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For an odd prime $p$ and $p\nmid a$, Euler's criterion identifies the Legendre symbol by modular exponentiation:

Thus a quadratic residue gives one and a quadratic nonresidue gives minus one, represented by $p-1$ if a nonnegative remainder is used. The formula also holds when $p\mid a$, because $(p-1)/2>0$ and both sides are zero modulo $p$.^[ref]

For example, modulo seven the exponent is three: $2^3\equiv 1$ confirms that two is a quadratic residue, while $3^3\equiv 6\equiv -1$ confirms that three is a quadratic nonresidue. Modulo eleven the exponent is five, and $2^5=32\equiv 10\equiv -1$ confirms that two is a quadratic nonresidue. The test answers whether a square root exists; it does not itself return the roots $3,4$ of $x^2\equiv 2(\mathrm{mod}7)$.

A short proof needs only Fermat's little theorem and the root bound for a polynomial. Put $m=(p-1)/2$. If $a=x^2\ne 0$ in the prime field, then $a^m=x^{p-1}=1$. There are already $m$ such quadratic residues. The nonzero polynomial $T^m-1$ has polynomial degree $m$, so it has no other polynomial roots. Finally, for every nonzero $a$, Fermat's little theorem gives $(a^m{)}^2=a^{p-1}=1$. Factoring $(a^m-1)(a^m+1)=0$ in the field shows $a^m\in \{1,-1\}$; every quadratic nonresidue must take the remaining value $-1$. This proves the criterion without first finding a primitive root.

For a larger original example, decide whether seven is a quadratic residue modulo the prime number $101$. Exponentiation by squaring computes the required power without listing fifty squares:Therefore $(7/101)=-1$ and $x^2\equiv 7(\mathrm{mod}101)$ has no solution. Each intermediate product is reduced modulo $101$, as in modular exponentiation. The advantage is that exponentiation by squaring uses a number of multiplications proportional to the binary digit length of the exponent rather than testing every potential square root.

The odd prime hypothesis is essential. For the composite modulus nine, $2^{(9-1)/2}=2^4\equiv 7(\mathrm{mod}9)$, which is neither one nor minus one. Although four is a nonzero square modulo nine, $4^4\equiv 4(\mathrm{mod}9)$, so even the square-to-one implication fails. The Jacobi symbol and Euler pseudoprime discuss different questions for odd composite moduli; they do not make this a universal square test.

Every odd composite integer has an Euler witness. For a nonsquarefree integer, a base congruent to $1+p$ modulo a repeated prime factor $p^2$ fails the Euler congruence by the binomial theorem. For a squarefree integer, use the Chinese remainder theorem to choose a base that is a quadratic nonresidue modulo one prime factor and $1$ modulo another. Its Jacobi symbol is $-1$, while its power is $1$ modulo the second factor.

For an odd positive nonsquare integer $a$, there are infinitely many primes $p$ with Legendre symbol $(a/p)=-1$. Choose a residue $n$ modulo $a$ with Jacobi symbol $(n/a)=-1$ by making it a quadratic nonresidue at one prime occurring to odd multiplicity and a square at every other prime. The Chinese remainder theorem also imposes $n\equiv 1(\mathrm{mod}4)$ and $n\equiv 1$ at any finite list of previously found primes. Then Jacobi reciprocity gives $(a/n)=-1$, so a prime factor of $n$ has $(a/p)=-1$ and is outside that list.

For an odd prime $p$, the least quadratic nonresidue $n(p)$ is the smallest positive integer whose nonzero residue class is a quadratic nonresidue modulo $p$.

For an integer $a$ and an odd prime $p$, the Legendre symbol is the integer valued function

Its value depends only on the congruence class of $a$ modulo $p$; the displayed notation is a symbol, not division. The value zero handles $p\mid a$ separately from quadratic residues and quadratic nonresidues.^[ref]

For example, $(2/7)=1$, $(3/7)=-1$, $(14/7)=0$, and $(2/11)=-1$. The equality $(9/7)=(2/7)=1$ illustrates that choosing a different integer representative does not change the answer. An arbitrary odd composite denominator uses the Jacobi symbol, not the Legendre symbol. For instance,but two is not a square modulo fifteen: any such square root would also give a square root modulo three, where the only nonzero square is one. A Jacobi symbol of one therefore cannot replace the Legendre symbol's square test.

