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Suppose $X^2\equiv -1(\mathrm{mod}p)$. Then $X$ is a unit modulo n for modulus $p$, and Fermat's little theorem impliesSince $p$ is an odd prime number, $1$ and $-1$ are distinct residue classes. Therefore $(p-1)/2$ is even, and $p\equiv 1(\mathrm{mod}4)$.

Conversely, put $q=(p-1)/2$. Pair each integer $j\in \{1,\dots ,q\}$ with $p-j$ in the factorial. Wilson theorem yieldsWhen $p\equiv 1(\mathrm{mod}4)$, $q$ is even, so $X=q!$ satisfies $X^2\equiv -1(\mathrm{mod}p)$. This proves the first supplementary law for quadratic reciprocity without assuming a primitive root. HenceThe two solutions are $q!$ and $-q!$ modulo $p$: their difference is nonzero, and factoring $X^2-Y^2$ shows that a degree-two polynomial over the field of residues modulo $p$ has no further roots.

Gauss lemma states that if $p\nmid a$ and $m$ of the least positive residues ofexceed $p/2$, then $(a/p)=(-1{)}^m$.

To prove it, replace each residue exceeding $p/2$ by its negative, obtaining numbers $r_1,\dots ,r_{(p-1)/2}$ in $\{1,\dots ,(p-1)/2\}$. No two of them are equal modulo $p$: such an equality would give $ia\equiv \pm ja(\mathrm{mod}p)$, hence $i=j$ in the allowed range. They therefore permute $1,\dots ,(p-1)/2$, and multiplication givesCancel the nonzero factorial and apply Euler's criterion to obtain the lemma.

For $a=-1$, all $(p-1)/2$ least positive residues $p-1,p-2,\dots ,(p+1)/2$ exceed $p/2$. Hence the first supplementary law for quadratic reciprocity is

For an odd prime $p$, the Legendre symbol isEuler criterion states thatIf $p\mid a$, both sides vanish. Otherwise choose a primitive root $g$ and write $a\equiv g^m$. Since $g^{(p-1)/2}\equiv -1$,The power $g^m$ is a square exactly when $m$ is even, proving the criterion. Taking $a=-1$ gives the first supplementary law for quadratic reciprocity

Write $\chi (a)=(\frac{a}{p})$. When $p\equiv 1(\mathrm{mod}4)$, the first supplementary law for quadratic reciprocity gives $\chi (-1)=1$, so $\chi (p-a)=\chi (a)$. Pairing $a$ with $p-a$ givesThe sum of the nontrivial character $\chi$ over all nonzero residue classes is zero. Because its two half-sums are equal when $\chi (-1)=1$,Hence

The first supplementary law for quadratic reciprocity isIt also follows immediately from Euler's criterion applied to $-1$.

The first supplementary law for quadratic reciprocity isIt also follows immediately from Euler's criterion applied to $-1$.

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 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

Quadratic reciprocity compares two questions: is the prime number $p$ a quadratic residue modulo $q$, and is $q$ a quadratic residue modulo $p$? Let $p,q$ be distinct odd primes. In terms of the Legendre symbol,The two Legendre symbols have opposite signs exactly when both primes are $3$ modulo $4$; otherwise they are equal. Their values are nonzero because the primes are distinct. Equal signs can mean two quadratic residues or two quadratic nonresidues: reciprocity compares the answers rather than declaring either answer positive.^[ref]

For a direct computation, the nonzero quadratic residues modulo $3,5,7$ are respectively $\{1\}$, $\{1,4\}$, and $\{1,2,4\}$. Squaring just $1,\dots ,(p-1)/2$ suffices because $x$ and $p-x$ have the same square modulo an odd prime $p$.The pair $3,7$ has a sign flip, whereas $5,7$ has two negative signs. These small quadratic residue tables make the distinction tangible.

The first supplementary law for quadratic reciprocity handles numerator $-1$, and the second supplementary law for quadratic reciprocity handles numerator $2$:For an odd prime $p$, $-1$ is a quadratic residue exactly when $p\equiv 1(\mathrm{mod}4)$, while $2$ is a quadratic residue exactly when $p\equiv 1,7(\mathrm{mod}8)$. The remaining odd congruence classes modulo $8$, namely $3,5$, make $2$ a quadratic nonresidue. These laws fill the cases excluded from swapping distinct odd primes.^[ref]