Difference between revisions of "Wilson's Theorem"

Revision as of 12:29, 28 March 2009

Wilson's Theorem is a result in elementary number theory. It states that if $p > 1$ is in integer, then $(p-1)! + 1$ is a multiple of $p$ if and only if $p$ is a prime.

Proofs

Suppose first that $p$ is composite. Then $p$ has a factor $d > 1$ that is less than or equal to $p-1$ . Then $d$ divides $(p-1)!$ , so $d$ does not divide $(p-1)! + 1$ . Therefore $p$ does not divide $(p-1)! + 1$ .

The converse of this statement is more interesting. We provide two proofs: an elementary one that rests close to basic principles of modular arithmetic, and an elegant method that relies on more powerful algebraic tools.

Elementary proof

Suppose $p$ is a prime. Then each of the integers $1, \dotsc, p-1$ has an inverse modulo $p$ . (Indeed, if one such integer $a$ does not have an inverse, then for some distinct $b$ and $c$ modulo $p$ , $ab \equiv ac \pmod{p}$ , so that $a(b-c)$ is a multiple of $p$ , when $p$ does not divide $a$ or $b-c$ —a contradiction.) This inverse is unique, and each number is the inverse of its inverse. If one integer $a$ is its own inverse, then $0 \equiv a^2 - 1 \equiv (a-1)(a+1) \pmod{p} ,$ so that $a \equiv 1$ or $a \equiv p-1$ . Thus we can partition the set $\{ 2 ,\dotsc, p-2\}$ into pairs $\{a,b\}$ such that $ab \equiv 1 \pmod{p}$ . It follows that $(p-1)$ is the product of these pairs times $1 \cdot (-1)$ . Since the product of each pair is conguent to 1 modulo $p$ , we have $(p-1)! \equiv 1\cdot 1 \cdot (-1) \equiv -1 \pmod{p},$ as desired. $\blacksquare$

Algebraic proof

Let $p$ be a prime. Consider the field of integers modulo $p$ . By Fermat's Little Theorem, every nonzero element of this field is a root of the polynomial $P(x) = x^p - 1 .$ Since this field has only $p-1$ nonzero elements, it follows that $x^p - 1 = \prod_{r=1}^{p-1}(x-r) .$ Now, either $p=2$ , in which case $a \equiv -a \pmod 2$ for any integer $a$ , or $p-1$ is even. In either case, $(-1)^{p-1} \equiv 1 \pmod{p}$ , so that $x^p - 1 = \prod_{r=1}^{p-1}(x-r) = \prod_{r=1}^{p-1}(-x + r) .$ If we set $x$ equal to 0, the theorem follows. $\blacksquare$

Problems

Introductory

Let $a$ be an integer such that $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{23}=\frac{a}{23!}$ . Find the remainder when $a$ is divided by $13$ .

Advanced

If ${p}$ is a prime greater than 2, define $p=2q+1$ . Prove that $(q!)^2 + (-1)^q$ is divisible by ${p}$ . <url>Forum/viewtopic.php?&t=21733 Solution</url>

Let ${p}$ be a prime number such that dividing ${p}$ by 4 leaves the remainder 1. Show that there is an integer ${n}$ such that $n^2 + 1$ is divisible by ${p}$ .

