Difference between revisions of "2016 USAMO Problems/Problem 2"

Revision as of 03:24, 29 June 2025

Problem

Prove that for any positive integer $k,$ $\left(k^2\right)!\cdot\prod_{j=0}^{k-1}\frac{j!}{\left(j+k\right)!}$ is an integer.

Solution 1

Define $v_p(N)$ for all rational numbers $N$ and primes $p$ , where if $N=\frac{x}{y}$ , then $v_p(N)=v_p(x)-v_p(y)$ , and $v_p(x)$ is the greatest power of $p$ that divides $x$ for integer $x$ . Note that the expression(that we're trying to prove is an integer) is clearly rational, call it $N$ .

$v_p(N)=\sum_{i=1}^\infty \left\lfloor \frac{k^{2}}{p^{i}} \right\rfloor+\sum_{j=0}^{k-1} \sum_{i=1}^\infty \left\lfloor \frac{j}{p^{i}}\right\rfloor-\sum_{j=k}^{2k-1} \sum_{i=1}^\infty \left\lfloor \frac{j}{p^{i}} \right\rfloor$ , by Legendre. Clearly, $\left\lfloor{\frac{x}{p}}\right\rfloor={\frac{x-r(x,p)}{p}}$ , and $\sum_{i=0}^{k-1} r(i,m)\leq \sum_{i=k}^{2k-1} r(i,m)$ , where $r(i,m)$ is the remainder function(we take out groups of $m$ which are just permutations of numbers $1$ to $m$ until there are less than $m$ left, then we have $m$ distinct values, which the minimum sum is attained at $0$ to $k-1$ ). Thus, $v_p(N)=\sum_{m=p^{i}, i\in \mathbb{N}_{+}}-\frac{k^{2}}{m}+\left\lfloor{\frac{k^{2}}{m}}\right\rfloor-\frac{\sum_{i=0}^{k-1} r(i,m)-\sum_{i=k}^{2k-1} r(i,m)}{m} \geq \sum_{m=p^{i}, i\in \mathbb{N}} \left\lceil -\frac{k^{2}}{m}+\lfloor{\frac{k^{2}}{m}}\rfloor\right\rceil \geq 0$ , as the term in each summand is a sum of floors also and is clearly an integer.

Solution 2 (Controversial)

Consider an $k\times k$ grid, which is to be filled with the integers $1$ through $k^2$ such that the numbers in each row are in increasing order from left to right, and such that the numbers in each column are in increasing order from bottom to top. In other words, we are creating an $k\times k$ standard Young tableaux.

The Hook Length Formula is the source of the controversy, as it is very powerful and trivializes this problem. The Hook Length Formula states that the number of ways to create this standard Young tableaux (call this $N$ for convenience) is: $N = \frac{\left(k^2\right)!}{\prod_{1\le i, j\le k}(i+j-1)}.$ Now, we do some simple rearrangement: $N = \left(k^2\right)!\cdot\prod_{j=1}^{k}\prod_{i=1}^{k}\frac{1}{i+j-1} = \left(k^2\right)!\cdot\prod_{j=1}^{k}\frac{\left(j-1\right)!}{\left(j+k-1\right)!}$ $= \left(k^2\right)!\cdot\prod_{j=0}^{k-1}\frac{j!}{\left(j+k\right)!}.$ This is exactly the expression given in the problem! Since the expression given in the problem equals the number of distinct $k\times k$ standard Young tableaux, it must be an integer, so we are done.

