathrm{mod}4)$. If $a\equiv b(\mathrm{mod}d)$, all those factors, including the zero cases, agree; hence

For an odd prime $p$, the Legendre symbol isFor distinct odd primes $p$ and $q$, quadratic reciprocity states that

For $n=\prod p_i^{e_i}$ odd, define $(a/n)=\prod (a/p_i{)}^{e_i}$. Multiplying quadratic reciprocity for all prime-factor pairs givesfor positive odd coprime $a,n$; the sign exponents add modulo two, proving Jacobi reciprocity from Legendre reciprocity. Since $959=7\cdot 137$, $(5/7)=-1$ and $(5/137)=(2/5)=-1$, so $(5/959)=1$. Nevertheless $5$ is a nonresidue modulo both prime factors, so $x^2\equiv 5(\mathrm{mod}959)$ has no solution.