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 conguent to 1 modulo $p$, we have [(p-1)! equiv 1cdot 1 cdot (-1) equiv -1 pmod{p},] as desired. $lacksquare$
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. $lacksquare$
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 $13midfrac{23!}{k}$ for all $k
eq13$. Thus we are left with [aequivfrac{23!}{13}equiv12!cdot14cdot15cdot16cdot...cdot23equiv(-1)(1)(2)(3)(...)(10)equivoxed{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 ${p}$.
Problem
For how many integers $n$ between $1$ and $50$, inclusive, is [frac{(n^2-1)!}{(n!)^n}] an integer? (Recall that $0! = 1$.)
$ extbf{(A) } 31 qquad extbf{(B) } 32 qquad extbf{(C) } 33 qquad extbf{(D) } 34 qquad extbf{(E) } 35$
Solution 1
The main insight is that
[frac{(n^2)!}{(n!)^{n+1}}]
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,
[frac{(n^2-1)!}{(n!)^n}=frac{(n^2)!}{(n!)^{n+1}}cdotfrac{n!}{n^2}]
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
[frac{(n^2-1)!}{(n!)^{n}}]
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 $50-16=oxed{mathbf{(D)} 34}$.
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}
floor]
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 $amid b$ 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}
floor = (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}
floor = 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 $50-16 = oxed{mathbf{(D)} 34}$ values that work.