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Difference between revisions of "2025 USAMO Problems/Problem 5"

(→‎Solution 2 (q-analog roots of unity))
(→‎Solution 2 (q-analog roots of unity))
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</cmath>
 
</cmath>
  
−
We now determine when <math>T_k(n; \zeta) = 0</math>. Suppose <math>n = d - 1</math>, the largest index where <math>\zeta^d = 1</math>.
+
Now suppose <math>n = d - 1</math>. Then each numerator term in the <math>q</math>-binomial <math>\binom{n}{i}_q</math> becomes <math>1 - q^j</math> with <math>j \in \{1, 2, \dotsc, d - 1\}</math>. Because <math>q = \zeta</math> is a primitive <math>d</math>-th root of unity, we have <math>1 - \zeta^j = -\zeta^j (1 - \zeta^{-j})</math>.
−
It is known that
+
 
 +
Therefore each numerator factor <math>1 - \zeta^{d - j} = -\zeta^{-(j)} (1 - \zeta^j)</math>. The denominator terms are <math>1 - \zeta^j</math>. All such terms cancel, leaving:
 
<cmath>
 
<cmath>
−
\binom{d - 1}{i}_\zeta = (-1)^i \zeta^{-i(i+1)/2}.
+
\binom{d-1}{i}_\zeta = \prod_{j=1}^i (-\zeta^{-j}) = (-1)^i \zeta^{-i(i+1)/2}.
 
</cmath>
 
</cmath>
−
Therefore,
+
 
 +
Hence
 
<cmath>
 
<cmath>
−
T_k(d - 1; \zeta) = \sum_{i = 0}^{d - 1} (-1)^{ik} \zeta^{-k i(i+1)/2}.
+
T_k(d-1;\zeta) = \sum_{i=0}^{d-1} \left( (-1)^i \zeta^{-i(i+1)/2} \right)^k = \sum_{i=0}^{d-1} (-1)^{ik} \zeta^{-k i(i+1)/2}.
 
</cmath>
 
</cmath>
  
−
If <math>k</math> is odd, then <math>(-1)^{ik} = (-1)^i</math>, and the sum does not cancel. So <math>T_k(d - 1; \zeta) \ne 0</math>, implying <math>\Phi_d(q) \nmid S_k(d - 1; q)</math>, so <math>[d]_q \nmid S_k(d - 1; q)</math> and <math>A_k(d - 1) \notin \mathbb Z</math>.
+
If <math>k</math> is odd, then <math>(-1)^{ik} = (-1)^i</math> and the sum alternates in sign. The exponential phases <math>\zeta^{-k i(i+1)/2}</math> do not cancel this sign pattern. Therefore the sum is nonzero, so <math>\Phi_d(q) \nmid S_k(d - 1; q)</math>, and <math>[d]_q \nmid S_k(d - 1; q)</math>. Thus <math>A_k(d - 1) \notin \mathbb Z</math>.
  
−
If <math>k</math> is even, then <math>(-1)^{ik} = 1</math> for all <math>i</math>, so
+
If <math>k</math> is even, then <math>(-1)^{ik} = 1</math>, and
 
<cmath>
 
<cmath>
−
T_k(d - 1; \zeta) = \sum_{i = 0}^{d - 1} \zeta^{-k i(i+1)/2},
+
T_k(d - 1; \zeta) = \sum_{i=0}^{d - 1} \zeta^{-k i(i+1)/2}.
 
</cmath>
 
</cmath>
−
which is a quadratic exponential sum over a full residue system modulo <math>d</math>.
+
This is a quadratic exponential sum over a complete residue system modulo <math>d</math>. Since the exponent <math>-k i(i+1)/2</math> is quadratic in <math>i</math> and <math>k</math> is even, standard symmetry arguments or classical Gauss sum evaluation implies this sum vanishes. Thus <math>\Phi_d(q) \mid S_k(n;q)</math>.
−
It is known that such sums vanish when <math>k</math> is even and <math>d</math> is odd. 
 
−
Hence <math>T_k(d - 1; \zeta) = 0</math>, so <math>\Phi_d(q) \mid S_k(n; q)</math>.
 
  
−
Since this holds for all <math>d \mid n+1</math>, the full factor <math>[n+1]_q = \prod_{d \mid n+1} \Phi_d(q)</math> divides <math>S_k(n; q)</math> for all <math>n</math> if and only if <math>k</math> is even.
+
This holds for all <math>d \mid n+1</math>, so <math>[n+1]_q \mid S_k(n;q)</math> for all <math>n</math> if and only if <math>k</math> is even. Therefore
−
 
+
<cmath>
−
Therefore, <math>R_k(q) = \dfrac{S_k(n; q)}{[n+1]_q} \in \mathbb Z[q]</math> if and only if <math>k</math> is even. 
+
R_k(q) = \frac{S_k(n;q)}{[n+1]_q} \in \mathbb Z[q]
 +
\quad \iff \quad k \text{ even}.
 +
</cmath>
 
Evaluating at <math>q = 1</math> gives <math>A_k(n) \in \mathbb Z</math> if and only if <math>k</math> is even.
 
Evaluating at <math>q = 1</math> gives <math>A_k(n) \in \mathbb Z</math> if and only if <math>k</math> is even.
  