Solution 3 (Induction)

Define $A(k) = \left(k^2\right)!\cdot\prod_{j=0}^{k-1}\frac{j!}{\left(j+k\right)!}.$ Clearly, $A(1) = 1$ and $A(2) = 2.$

Then $\frac{A(k+1)}{A(k)} = \frac{\left(k^2+2k+1\right)!\cdot\prod_{j=0}^{k}\frac{j!}{\left(j+k+1\right)!}}{\left(k^2\right)!\cdot\prod_{j=0}^{k-1}\frac{j!}{\left(j+k\right)!}}.$ Lots of terms cancel, and we are left with $\frac{A(k+1)}{A(k)} = \frac{(k^2+1)(k^2+2)\cdots(k^2+2k+1)}{2(2k+1)}.$ The numerator has $2k+1$ consecutive positive integers, so one of them must be divisible by $(2k+1).$ Also, there are $2k$ terms left, $k$ of which are even. We can choose one of these to cancel out the $2$ in the denominator. Therefore, the ratio between $A(k+1)$ and $A(k)$ is an integer. By our inductive hypothesis, $A(k)$ is an integer. Therefore, $A(k+1)$ is as well, and we are done.

Note: This is incorrect.

Solution 4

Let us work in the ring of polynomials $\mathbb{Z}[q]$ . Define the $q$ -integer and $q$ -factorial by $[n]_q := 1 + q + q^2 + \cdots + q^{n-1}, \qquad [n]_q! := \prod_{i=1}^n [i]_q.$ Each $[i]_q$ is a monic degree $i - 1$ polynomial with integer coefficients, so $[n]_q! \in \mathbb{Z}[q]$ . Furthermore, $\lim_{q \to 1} [n]_q! = n!.$

Define the expression $P_k(q) := [k^2]_q! \cdot \prod_{j = 0}^{k - 1} \frac{[j]_q!}{[j + k]_q!}.$

Let $\nu_d(F)$ denote the exponent of the cyclotomic polynomial $\Phi_d(q)$ in the factorization of $F \in \mathbb{Z}[q]$ . We use the identity: $\nu_d([n]_q!) = \left\lfloor \frac{n}{d} \right\rfloor$ which follows from the factorization $[n]_q! = \prod_{i = 1}^n (1 - q^i) = \prod_{d = 1}^n \Phi_d(q)^{\left\lfloor \frac{n}{d} \right\rfloor}.$

Applying this to $P_k(q)$ , we compute: $\nu_d(P_k(q)) = \nu_d([k^2]_q!) + \sum_{j = 0}^{k - 1} \nu_d([j]_q!) - \sum_{j = 0}^{k - 1} \nu_d([j + k]_q!).$

Expanding each term: $\nu_d(P_k(q)) = \left\lfloor \frac{k^2}{d} \right\rfloor + \sum_{j = 0}^{k - 1} \left\lfloor \frac{j}{d} \right\rfloor - \sum_{j = 0}^{k - 1} \left\lfloor \frac{j + k}{d} \right\rfloor.$

We simplify the last two sums: $\sum_{j = 0}^{k - 1} \left( \left\lfloor \frac{j}{d} \right\rfloor - \left\lfloor \frac{j + k}{d} \right\rfloor \right) = - \sum_{j = 0}^{k - 1} \sum_{i = j + 1}^{j + k} \delta(i \equiv 0 \bmod d)$

Thus, $\nu_d(P_k(q)) = \sum_{i = 1}^{k^2} \left\lfloor \frac{i}{d} \right\rfloor - \sum_{j = 0}^{k - 1} \sum_{i = j + 1}^{j + k} \left\lfloor \frac{i}{d} \right\rfloor.$

Note that the second double sum runs over exactly $k^2$ total terms, all in the range $[1, k^2]$ , with each $i$ appearing at most once. Therefore, $\sum_{j = 0}^{k - 1} \sum_{i = j + 1}^{j + k} \left\lfloor \frac{i}{d} \right\rfloor \le \sum_{i = 1}^{k^2} \left\lfloor \frac{i}{d} \right\rfloor,$ which implies $\nu_d(P_k(q)) \ge 0 \qquad \text{for all } d \ge 1.$

Hence, every cyclotomic factor $\Phi_d(q)$ appears with nonnegative multiplicity in $P_k(q)$ , so $P_k(q) \in \mathbb{Z}[q]$ .

