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Proof

(related to Theorem: Properties of the Jacobi Symbol)

By hypothesis, $n,m$ are odd positive integers, while $a,b$ are any integers. Moreover, in the following, we assume that $n=p_1\cdots p_r$ and $m=q_1\cdots q_s$ are the factorizations of $n$ and $m,$ in which we allow the prime numbers $p_i$ to repeat, as $i$ runs through the values $1,\ldots,r.$ The same holds for the prime numbers $q_j$ for $j=1,\ldots, s.$ This will keep the formulae below clearer and simpler, but please keep this in mind when reading the proof below.

Ad 1

(Generalization of Legendre symbols of equal residues)

• Let the congruence $a(n)\equiv b(n)$ be fulfilled.
• From congruence modulo a divisor, it follows that $a(p_i)\equiv b(p_i)$ for every prime dividing $n.$
• According to Legendre symbols of equal residues, it follows that $\left(\frac{a}{p_i}\right)=\left(\frac{b}{p_i}\right)$ for each Legendre symbol.
• By the definition of the Jacobi symbol, it follows that $\left(\frac{a}{n}\right)=\left(\frac{b}{n}\right).$

Ad 2

(Generalization of the multiplicativity of the Legendre symbol)

• For every prime $p_i$ dividing $n$ it follows from the multiplicativity of the Legendre symbol that $\left(\frac{ab}{p_i}\right)=\left(\frac{a}{p_i}\right)\cdot \left(\frac{b}{p_i}\right).$
• By the definition of the Jacobi symbol, it follows that $\left(\frac{ab}{n}\right)=\left(\frac{a}{n}\right)\cdot \left(\frac{b}{n}\right).$

Ad 3

• Case 1) $a$ is co-prime to $n$ and $m.$
  • By the definition of the Jacobi symbol, it follows $$\left(\frac{a}{n}\right)\cdot \left(\frac{a}{m}\right)=\prod_{i=1}^r\left(\frac{a}{p_i}\right)\cdot \prod_{j=1}^s\left(\frac{a}{q_j}\right)=\left(\frac{a}{nm}\right),$$ since if a prime $p$ divides both numbers $n$ and $m$ at once, then its multiplicities sum up in both products, otherwise the multiplicities remain the same if a prime either divides the number $n$ or the number $m.$
• Case 2) $a$ is not co-prime to $n$ or not co-prime to $m$.
  • Therefore, $a$ is not co-prime to $nm$, and we have $\left(\frac{a}{nm}\right)=0.$
  • But then, either $\left(\frac{a}{n}\right)=0,$ or $\left(\frac{a}{m}\right)=0,$ or both.
  • Therefore, again $\left(\frac{a}{n}\right)\cdot \left(\frac{a}{m}\right)=\left(\frac{a}{nm}\right).$

Ad 4

(Generalization of the quadratic reciprocity law)

• By hypothesis, $n$, and $m$ are co-prime.
• Therefore, all prime numbers $p_i$ are distinct from the prime numbers $q_j.$
• By the definition of the Jacobi symbol, we have $$\left(\frac{n}{m}\right)\cdot \left(\frac{m}{n}\right)=\prod_{j=1}^s\left(\frac{n}{q_j}\right)\cdot \prod_{i=1}^r \left(\frac{m}{p_i}\right)=\prod_{i=1}^r\prod_{j=1}^s\left(\frac{p_i}{q_j}\right)\cdot \left(\frac{q_j}{p_i}\right).$$
• Applying the quadratic reciprocity law to the right side of the equation, we get $\left(\frac{n}{m}\right)\cdot \left(\frac{m}{n}\right)=(-1)^U$ with a sum $$U=\sum_{i=1}^r\sum_{j=1}^s\frac{p_i-1}{2}\frac{q_j-1}{2}.$$
• By the generalized distributivity rule, we get $$U=\left(\sum_{i=1}^r\frac{p_i-1}{2}\right)\cdot\left(\sum_{j=1}^s\frac{q_j-1}{2}\right).$$
• For the theorem, it is sufficient to show that $U\equiv\frac{n-1}2 \frac{m-1}2\mod 2.$
• We get this result by applying a general rule repeatedly to all primes $p_i$ and $q_j$ which are all odd. The rule states that for arbitrary two odd integers $a,b$ we have $$\frac{a-1}2 +\frac{b-1}2\equiv \frac{ab-1}2\mod 2.\label{eq:E18802}\tag{1}$$
• To complete the proof, we note that $$0\equiv\frac{(a-1)(b-1)}2\equiv \frac{ab-a-b+1}2\equiv\frac{ab-1}2-\left(\frac{a-1}2+\frac{b-1}2\right) \mod 2.$$

Ad 5

(Generalization of the first supplementary law to the quadratic reciprocity law)

• By the definition of the Jacobi symbol, and applying the first supplementary law to the quadratic reciprocity law, we get $$\left(\frac{-1}{n}\right)=\prod_{i=1}^r\left(\frac{-1}{p_i}\right)=\prod_{i=1}^r(-1)^{\frac{p_i-1}{2}}=(-1)^{V},$$ with $$V:=\sum_{i=1}^r\frac{p_i-1}{2}.$$
• Since $n$ is odd, we can apply $(\ref{eq:E18802})$ repeatedly, and get $$V\equiv \frac{n-1}{2}\mod 2.$$

Ad 6

(Generalization of the second supplementary law to the quadratic reciprocity law)

• By the definition of the Jacobi symbol, and applying the second supplementary law to the quadratic reciprocity law, we get $$\left(\frac{2}{n}\right)=\prod_{i=1}^r\left(\frac{2}{p_i}\right)=\prod_{i=1}^r(-1)^{\frac{p_i^2-1}{8}}=(-1)^{W},$$ with $$W:=\sum_{i=1}^r\frac{p_i^2-1}{8}.$$
• For the theorem, it is sufficient to show that $W\equiv\frac{n^2-1}8 \mod 2.$
• We get this result by applying a general rule repeatedly to all primes $p_i$ which are all odd. The rule states that for arbitrary two odd integers $a,b$ we have $$\frac{a^2-1}8 +\frac{b^2-1}8\equiv \frac{(ab)^2-1}8\mod 2.$$
• To complete the proof, we note that $$0\equiv\frac{(a^2-1)(b^2-1)}8\equiv \frac{(ab)^2-a^2-b^2+1}8\equiv\frac{(ab)^2-1}8-\left(\frac{a^2-1}8+\frac{b^2-1}8\right) \mod 2.$$
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References

Bibliography

1. Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927
2. Blömer, J.: "Lecture Notes Algorithmen in der Zahlentheorie", Goethe University Frankfurt, 1997