Law of Quadratic Reciprocity

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Theorem

Let $p$ and $q$ be distinct odd primes.

Then:

$\paren {\dfrac p q} \paren {\dfrac q p} = \paren {-1}^{\dfrac {\paren {p - 1} \paren {q - 1} } 4}$

where $\paren {\dfrac p q}$ and $\paren {\dfrac q p}$ are defined as the Legendre symbol.

An alternative formulation is: $\paren {\dfrac p q} = \begin{cases} \quad \paren {\dfrac q p} & : p \equiv 1 \lor q \equiv 1 \pmod 4 \\ -\paren {\dfrac q p} & : p \equiv q \equiv 3 \pmod 4 \end{cases}$


The fact that these formulations are equivalent is immediate.


This fact is known as the Law of Quadratic Reciprocity, or LQR for short.


First Supplement to Law of Quadratic Reciprocity

$\paren {\dfrac {-1} p} = \paren {-1}^{\paren {p - 1} / 2} = \begin{cases} +1 & : p \equiv 1 \pmod 4 \\ -1 & : p \equiv 3 \pmod 4 \end{cases}$


Second Supplement to Law of Quadratic Reciprocity

$\paren {\dfrac 2 p} = \paren {-1}^{\paren {p^2 - 1} / 8} = \begin{cases} +1 & : p \equiv \pm 1 \pmod 8 \\ -1 & : p \equiv \pm 3 \pmod 8 \end{cases}$


Proof

Let $p$ and $q$ be distinct odd primes.

Consider the rectangle in the $x y$ plane with vertices at $\tuple {0, 0}, \tuple {\dfrac p 2, 0}, \tuple {\dfrac p 2, \dfrac q 2}, \tuple {0, \dfrac q 2}$.

The number of lattice points inside this rectangle is $\dfrac {p - 1} 2 \times \dfrac {q - 1} 2$.


LQR.png


Consider the diagonal from $\tuple {0, 0}$ to $\tuple {\dfrac p 2, \dfrac q 2}$.

This has the equation:

$y = \dfrac q p x$


Aiming for a contradiction, suppose there were a lattice point $\tuple {a, b}$ on the diagonal.

Then:

$b = \dfrac q p a$

But as $p$ and $q$ are distinct primes, this would mean:

$p$ divides $a$

and:

$q$ divides $b$

This means that $\tuple {a, b}$ has to be outside the rectangle.

So there are no lattice points on the diagonal inside the rectangle.


Let $A$ and $B$ be the triangular regions inside the rectangle lying respectively above and below the diagonal.

Let $k \in \Z: 0 < k < \dfrac p 2$.

The number of lattice points in $B$ which lie directly above the point $\tuple {k, 0}$ is:

$\floor {\dfrac q p k}$

where $\floor {\dfrac q p k}$ is the floor of $\dfrac q p k$.

So the total number of lattice points in $B$ is given by:

$\ds N_B = \sum_{k \mathop = 1}^{\frac{p - 1} 2} \floor {\dfrac q p k}$

Let $\map \alpha {q, p}$ be defined as:

$\ds \map \alpha {q, p} = \sum_{k \mathop = 1}^{\frac {p - 1} 2} \floor {\dfrac q p k}$

In the same way, by counting the lattice points to the right of $\tuple {0, k}$, the total number of lattice points in $A$ is given by:

$N_A = \map \alpha {p, q}$


Now the total number of lattice points in the rectangle is $N_A + N_B$.

But this is also equal to:

$\dfrac {p - 1} 2 \times \dfrac {q - 1} 2 = \dfrac {\paren {p - 1} \paren {q - 1} } 4$

So we have that:

$\map \alpha {p, q} + \map \alpha {q, p} = \dfrac {\paren {p - 1} \paren {q - 1} } 4$

The fact that these counts are equal relies upon the fact that there are no lattice points on the diagonal, as demonstrated above.


Now we invoke Eisenstein's Lemma:

$\paren {\dfrac a p} = \paren {-1}^{\map \alpha {a, p} }$

So this gives us:

\(\ds \paren {\frac p q} \paren {\frac q p}\) \(=\) \(\ds \paren {-1}^{\map \alpha {p, q} } \times \paren {-1}^{\map \alpha {q, p} }\)
\(\ds \) \(=\) \(\ds \paren {-1}^{\map \alpha {p, q} + \map \alpha {q, p} }\)
\(\ds \) \(=\) \(\ds \paren {-1}^{\frac {\paren {p - 1} \paren {q - 1} } 4}\)

which is LQR.

$\blacksquare$


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Historical Note

The Law of Quadratic Reciprocity was investigated by Leonhard Paul Euler who stated it imperfectly and failed to find a proof.

Adrien-Marie Legendre first stated it correctly, but his attempted $\text {1785}$ proof of it was incorrect.

Carl Friedrich Gauss first became aware of this result, and found a correct proof. He wrote this up in $\text {1798}$ and published it in his Disquisitiones Arithmeticae in $\text {1801}$.

The proof as given here was found by Ferdinand Eisenstein and published in $\text {1844}$.


Sources

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