Finally, evaluating at $q = 1$ gives $P_k(1) = (k^2)! \cdot \prod_{j = 0}^{k - 1} \frac{j!}{(j + k)!} \in \mathbb{Z},$ as desired. ~Lopkiloinm

Note: This solution is fundamentally different from the first, which works purely in the integers. Here, we work in the integer polynomial ring $\mathbb{Z}[q]$ —a graded algebra—which allows us to test integrality by tracking the multiplicities of reducible elements like $1 - q^d$ for all integers $1 \le d \le k^2$ . In contrast, working in $\mathbb{Z}$ —a non-graded algebra—requires using $p$ -adic valuations via Legendre’s formula, which only considers irreducibles (primes). The graded structure of $\mathbb{Z}[q]$ simplifies the analysis, making it a powerful strategy to lift problems into a graded algebra whenever possible.

@@ Line 26: / Line 26: @@
 ==Solution 4==
-Throughout this proof we work in the polynomial ring <math>\mathbb Z[q]</math>.
+Let us work in the ring of polynomials <math>\mathbb{Z}[q]</math>. Define the <math>q</math>-integer and <math>q</math>-factorial by
-For any positive integer <math>n</math>, define the <math>q</math>-integer and <math>q</math>-factorial by
 <cmath>
 [n]_q := 1 + q + q^2 + \cdots + q^{n-1}, \qquad [n]_q! := \prod_{i=1}^n [i]_q.
 </cmath>
-Each <math>[i]_q</math> is a degree <math>i-1</math> polynomial in <math>\mathbb Z[q]</math>, so <math>[n]_q! \in \mathbb Z[q]</math>.
+Each <math>[i]_q</math> is a monic degree <math>i - 1</math> polynomial with integer coefficients, so <math>[n]_q! \in \mathbb{Z}[q]</math>. Furthermore,
-Evaluating at <math>q = 1</math> gives <math>\lim_{q \to 1} [n]_q! = n!</math>.
+<cmath>
+\lim_{q \to 1} [n]_q! = n!.
+</cmath>
 Define the expression
@@ Line 40: / Line 40: @@
 </cmath>
-Let <math>\nu_d(F)</math> denote the exponent of <math>(1 - q^d)</math> in a polynomial <math>F</math>.
+Let <math>\nu_d(F)</math> denote the exponent of the cyclotomic polynomial <math>\Phi_d(q)</math> in the factorization of <math>F \in \mathbb{Z}[q]</math>. We use the identity:
-We use the identity
 <cmath>
-\nu_d([n]_q!) = \sum_{i = 1}^{n} \left\lfloor \frac{i}{d} \right\rfloor - n \cdot \delta_{d, 1},
+\nu_d([n]_q!) = \left\lfloor \frac{n}{d} \right\rfloor
+</cmath>
+which follows from the factorization
+<cmath>
+[n]_q! = \prod_{i = 1}^n (1 - q^i) = \prod_{d = 1}^n \Phi_d(q)^{\left\lfloor \frac{n}{d} \right\rfloor}.
 </cmath>
-where <math>\delta_{d,1} = 1</math> if <math>d = 1</math> and <math>0</math> otherwise.
-Apply this to <math>P_k(q)</math>:
+Applying this to <math>P_k(q)</math>, we compute:
 <cmath>
 \nu_d(P_k(q)) = \nu_d([k^2]_q!)
@@ Line 54: / Line 56: @@
 </cmath>
-Now expand each term using the formula:
+Expanding each term:
 <cmath>
 \nu_d(P_k(q)) =
-\left( \sum_{i = 1}^{k^2} \left\lfloor \frac{i}{d} \right\rfloor - k^2 \delta_{d,1} \right)