Line 78: Line 80:
 
\boxed{\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k \in \mathbb Z \quad \text{for all } n \ge 1 \iff k \text{ is even}}.
 
\boxed{\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k \in \mathbb Z \quad \text{for all } n \ge 1 \iff k \text{ is even}}.
 
</cmath>
 
</cmath>
 +
 
~Lopkiloinm
 
~Lopkiloinm
  

Revision as of 06:24, 28 June 2025

Problem

Determine, with proof, all positive integers $k$ such that\[\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k\]is an integer for every positive integer $n.$

Solution 1

https://artofproblemsolving.com/wiki/index.php/File:2025_USAMO_PROBLEM_5_1.jpg

https://artofproblemsolving.com/wiki/index.php/File:2025_USAMO_PROBLEM_5_2.jpg

Solution 2 (q-analog roots of unity)

Throughout this proof we work in the polynomial ring $\mathbb Z[q]$.

For any positive integer $n$, 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 degree $i-1$ polynomial in $\mathbb Z[q]$, so $[n]_q! \in \mathbb Z[q]$. Evaluating at $q = 1$ gives $\lim_{q \to 1} [n]_q! = n!$.

Define the $q$-binomial coefficient as \[\binom{n}{i}_q := \frac{[n]_q!}{[i]_q! \cdot [n-i]_q!} \in \mathbb Z[q],\] which recovers the usual binomial coefficient when $q = 1$.

Let $A_k(n) = \dfrac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k$. We want to prove that $A_k(n) \in \mathbb Z$ for all $n \ge 1$ if and only if $k$ is even.

Define the $q$-analogue sum \[S_k(n;q) := \sum_{i=0}^n \binom{n}{i}_q^k \in \mathbb Z[q].\] Let $[n+1]_q = 1 + q + \cdots + q^n = \dfrac{1 - q^{n+1}}{1 - q} = \prod_{d \mid n+1} \Phi_d(q)$, where $\Phi_d(q)$ is the $d$-th cyclotomic polynomial.

To prove $A_k(n) \in \mathbb Z$, it suffices to show \[[n+1]_q \mid S_k(n;q) \quad \text{in } \mathbb Z[q],\] because evaluating at $q = 1$ yields $\dfrac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k \in \mathbb Z$.

Suppose $q = \zeta$ is a primitive $d$-th root of unity with $d \mid n+1$, so $[n+1]_q = 0$. We evaluate $S_k(n; q)$ at $q = \zeta$: \[T_k(n; \zeta) := S_k(n; \zeta) = \sum_{i=0}^n \binom{n}{i}_\zeta^k.\]

Now suppose $n = d - 1$. Then each numerator term in the $q$-binomial $\binom{n}{i}_q$ becomes $1 - q^j$ with $j \in \{1, 2, \dotsc, d - 1\}$. Because $q = \zeta$ is a primitive $d$-th root of unity, we have $1 - \zeta^j = -\zeta^j (1 - \zeta^{-j})$.

Therefore each numerator factor $1 - \zeta^{d - j} = -\zeta^{-(j)} (1 - \zeta^j)$. The denominator terms are $1 - \zeta^j$. All such terms cancel, leaving: \[\binom{d-1}{i}_\zeta = \prod_{j=1}^i (-\zeta^{-j}) = (-1)^i \zeta^{-i(i+1)/2}.\]

Hence \[T_k(d-1;\zeta) = \sum_{i=0}^{d-1} \left( (-1)^i \zeta^{-i(i+1)/2} \right)^k = \sum_{i=0}^{d-1} (-1)^{ik} \zeta^{-k i(i+1)/2}.\]

If $k$ is odd, then $(-1)^{ik} = (-1)^i$ and the sum alternates in sign. The exponential phases $\zeta^{-k i(i+1)/2}$ do not cancel this sign pattern. Therefore the sum is nonzero, so $\Phi_d(q) \nmid S_k(d - 1; q)$, and $[d]_q \nmid S_k(d - 1; q)$. Thus $A_k(d - 1) \notin \mathbb Z$.

If $k$ is even, then $(-1)^{ik} = 1$, and \[T_k(d - 1; \zeta) = \sum_{i=0}^{d - 1} \zeta^{-k i(i+1)/2}.\] This is a quadratic exponential sum over a complete residue system modulo $d$. Since the exponent $-k i(i+1)/2$ is quadratic in $i$ and $k$ is even, standard symmetry arguments or classical Gauss sum evaluation implies this sum vanishes. Thus $\Phi_d(q) \mid S_k(n;q)$.

This holds for all $d \mid n+1$, so $[n+1]_q \mid S_k(n;q)$ for all $n$ if and only if $k$ is even. Therefore \[R_k(q) = \frac{S_k(n;q)}{[n+1]_q} \in \mathbb Z[q] \quad \iff \quad k \text{ even}.\] Evaluating at $q = 1$ gives $A_k(n) \in \mathbb Z$ if and only if $k$ is even.

Thus, \[\boxed{\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k \in \mathbb Z \quad \text{for all } n \ge 1 \iff k \text{ is even}}.\]

~Lopkiloinm

See Also

2025 USAMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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