The Legendre symbol also counts square roots compactly. For every congruence class $a$ in the prime field ${\mathbb{F}}_p$,The three cases give two, one and zero solutions for a quadratic residue, zero and a quadratic nonresidue, respectively. This is useful when counting solutions of modular congruences involving polynomials: a solvability question becomes a sum of Legendre symbols, as in the complete Legendre-symbol sum and quadratic Legendre-symbol correlation.

For an odd prime $p$ and any integers $a,b$, the Legendre symbol obeysIn particular, if $p\nmid a$, then $(a^{-1}/p)=(a/p)$, where $a^{-1}$ is a modular multiplicative inverse. Multiplication by a nonzero square also leaves the Legendre symbol unchanged: $(a{r}^2/p)=(a/p)$ for $p\nmid r$.^[ref]

The subgroup and coset arguments under quadratic residue and quadratic nonresidue prove the sign rule without computing a large power: square times square is square, square times nonsquare is nonsquare, and nonsquare times nonsquare is square. If either factor is zero modulo $p$, both sides of the identity vanish. On ${\mathbb{F}}_p^{\times }$, the Legendre symbol is thus a group homomorphism to $\{1,-1\}$ whose kernel of a group homomorphism is the quadratic residue subgroup. Viewed on all integers, it is a completely multiplicative function and a Dirichlet character modulo $p$.

For an original calculation, factor the numerator $30=2\cdot 3\cdot 5$. Modulo seven, $(30/7)=(2/7)(3/7)(5/7)=1\cdot (-1)\cdot (-1)=1$, agreeing with $30\equiv 2(\mathrm{mod}7)$. Modulo eleven, the signs are instead $-1,1,1$, giving $(30/11)=-1$, agreeing with $30\equiv 8(\mathrm{mod}11)$. The same factorization can therefore certify different square behavior at different prime numbers.

Alternatively, Euler's criterion gives $(ab{)}^{(p-1)/2}\equiv a^{(p-1)/2}{b}^{(p-1)/2}$ modulo $p$. For nonzero factors both sides correspond to signs in $\{1,-1\}$. Since $p$ is odd, these signs are distinct modulo $p$, so the modular congruence proves the asserted equality of integers, not merely an equality of representatives.

For positive odd coprime integers $m,n$, the Jacobi reciprocity law isFactor $a={\prod }_j{q}_j^{e_j}$. Since $a$ is not a square, some $e_j$ is odd. Choose a quadratic nonresidue modulo that $q_j$, and choose residue one modulo each other $q_i$. The Chinese remainder theorem provides a positive $n$ with these residues and with $n\equiv 1(\mathrm{mod}4)$. Multiplicativity of the Jacobi symbol then gives $(n/a)=-1$.

Suppose only finitely many primes $p_1,\dots ,p_s$ satisfy $(a/p_i)=-1$. They are coprime to $a$. Repeat the preceding Chinese remainder theorem construction, additionally imposing $n\equiv 1(\mathrm{mod}{p}_i)$ at every odd $p_i$ and taking $n\equiv 1(\mathrm{mod}4)$ to avoid two. Since $n\equiv 1(\mathrm{mod}4)$, Jacobi reciprocity gives $(a/n)=(n/a)=-1$. Factoring this Jacobi symbol over the prime factors of $n$ shows that at least one such prime $p$ has $(a/p)=-1$. But none of the listed primes divides $n$, a contradiction. This proves the infinitely many prime quadratic nonresidues of a nonsquare integer assertion.

Euler's criterion gives $\chi (-1)=(-1{)}^{(p-1)/2}=1$ in this case. There are equally many nonzero quadratic residues and quadratic nonresidues, so ${\sum }_{n=1}^{p-1}\chi (n)=0$. Replacing $n$ by $p-n$ and using the multiplicativity of the Legendre symbol givesConsequently $\boxed{S=0}$.

