Wilson's Theorem

时间:2020-01-15 00:09:23   收藏:0   阅读:142

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 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 \[P(x) = x^{p-1} - 1 .\] 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. $\blacksquare$

(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 $13$.
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 \[a\equiv\frac{23!}{13}\equiv12!\cdot14\cdot15\cdot16\cdot...\cdot23\equiv(-1)(1)(2)(3)(...)(10)\equiv\boxed{7}\mod13\]
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 ${p}$.


For how many integers $n$ between $1$ and $50$, inclusive, is \[\frac{(n^2-1)!}{(n!)^n}\] an integer? (Recall that $0! = 1$.)

$\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35$

Solution 1
The main insight is that

is always an integer. This is true because it is precisely the number of ways to split up $n^2$ objects into $n$ unordered groups of size $n$. Thus,

is an integer if $n^2 \mid n!$, or in other words, if $n \mid (n-1)!$. This condition is false precisely when $n=4$ or $n$ is prime, by Wilson‘s Theorem. There are $15$ primes between $1$ and $50$, inclusive, so there are 15 + 1 = 16 terms for which

is potentially not an integer. It can be easily verified that the above expression is not an integer for $n=4$ as there are more factors of 2 in the denominator than the numerator. Similarly, it can be verified that the above expression is not an integer for any prime $n=p$, as there are more factors of p in the denominator than the numerator. Thus all 16 values of n make the expression not an integer and the answer is $50-16=\boxed{\mathbf{(D)}\ 34}$.

Solution 2
We can use the P-Adic Valuation of n to solve this problem (recall the P-Adic Valuation of ‘n‘ is denoted by $v_p (n)$ and is defined as the greatest power of some prime ‘p‘ that divides n. For example, $v_2 (6)=1$ or $v_7 (245)=2$ .) Using Legendre‘s formula, we know that :

\[v_p (n!)= \sum_{i=1}^\infty \lfloor \frac {n}{p^i} \rfloor\]
Seeing factorials involved in the problem, this prompts us to use Legendre‘s formula where n is a power of a prime.

We also know that , $v_p (m^n) = n \cdot v_p (m)$ . Knowing that $a\mid b$ if $v_p (a) \le v_p (b)$ , we have that :

\[n \cdot v_p (n!) \le v_p ((n^2 -1 )!)\] and we must find all n for which this is true.

If we plug in $n=p$, by Legendre‘s we get two equations:

\[v_p ((n^2 -1)!) = \sum_{i=1}^\infty \lfloor \frac {n^2 -1}{p^i} \rfloor = (p-1)+0+...+0 = p-1\]
And we also get :

\[v_p ((n!)^n) = n \cdot v_p (n!)= n \cdot \sum_{i=1}^\infty \lfloor \frac {n}{p^i} \rfloor = p \cdot ( 1+0+...0) = p\]
But we are asked to prove that $n \cdot v_p (n!) \le v_p ((n^2 -1 )!) \Longrightarrow p \le p-1$ which is false for all ‘n‘ where n is prime.

Now we try the same for $n=p^2$ , where p is a prime. By Legendre we arrive at:

\[v_p ((p^4 -1)!) = p^3 + p^2 + p -3\] and \[p^2 \cdot v_p (p^2 !) = p^3 + p^2\]
Then we get:

\[p^2 \cdot v_p (p!) \le v_p ((n^4 -1)!) \Longrightarrow p^3 + p^2 \le p^3 + p^2 + p -3\] Which is true for all primes except for 2, so $2^2 = 4$ doesn‘t work. It can easily be verified that for all $n=p^i$ where $i$ is an integer greater than 2, satisfies the inequality :\[n \cdot v_p (n!) \le v_p ((n^2 -1 )!)\].

Therefore, there are 16 values that don‘t work and $50-16 = \boxed{\mathbf{(D)}\ 34}$ values that work.

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