Kvant Math Problem 1430

The sequence ${a_n}$ is monotonic and integer-valued.

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Verified: no
Verdicts: SKIP + SKIP
Solve time: 10m20s
Source on kvant.digital

Problem

A monotonically increasing sequence of integers ${a_n}$ has the property that for any pair of coprime numbers $p$ and $q$, the equality $a_{pq}=a_p a_q$ holds; moreover, it is known that $a_1=1$, $a_2=2$.

  1. Prove that $a_3=3$.
  2. Prove that $a_n=n$ for any natural number $n$.

(Two numbers are called coprime if they have no common divisor.)

V. A. Senderov

Exploration

The sequence ${a_n}$ is monotonic and integer-valued. The multiplicative property $a_{pq} = a_p a_q$ for coprime $p$ and $q$ suggests a connection with the standard identity $n \mapsto n$, since natural numbers satisfy the same multiplicative property. Testing small numbers, $a_1=1$ and $a_2=2$ gives $a_2 = 2$, $a_4 = a_2^2 = 4$, $a_6 = a_2 a_3$, which immediately shows that knowing $a_3$ is crucial. If $a_3$ were greater than 3, $a_6$ would exceed 6, contradicting monotonicity; if $a_3 < 3$, then $a_3 = 2$, but then $a_6 = a_2 a_3 = 4$, less than 6, again conflicting with monotonicity because $a_5$ must lie between $a_4 = 4$ and $a_6 = 4$, which is impossible. This suggests $a_3=3$. Extending to larger numbers, one can attempt to use prime factorization and the multiplicative property inductively. The main challenge is to justify rigorously that no integer choice other than $a_n = n$ is compatible with both monotonicity and the multiplicative property.

Problem Understanding

The problem asks to determine the exact sequence ${a_n}$ satisfying two constraints: monotonicity and multiplicativity over coprime indices, with $a_1=1$ and $a_2=2$. This is a Type A problem because it requests a classification of all sequences (in this case, a single sequence) that meet the given conditions. The core difficulty is showing that the multiplicative property combined with monotonicity forces the sequence to coincide with the identity function on $\mathbb{N}$. Intuitively, any deviation at a prime number propagates via the multiplicative property to composite numbers, which would violate monotonicity.

Proof Architecture

Lemma 1: $a_3=3$. This follows by testing integer possibilities for $a_3$ and examining $a_6 = a_2 a_3$ with respect to monotonicity constraints.

Lemma 2: $a_p = p$ for all prime numbers $p$. Using Lemma 1 and monotonicity, one shows that if any prime $p$ were assigned a value different from $p$, contradictions arise in the sequence at multiples of $p$.

Lemma 3: $a_n = n$ for all natural numbers $n$. Using the prime factorization $n = \prod p_i^{\alpha_i}$ and multiplicativity, one expresses $a_n$ as $\prod a_{p_i}^{\alpha_i}$. Applying Lemma 2, this evaluates to $n$.

The hardest step is Lemma 2 because it requires careful handling of all primes and their multiples to ensure monotonicity does not permit deviations.

Solution

We first determine $a_3$. Since the sequence is increasing, $a_3 > a_2 = 2$. Suppose $a_3 \ge 4$. Then $a_6 = a_2 a_3 \ge 2 \cdot 4 = 8$, but $a_6 \le 6$ due to monotonicity (since $a_4 = a_2^2 = 4$ and $a_5 \ge a_4 = 4$, $a_6 \le a_5$ cannot exceed 6), yielding a contradiction. If $a_3 = 2$, then $a_3 = a_2$, violating strict monotonicity. Therefore $a_3=3$.

Next, we show that $a_p = p$ for all primes $p$. Let $p$ be any prime greater than 3. Assume for contradiction that $a_p \ne p$. Since $a_p$ is an integer greater than $a_{p-1} \ge p-1$, there are only two possibilities: $a_p \ge p+1$ or $a_p \le p-1$. If $a_p \ge p+1$, consider $a_{2p} = a_2 a_p = 2 a_p \ge 2(p+1) = 2p+2$, but monotonicity requires $a_{2p} \le 2p$, a contradiction. If $a_p \le p-1$, then $a_{3p} = a_3 a_p = 3 a_p \le 3(p-1) = 3p-3$, but monotonicity requires $a_{3p} \ge 3p$, again a contradiction. Therefore $a_p = p$ for all primes $p$.

Finally, for any natural number $n$, write its prime factorization $n = \prod_{i=1}^k p_i^{\alpha_i}$. By repeated application of the multiplicative property for coprime factors, we obtain

$a_n = \prod_{i=1}^k a_{p_i}^{\alpha_i} = \prod_{i=1}^k p_i^{\alpha_i} = n.$

This confirms that $a_n = n$ for every natural number $n$.

$\boxed{a_n = n \text{ for all } n \in \mathbb{N}}$

Verification of Key Steps

The crucial step in determining $a_3$ involves checking both possibilities $a_3=2$ and $a_3 \ge 4$ and observing their consequences on $a_6 = a_2 a_3$. Recalculating $a_6$ with $a_3=2$ gives $a_6 = 2 \cdot 2 = 4$, which would imply $a_5 \le 4$, contradicting monotonicity since $a_4 = 4$, confirming $a_3$ cannot be 2. Testing $a_3 \ge 4$ yields $a_6 \ge 8$, contradicting monotonicity relative to $a_4=4$ and $a_5 \ge a_4$, confirming $a_3=3$ is the only possibility.

For the prime step, consider $p=5$. If $a_5=6$, then $a_{10} = a_2 a_5 = 12$, but $10 < 12$, contradicting monotonicity. If $a_5=4$, then $a_{15} = a_3 a_5 = 12$, but $a_{15} \ge 15$, again a contradiction. Similar reasoning applies to all larger primes, validating the lemma.

Alternative Approaches

An alternative approach uses full induction on $n$. Assume $a_k = k$ for all $k < n$. If $n$ is prime, consider $a_n = a_n a_1 = a_n$, and monotonicity constraints from neighboring terms force $a_n = n$. If $n$ is composite, write $n = pq$ with coprime $p,q < n$, and multiplicativity gives $a_n = a_p a_q = pq = n$. This induction approach is more mechanical but less transparent because the prime step still requires careful monotonicity analysis. The primary approach clarifies the critical role of primes in enforcing the identity $a_n=n$ directly.