Factor $n={\prod }_j{q}_j^{e_j}$. Since it is not a square, one exponent, say $e_1$, is odd. Choose a nonzero quadratic nonresidue $a$ modulo $q_1$; one exists because the squaring map on the nonzero residues is two-to-one and hence has only $(q_1-1)/2$ images. The Chinese remainder theorem supplies a positive odd integer $m>1$ satisfyingThen $\gcd (m,n)=1$ and $(m/n)=(-1{)}^{e_1}=-1$. The Jacobi reciprocity law proved in part (i), together with $m\equiv 1(\mathrm{mod}4)$, gives $(n/m)=-1$. Factoring $m={\prod }_{\ell }{p}_{\ell }^{r_{\ell }}$ givesAll $p_{\ell }$ are odd and none divides $n$. At least one factor must be negative, yielding an odd prime number $\boxed{p\text{ with }(n/p)=-1}$. No theorem about primes in arithmetic progressions is needed.

Suppose $x$ is a quadratic nonresidue. On the nonzero residue classes modulo $p$, define the involutionIt is an involution because $T(T(a))=a$. A fixed point would satisfy $a^2=x$, which is excluded by the quadratic nonresidue hypothesis. Thus it partitions the $p-1$ classes into $(p-1)/2$ disjoint pairs $\{a,x{a}^{-1}\}$, each with product $x$. ConsequentlyNow evaluate the same product by pairing every class with its own modular inverse. The only classes equal to their modular inverses solve $(a-1)(a+1)=0$ and hence are $1$ and $-1$. All other pairs have product one. The complete product is therefore $-1$, soTogether with the preceding part, this is Euler's criterion; the pairing argument establishes the converse without assuming it.

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.

True. Since $N$ is not a square, some exponent $e_i$ is odd. Choose a quadratic nonresidue modulo that $p_i$ and choose residue one at every other prime. The Chinese remainder theorem produces a unit $b$ modulo $N$ with $(b/N)=-1$. Multiplication by $b$ permutes the residue classes modulo $N$. Thus, by multiplicativity of the Jacobi symbol,The sum is therefore $\boxed{0}$. Nonunits contribute zero, so the permutation argument covers every term.

True. First suppose $p^e\mid N$ with $e\ge 2$ maximal. By the Chinese remainder theorem, choose $a\equiv 1+p^{e-1}(\mathrm{mod}{p}^e)$ and $a\equiv 1$ modulo all other prime-power factors. This is a unit, with Jacobi symbol one, because it is one modulo every prime. For $m=(N-1)/2$, the binomial theorem givesbecause $p\nmid m$ and all higher terms contain $p^e$. It is therefore a witness to the desired failure.

If $N$ is square-free and composite, it has at least two prime factors. Choose $a$ to be a quadratic nonresidue at one prime $p$ and one at every other prime, again using the Chinese remainder theorem. Then $(a/N)=-1$. At a different odd prime $q\mid N$, however, $a^m\equiv 1\not\equiv -1(\mathrm{mod}q)$. These two cases prove that every odd composite has an Euler-Jacobi witness, unlike the situation for the ordinary Fermat primality test.

If $p\mid D$, square-freeness gives the cuspidal reduction $y^2=x^3$. Its nonsingular group is isomorphic to the additive group ${\mathbb{F}}_p$, so it is cyclic of order $p$.

Suppose $p\nmid D$. The reduction is an elliptic curve. Since $p\equiv 3(\mathrm{mod}4)$, $-1$ is a quadratic nonresidue. Pairing $x$ with $-x$ in the quadratic-character sum and usingshows that the two contributions cancel. Therefore $\#\tilde{E}({\mathbb{F}}_p)=p+1$. The three roots $0,D,-D$ are distinct, so the full 2-torsion is rational. A cyclic group has at most two elements killed by $2$, hence $\tilde{E}({\mathbb{F}}_p)$ is noncyclic. Thus

If the positive odd integer $n$ has prime factorizationits Jacobi symbol iswhere each factor is a Legendre symbol; by convention $(a/1)=1$. Thus the symbol is zero exactly when $\gcd (a,n)>1$.

The quadratic reciprocity law for the Jacobi symbol says that for coprime positive odd integers $m,n$,To prove it, write $m={\prod }_i{p}_i^{e_i}$ and $n={\prod }_j{q}_j^{f_j}$. The quadratic reciprocity law for distinct odd primes statesMultiplying this identity over $i,j$ with multiplicities $e_i{f}_j$ gives the required product of Jacobi symbols. Its sign exponent modulo two isbecause multiplication of odd integers adds the quantities $(r-1)/2$ modulo two. This proves the law.