+\left\lfloor \frac{k^2}{d} \right\rfloor
-+ \sum_{j = 0}^{k - 1} \left( \sum_{i = 1}^{j} \left\lfloor \frac{i}{d} \right\rfloor - j \delta_{d,1} \right)
++ \sum_{j = 0}^{k - 1} \left\lfloor \frac{j}{d} \right\rfloor
-- \sum_{j = 0}^{k - 1} \left( \sum_{i = 1}^{j + k} \left\lfloor \frac{i}{d} \right\rfloor - (j + k) \delta_{d,1} \right).
+- \sum_{j = 0}^{k - 1} \left\lfloor \frac{j + k}{d} \right\rfloor.
 </cmath>
-Handle the <math>\delta_{d,1}</math> terms separately:
+We simplify the last two sums:
 <cmath>
--k^2 - \sum_{j=0}^{k-1} j + \sum_{j=0}^{k-1} (j + k) = -k^2 - \frac{k(k - 1)}{2} + \left( \frac{k(k - 1)}{2} + k^2 \right) = 0.
+\sum_{j = 0}^{k - 1} \left( \left\lfloor \frac{j}{d} \right\rfloor
-</cmath>
+- \left\lfloor \frac{j + k}{d} \right\rfloor \right)
+= - \sum_{j = 0}^{k - 1} \sum_{i = j + 1}^{j + k} \delta(i \equiv 0 \bmod d)
-So all the <math>\delta_{d,1}</math> terms cancel.
-We are left with:
-<cmath>
-\nu_d(P_k(q)) = \sum_{i = 1}^{k^2} \left\lfloor \frac{i}{d} \right\rfloor
-+ \sum_{j = 0}^{k - 1} \sum_{i = 1}^{j} \left\lfloor \frac{i}{d} \right\rfloor
-- \sum_{j = 0}^{k - 1} \sum_{i = 1}^{j + k} \left\lfloor \frac{i}{d} \right\rfloor.
-</cmath>
-We combine the last two double sums:
-<cmath>
-\sum_{j = 0}^{k - 1} \sum_{i = j + 1}^{j + k} \left\lfloor \frac{i}{d} \right\rfloor.
 </cmath>
@@ Line 86: / Line 77: @@
 </cmath>
-Now observe: for each <math>j</math>, the inner sum covers a block of <math>k</math> consecutive integers.
+Note that the second double sum runs over exactly <math>k^2</math> total terms, all in the range <math>[1, k^2]</math>, with each <math>i</math> appearing at most once. Therefore,
-As <math>j</math> runs from <math>0</math> to <math>k - 1</math>, the total collection of <math>i</math>'s lies within <math>[1, k^2]</math>, and each <math>i</math> appears at most once.
-So
 <cmath>
 \sum_{j = 0}^{k - 1} \sum_{i = j + 1}^{j + k} \left\lfloor \frac{i}{d} \right\rfloor \le \sum_{i = 1}^{k^2} \left\lfloor \frac{i}{d} \right\rfloor,
@@ Line 97: / Line 86: @@
 </cmath>
-Since every factor <math>(1 - q^d)</math> appears with non-negative exponent, <math>P_k(q) \in \mathbb Z[q]</math>.
+Hence, every cyclotomic factor <math>\Phi_d(q)</math> appears with nonnegative multiplicity in <math>P_k(q)</math>, so <math>P_k(q) \in \mathbb{Z}[q]</math>.
-Evaluating at <math>q = 1</math> gives
+Finally, evaluating at <math>q = 1</math> gives
 <cmath>
-P_k(1) = (k^2)! \cdot \prod_{j = 0}^{k - 1} \frac{j!}{(j + k)!} \in \mathbb Z,
+P_k(1) = (k^2)! \cdot \prod_{j = 0}^{k - 1} \frac{j!}{(j + k)!} \in \mathbb{Z},
 </cmath>
-as required.
+as desired.
 ~Lopkiloinm

2016 USAMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
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