Difference between revisions of "Wilson's Theorem"

Latest revision as of 00:53, 2 February 2023

In number theory, Wilson's Theorem states that if integer $p > 1$ , then $(p-1)! + 1$ is divisible by $p$ if and only if $p$ is prime. It was stated by John Wilson. The French mathematician Lagrange proved it in 1771.

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 $(p-1)! + 1$ .

Two proofs of the converse are provided: 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 congruent to $1 \bmod p$ , 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 $P(x) = x^{p-1} - 1 .$ Since this field has only $p-1$ nonzero elements, it follows that $x^{p-1} - 1 = \prod_{r=1}^{p-1}(x-r) .$ 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 $x^{p-1} - 1 = \prod_{r=1}^{p-1}(x-r) = \prod_{r=1}^{p-1}(-x + r) .$ If we set $x$ equal to 0, the theorem follows. $\blacksquare$

Problems

Introductory

(Source: ARML 2002) 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 $13$ .

Solution

Multiplying both sides by $23!$ yields $\frac{23!}{1}+\frac{23!}{2}+...+\frac{23!}{23}=a$ Note that $13\mid\frac{23!}{k}$ for all $k\neq13$ . Thus we are left with $a\equiv\frac{23!}{13}\equiv12!\cdot14\cdot15\cdot16\cdot...\cdot23\equiv(-1)(1)(2)(3)(...)(10)\equiv\boxed{7}\mod13$

Advanced

If ${p}$ is a prime greater than 2, define $p=2q+1$ . Prove that $(q!)^2 + (-1)^q$ is divisible by ${p}$ . Solution.

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 7: / Line 7: @@
 Two proofs of the converse are provided: an elementary one that rests close to basic principles of [[modular arithmetic]], and an elegant method that relies on more powerful [[algebra]]ic tools.
-==Elementary proof ==
+===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
+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 congruent to <math>1 \bmod p</math>, we have
 <cmath> (p-1)! \equiv 1\cdot 1 \cdot (-1) \equiv -1 \pmod{p}, </cmath>
 as desired.  <math>\blacksquare</math>
@@ Line 18: / Line 18: @@
 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>
+<cmath> P(x) = x^{p-1} - 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>
+<cmath> x^{p-1} - 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>
+<cmath> x^{p-1} - 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 ==
 === Introductory ===
-* Let <math>a</math> be an integer such that <math>\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{23}=\frac{a}{23!}</math>. Find the remainder when <math>a</math> is divided by <math>13</math>.
+* (Source: ARML 2002) Let <math>a</math> be an integer such that <math>\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{23}=\frac{a}{23!}</math>. Find the remainder when <math>a</math> is divided by <math>13</math>.
+==== Solution ====
+Multiplying both sides by <math>23!</math> yields
+<cmath>\frac{23!}{1}+\frac{23!}{2}+...+\frac{23!}{23}=a</cmath>
+Note that <math>13\mid\frac{23!}{k}</math> for all <math>k\neq13</math>. Thus we are left with
+<cmath>a\equiv\frac{23!}{13}\equiv12!\cdot14\cdot15\cdot16\cdot...\cdot23\equiv(-1)(1)(2)(3)(...)(10)\equiv\boxed{7}\mod13</cmath>
 === 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>. <url>Forum/viewtopic.php?&t=21733 Solution</url>
+* 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>. [https://artofproblemsolving.com/community/c6h21733 Solution.]
 * 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>.
@@ Line 43: / Line 50: @@
 * [[Wilson Prime]]
-[[Category:Number Theory]]
+[[Category:Number theory]]

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