For the requested value, two applications of reciprocity givesince $2019\equiv 7(\mathrm{mod}503)$, $503\equiv 6(\mathrm{mod}7)$, and $6$ is a quadratic nonresidue modulo $7$.

Write $\chi (x)=(x/p)$. Choose a quadratic nonresidue $a$. Multiplication by $a$ permutes ${\mathbb{F}}_p$, while multiplicativity of the Legendre symbol gives $\chi (ax)=-\chi (x)$. Thereforeso the complete Legendre-symbol sum is zero.

Since $4x(x+1)=(2x+1{)}^2-1$ and $\chi (4)=1$, the bijection $x\mapsto 2x+1$ reduces the next sum to ${\sum }_y\chi (y^2-1)$. Count pairs $(y,t)\in {\mathbb{F}}_p^2$ satisfying $t^2=y^2-1$. On one hand their number isOn the other hand $(y-t)(y+t)=1$, and every $u\in {\mathbb{F}}_p^{\times }$ gives exactly one pair by $y-t=u$, $y+t=u^{-1}$. There are therefore $p-1$ pairs, proving the quadratic Legendre-symbol correlation

For $1\le z\le p-2$, the indicator that both $z$ and $z+1$ are quadratic residues is $(1+\chi (z))(1+\chi (z+1))/4$. Summing and using the two identities above gives the number of consecutive nonzero quadratic residues modulo an odd prime

Put $e=(N-1)/2$. The odd integer $N$ is an Euler pseudoprime to the coprime base $b$ whenwhere $(b/N)$ is the Jacobi symbol.

Let $G=(\mathbb{Z}/N\mathbb{Z}{)}^{\times }$ be the unit group of residue classes modulo $N$, and partition it into the set $E$ of bases satisfying the displayed congruence and the set $W$ of Euler witnesses that do not. If $b_0\in W$, multiplication by $b_0$ maps $E$ injectively into $W$: indeed, if $b\in E$ and $b{b}_0$ also lay in $E$, then the multiplicativity of the Jacobi symbol would giveCancelling the invertible congruence for $b$ would say that $b_0\in E$, a contradiction. Hence $|W|\ge |E|$, so at least half of the coprime bases are witnesses.

It remains to construct one witness for every odd composite number $N$. If $N$ is not squarefree, choose an odd prime number $p$ with $p^2\mid N$. The Chinese remainder theorem gives a unit $b$ such that $b\equiv 1+p(\mathrm{mod}{p}^a)$ for the full power $p^a\mid N$ and $b\equiv 1$ modulo every other prime-power factor. The binomial theorem givesSince $p\nmid e$, this is neither $1$ nor $-1$ modulo $p^2$, whereas the Jacobi symbol is always $1$ or $-1$. Thus $b$ is a witness.

If $N$ is squarefree and composite, choose distinct primes $p,q\mid N$. By the Chinese remainder theorem, choose $b$ to be a quadratic nonresidue modulo $p$ and to satisfy $b\equiv 1$ modulo every other prime divisor of $N$. Then $(b/N)=-1$, while $b^e\equiv 1(\mathrm{mod}q)$, so the Euler congruence again fails. This proves the existence of an Euler witness for every odd composite integer.

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

Let $\alpha =\sqrt[3]{2}$. Its minimal polynomial is $f(X)=X^3-2$, whose discriminant is $-108$. Since $5\nmid 108$, the prime $5$ does not divide the index $[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]]$, so the Dedekind factorization theorem applies at $5$. In ${\mathbb{F}}_5[X]$,The quadratic factor has discriminant $3$, which is a quadratic nonresidue modulo $5$, so it is irreducible. ThereforewhereTheir residue-field degrees are one and two. Both factors of $f$ occur with multiplicity one, so both prime-ideal exponents are one. Hence neither ${\mathfrak{p}}_1$ nor ${\mathfrak{p}}_2$ is ramified.

Let $R$ be the sum of the nonzero quadratic residues. Each has two square roots by part (iii), sowhere part (ii) applies because $p>3$ makes $p-1$ not divide $2$. Since $2$ is invertible modulo $p$, $R\equiv 0$.

The sum of all nonzero classes is $p(p-1)/2\equiv 0(\mathrm{mod}p)$. Subtracting $R$ shows that the sum of quadratic nonresidues is also zero. Consequently both requested sums vanish modulo $p$.