Euler's Criterion

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Theorem

Let $p$ be an odd prime.

Let $a \not \equiv 0 \pmod p$.

Then:

\(\displaystyle a^{\frac {p-1} 2}\) \(\equiv\) \(\displaystyle 1\) \(\displaystyle \pmod p\) if and only if $a$ is a quadratic residue of $p$
\(\displaystyle a ^{\frac {p-1} 2}\) \(\equiv\) \(\displaystyle -1\) \(\displaystyle \pmod p\) if and only if $a$ is a quadratic non-residue of $p$.


Proof 1

Trivially, any $a \not \equiv 0 \pmod p$ is either a quadratic residue or a quadratic non-residue, modulo $p$.

Therefore, it suffices to check the sufficient condition for both of the equations (i.e., the if parts from the iffs).


So let $a$ be a quadratic non-residue of $p$.

Also, let $b \in \set {1, 2, \ldots, p - 1}$.

The congruence $b x \equiv a \pmod p$ has (modulo $p$) a unique solution $b'$ by Solution of Linear Congruence.

Note that $b' \not\equiv b$, because otherwise we would have $b^2 \equiv a \pmod p$ and $a$ would be a quadratic residue of $p$.

It follows that the residue classes $\set {1, 2, \ldots, p - 1}$ modulo $p$ fall into $\dfrac {p - 1} 2$ pairs $b, b'$ such that $b b' \equiv a \pmod p$.

Therefore, we have:

$\paren {p - 1}! = 1 \times 2 \times \cdots \times \paren {p - 1} \equiv a \times a \times \cdots \times a \equiv a^{\frac {p - 1} 2} \pmod p$

From Wilson's Theorem, we also have:

$\paren {p - 1}! \equiv -1 \pmod p$

And so, for any quadratic non-residue of $p$:

$a^{\frac {p - 1} 2} \equiv -1 \pmod p$


Subsequently, let $a$ be a quadratic residue of $p$.

By definition of a quadratic residue, the congruence $x^2 \equiv a \pmod p$ has a solution $x$.

Suppose also $y$ is a solution. Then we have:

\(\displaystyle x^2 - y^2\) \(\equiv\) \(\displaystyle 0\) \(\displaystyle \pmod p\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \paren {x + y} \paren {x - y}\) \(\equiv\) \(\displaystyle 0\) \(\displaystyle \pmod p\)


So either $x + y \equiv 0 \pmod p$ or $x - y \equiv 0 \pmod p$ from Product is Zero Divisor means Zero Divisor.

It follows that when $c \pmod p$ is one solution, $-c \pmod p$ is, too.

Also, these solutions are distinct as $p$ is odd.

Furthermore, we conclude that these are the only two solutions.


Now, remove $c$ and $p - c$ from $\set {1, 2, \ldots, p - 1}$.

The remaining integers fall, modulo $p$, into $\dfrac {p - 3} 2$ pairs $b, b'$ such that $b b' \equiv a \pmod p$ by Solution of Linear Congruence.

Therefore, we can compute the following:

\(\displaystyle \paren {p - 1}!\) \(=\) \(\displaystyle 1 \times 2 \times \cdots \times c \times \cdots \times \paren {p - c} \times \cdots \times \paren {p - 1}\)
\(\displaystyle \) \(\equiv\) \(\displaystyle a \times a \times \cdots \times a \times c \times \paren {-c}\) \(\displaystyle \pmod p\)
\(\displaystyle \) \(\equiv\) \(\displaystyle a^{\frac {p - 3} 2} \paren {-c^2}\) \(\displaystyle \pmod p\)
\(\displaystyle \) \(\equiv\) \(\displaystyle a^{\frac {p - 3} 2} \paren {-a}\) \(\displaystyle \pmod p\)
\(\displaystyle \) \(\equiv\) \(\displaystyle -a^{\frac {p - 1} 2}\) \(\displaystyle \pmod p\)


Again applying Wilson's Theorem, we conclude:

$-a^{\frac {p-1} 2} \equiv - 1 \pmod p$

The assertion for quadratic residues $a$ of $p$ follows.

$\blacksquare$


Proof 2

First note that the square roots of $1$ are $1, -1 \pmod p$.

Also, we have that $a^{p - 1} \equiv 1 \pmod p$ by Fermat's Little Theorem.

Combining these two observations, we find:

$a^{\frac {p - 1} 2} \equiv 1 \text{ or } -1 \pmod p$


The theorem is therefore equivalent to stating that $a$ is a quadratic residue modulo $p$ if and only if $a^{\frac{p - 1} 2} \equiv 1 \pmod p$.

Namely, considering the above, we see this also implies that all quadratic non-residues will be congruent to $-1 \pmod p$.


We prove each direction of the equivalent statement separately:


Sufficient Condition

Assume $a$ is a quadratic residue modulo $p$.

We pick $k$ such that $k^2 \equiv a \pmod p$.

Then by Congruence of Powers and Fermat's Little Theorem:

$a^{\frac{p-1} 2} \equiv k^{p-1} \equiv 1 \pmod p$


Necessary Condition

Now assume $a^{\frac{p - 1} 2} \equiv 1 \pmod p$.

Then let $y$ be a primitive root modulo $p$, so that $a$ can be written as $y^j$.

In particular:

$y^{j \frac {p - 1} 2} \equiv 1 \pmod p$

From the definition of $y$, it has order $p-1$.

It follows that $p - 1 \divides j \dfrac {p - 1} 2$ from Element to Power of Multiple of Order is Identity.

We conclude that $j$ is necessarily an even integer, and denote $j' = \dfrac j 2$.

Let $k$ be such that $k \equiv y^{j'} \pmod p$.

By construction, we have:

$k^2 \equiv y^{2 j'} \equiv y^j \equiv a \pmod p$

Hence $a$ is a quadratic residue modulo $p$.

$\blacksquare$


Source of Name

This entry was named for Leonhard Paul Euler.