@@ Line 1: / Line 1: @@
-== Statement ==
+'''Wilson's Theorem''' is a result in elementary [[number theory]].  It states that if <math>p > 1</math> is in [[integer]], then <math>(p-1)! + 1</math> is a multiple of <math>p</math> if and only if <math>p</math> is a prime.
-If and only if <math>{p}</math> is a [[prime]], then <math>(p-1)! + 1</math> is a multiple of <math>{p}</math>.  In other words <math>(p-1)! \equiv -1 \pmod{p}</math>.
-== Proof ==
+== Proofs ==
-Wilson's theorem is easily verifiable for 2 and 3, so let's consider <math>p>3</math>.  If <math>{p}</math> is composite, then its positive factors are among
-<center><math>1, 2, 3, \dots, p-1</math>.</center>  <br>
-Hence, <math>\gcd( (p - 1)!, p) > 1</math>, so  <math>(p-1)! \neq -1 \pmod{p}</math>.
-However, if <math>{p}</math> is prime, then each of the above integers are [[relatively prime]] to <math>{p}</math>. So, for each of these integers a, there is another <math>b</math> such that <math>ab \equiv 1 \pmod{p}</math>. It is important to note that this <math>b</math> is unique [[modulo]] <math>{p}</math>, and that since <math>{p}</math> is prime, <math>a = b</math> if and only if <math>{a}</math> is <math>1</math> or <math>p-1</math>. Now, if we omit 1 and <math>p-1</math>, then the others can be grouped into pairs whose product is congruent to one,
+Suppose first that <math>p</math> is composite.  Then <math>p</math> has a factor <math>d > 1</math> that is less than or equal to <math>p-1</math>.  Then <math>d</math> divides <math>(p-1)!</math>, so <math>d</math> does not divide <math>(p-1)! + 1</math>.  Therefore <math>p</math> does not divide <math>(p-1)! + 1</math>.
-<center><math>2\cdot3\cdot4\cdots(p-2) \equiv 1\pmod{p}</math></center>  <br>
-Finally, multiply this equality by <math>p-1</math> to complete the proof.
+The converse of this statement is more interesting.  We provide two proofs: an elementary one that rests close to basic principles of [[modular arithmetic]], and an elegant method that relies on more powerful algebraic tools.
+=== Elementary proof ===
+Suppose <math>p</math> is a prime.  Then each of the integers <math>1, \dotsc, p-1</math> has an inverse modulo <math>p</math>.  (Indeed, if one such integer <math>a</math> does not have an inverse, then for some distinct <math>b</math> and <math>c</math> modulo <math>p</math>, <math>ab \equiv ac \pmod{p}</math>, so that <math>a(b-c)</math> is a multiple of <math>p</math>, when <math>p</math> does not divide <math>a</math> or <math>b-c</math>&mdash;a contradiction.)  This inverse is unique, and each number is the inverse of its inverse.  If one integer <math>a</math> is its own inverse, then
+<cmath> 0 \equiv a^2 - 1 \equiv (a-1)(a+1) \pmod{p} , </cmath>
+so that <math>a \equiv 1</math> or <math>a \equiv p-1</math>.  Thus we can partition the set <math>\{ 2 ,\dotsc, p-2\}</math> into pairs <math>\{a,b\}</math> such that <math>ab \equiv 1 \pmod{p}</math>.  It follows that <math>(p-1)</math> is the product of these pairs times <math>1 \cdot (-1)</math>.  Since the product of each pair is conguent to 1 modulo <math>p</math>, we have
+<cmath> (p-1)! \equiv 1\cdot 1 \cdot (-1) \equiv -1 \pmod{p}, </cmath>
+as desired.  <math>\blacksquare</math>
+=== Algebraic proof ===
+Let <math>p</math> be a prime.  Consider the [[field]] of integers modulo <math>p</math>.  By [[Fermat's Little Theorem]], every nonzero element of this field is a root
+of the polynomial
+<cmath> P(x) = x^p - 1 . </cmath>
+Since this field has only <math>p-1</math> nonzero elements, it follows that
+<cmath> x^p - 1 = \prod_{r=1}^{p-1}(x-r) . </cmath>
+Now, either <math>p=2</math>, in which case <math>a \equiv -a \pmod 2</math> for any integer <math>a</math>,
+or <math>p-1</math> is even.  In either case, <math>(-1)^{p-1} \equiv 1 \pmod{p}</math>, so that
+<cmath> x^p - 1 = \prod_{r=1}^{p-1}(x-r) = \prod_{r=1}^{p-1}(-x + r) . </cmath>
+If we set <math>x</math> equal to 0, the theorem follows.  <math>\blacksquare</math>
 == Problems ==
@@ Line 17: / Line 32: @@
 === Advanced ===
-* If <math>{p}</math> is a prime greater than 2, define <math>p=2q+1</math>. Prove that <math>(q!)^2 + (-1)^q</math> is divisible by <math>{p}</math>. [http://www.mathlinks.ro/Forum/viewtopic.php?&t=21733 Solution]
+* If <math>{p}</math> is a prime greater than 2, define <math>p=2q+1</math>. Prove that <math>(q!)^2 + (-1)^q</math> is divisible by <math>{p}</math>. <url>Forum/viewtopic.php?&t=21733 Solution</url>
 * Let <math>{p}</math> be a prime number such that dividing <math>{p}</math> by 4 leaves the remainder 1. Show that there is an integer <math>{n}</math> such that <math>n^2 + 1</math> is divisible by <math>{p}</math>.
 == See also ==
 * [[Number theory]]
 * [[Modular arithmetic]]
+* [[Fermat's Little Theorem]]
 * [[Factorial]]
+[[Category:Number theory]]

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