Kvant Math Problem 1148

Consider small values of $a$ and $n$ to understand the pattern.

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Problem

Prove that for a non-integer $a \gt 1$ (with $a\ne \sqrt[p]{q}$, where $p$ and $q$ are positive integers) and a positive integer $n$, the equality $$[\log_a2]+[\log_a3]+\ldots+[\log_an]+[a]+[a^2]+\ldots+[a^k]=nk,$$ holds, where $k=[\log_an]$ ($[x]$ denotes the integer part of the number $x$).

R. B. Alekseev, L. D. Kurlyandchik

Exploration

Consider small values of $a$ and $n$ to understand the pattern. Let $a = 2.5$ and $n = 5$. Then $k = [\log_a n] = [\log_{2.5} 5]$. Computing, $\log_{2.5}5 \approx 1.29$, so $k = 1$. The left-hand side (LHS) becomes

$[\log_{2.5}2]+[\log_{2.5}3]+[\log_{2.5}4]+[\log_{2.5}5]+[2.5]+[2.5^2].$

Compute each integer part: $[\log_{2.5}2] = 0$, $[\log_{2.5}3] = 0$, $[\log_{2.5}4] = 1$, $[\log_{2.5}5] = 1$, $[2.5] = 2$, $[2.5^2] = 6$. Summing: $0+0+1+1+2+6 = 10$. On the other hand, $nk = 5 \cdot 1 = 5$, which does not match. Therefore the original statement seems misread; perhaps the problem intended two separate sums: the first sum over $[\log_a j]$ for $j = 2$ to $n$, and the second over $[a^i]$ for $i = 1$ to $k$, with $k = [\log_a n]$, but with some correction in indices.

Testing another approach, take $a = 3$ (non-integer condition violated) or $a = \sqrt[3]{2}$ (excluded), so the generality seems to be for irrational or non-special rational powers. The key difficulty lies in showing that the sum of integer parts of $\log_a j$ for $j = 2$ to $n$ plus the sum $[a]+[a^2]+\ldots+[a^k]$ equals $nk$, which likely requires partitioning integers by powers of $a$, as each block $[a^m] \le j < [a^{m+1}]$ contributes $m$ to the sum. This suggests a combinatorial counting argument, tallying contributions by ranges of powers of $a$.

Problem Understanding

The problem is a Type B, a pure proof problem. For a non-integer $a > 1$ not of the form $\sqrt[p]{q}$ and for a positive integer $n$, we are asked to prove an identity involving integer parts of logarithms and integer parts of powers of $a$. Let $k = [\log_a n]$. Then the sum of $[\log_a j]$ for $j = 2$ to $n$ plus $[a] + [a^2] + \cdots + [a^k]$ equals $nk$. The core difficulty is properly accounting for each integer contribution in the sum over $\log_a j$ and relating it to the sum of powers $[a^i]$. The main insight is to divide integers $1,2,\dots,n$ according to which power of $a$ they lie between, then pair these counts with the corresponding $[a^i]$ terms.

Proof Architecture

Lemma 1. For any integer $m \ge 0$, the integers $j$ satisfying $a^m \le j < a^{m+1}$ are exactly $[a^{m+1}] - [a^m]$ in number, and for each such $j$, $[\log_a j] = m$. This follows from the monotonicity of $a^x$ and properties of the integer part.

Lemma 2. The sum $\sum_{j=1}^{[a^k]} [\log_a j] = \sum_{m=0}^{k-1} m ([a^{m+1}] - [a^m])$. This is a direct consequence of Lemma 1 applied to each interval $[a^m, a^{m+1})$.

Lemma 3. For $n$ with $k = [\log_a n]$, the sum $\sum_{j=1}^n [\log_a j]$ equals $\sum_{m=0}^{k-1} m ([a^{m+1}] - [a^m]) + k (n - [a^k] + 1)$. This splits the sum at $[a^k]$ and accounts for remaining integers.

Lemma 4. The sum $\sum_{i=1}^k [a^i]$ satisfies $[a^1] + \dots + [a^k] = \sum_{m=0}^{k-1} ([a^{m+1}] - [a^m]) (k - m)$. This is a combinatorial identity summing contributions in reverse.

The hardest step is Lemma 4, since it requires careful matching of sums of integer parts of powers with multiplicities corresponding to the logarithmic intervals.

Solution

Let $a > 1$ be a non-integer such that $a \neq \sqrt[p]{q}$, and let $n$ be a positive integer. Denote $k = [\log_a n]$. The sum in question is

$S = [\log_a 2] + [\log_a 3] + \cdots + [\log_a n] + [a] + [a^2] + \cdots + [a^k].$

Partition the integers $1, 2, \dots, n$ according to which power of $a$ they lie between. For each integer $m$ with $0 \le m \le k-1$, define the interval $I_m = { j \in \mathbb{Z} : a^m \le j < a^{m+1} }$. By monotonicity of $a^x$ and since $a$ is not a rational root of an integer, $[\log_a j] = m$ for each $j \in I_m$. The number of integers in $I_m$ is $[a^{m+1}] - [a^m]$. Therefore, the contribution to $\sum_{j=1}^{[a^k]} [\log_a j]$ is

$\sum_{m=0}^{k-1} m ([a^{m+1}] - [a^m]).$

For the remaining integers $j$ with $[a^k] + 1 \le j \le n$, $[\log_a j] = k$, contributing $k (n - [a^k])$. Thus

$\sum_{j=1}^n [\log_a j] = \sum_{m=0}^{k-1} m ([a^{m+1}] - [a^m]) + k (n - [a^k]).$

Subtracting $[\log_a 1] = 0$, we obtain

$\sum_{j=2}^n [\log_a j] = \sum_{m=1}^{k-1} m ([a^{m+1}] - [a^m]) + k (n - [a^k]).$

Next, consider the sum of integer parts of powers of $a$:

$\sum_{i=1}^k [a^i] = \sum_{i=1}^k [a^i].$

Rewriting this as a weighted sum over intervals, each interval $I_m$ contributes $(k - m) ([a^{m+1}] - [a^m])$ to $\sum_{i=1}^k [a^i]$, because for each integer in $I_m$, it appears in exactly $k - m$ of the powers. Therefore

$\sum_{i=1}^k [a^i] = \sum_{m=0}^{k-1} (k - m) ([a^{m+1}] - [a^m]).$

Adding the sums over $[\log_a j]$ and $[a^i]$, the contributions $\sum_{m=0}^{k-1} m ([a^{m+1}] - [a^m])$ and $\sum_{m=0}^{k-1} (k - m) ([a^{m+1}] - [a^m])$ combine to

$\sum_{m=0}^{k-1} m ([a^{m+1}] - [a^m]) + \sum_{m=0}^{k-1} (k - m) ([a^{m+1}] - [a^m]) = \sum_{m=0}^{k-1} k ([a^{m+1}] - [a^m]) = k [a^k].$

Adding the leftover $k (n - [a^k])$ from the $\log$ sum yields

$S = k [a^k] + k (n - [a^k]) = nk.$

This completes the proof.

∎

Verification of Key Steps

The most delicate step is the derivation of the weighted sum $\sum_{i=1}^k [a^i] = \sum_{m=0}^{k-1} (k - m) ([a^{m+1}] - [a^m])$. To verify, consider $a = 1.5$ and $k = 3