Este o exprimare făcută cu dragoste pentru EA, deoarece când am terminat de scanat aceste Notițe de Curs, am scris pe ultima pagină despre tine... Începând de atunci, în orice postare, fie că se potrivește fie că nu, te pomenesc cu dor...
There is a lot to say about Dumitru BATINEȚU... I met him mostly in the pages of Gazeta Matematica, series A and B. But his wife, Maria GIURGIU - mathematics professor at the Military (Technical) Academy in Bucharest, was on an inspection with us at the Academy in SIBIU, when I was also teaching there... I dedicate this post to them.
"O : 60. Let $x\in\mathbb{N}^*\setminus \{1\}$ and $1=d_0<d_1<\dots<d_{n-1}<d_n=x$ all
natural divisors of $x.$ Prove that
$$2\sum_{i=1}^{n-1}\frac{1}{\log_{d_i}x} \tag{E}$$
is a natural number.
D. M. BĂTINEȚU, professor Bucharest"
(at page 292)
ANSWER CiP
Denoting $\tau (x)$ the number of divisors of $x\in\mathbb{N}^*\setminus\{1\}$ , including $1\; and\;x$ , we have
$$\sum_{i=1}^{n-1}\frac{1}{\log_{d_i}x}=\frac{\tau(x)}{2}-1$$
Example 1 CiP $x$ not a perfect square
\begin{array}{c|c|}2024&d_0=1&d_1=2&d_2=4&d_3=8&d_4=11&d_5=22&d_6=23&d_7=44\\\hline &d_{15}=2024&d_{14}=1012&d_{13}=506&d_{12}=253&d_{11}=184&d_{10}=92&d_9=88&d_8=46\\\end{array}
$2024=2\times 1012=4\times 506=8\times 253=11\times 184=22\times 92=23\times 88=44\times 46$
half of (E)=
$$=\sum_{i=1}^{14}\frac{1}{\log_{d_i}2024}=\sum_{i=1}^{14}\log_{2024} d_i=(log_{2024}d_1+\log_{2024}d_{14})+(\log_{2024}d_2+\log_{2024} d_{13})+(\log_{2024}d_3+\log_{2024}d_{12})+$$
$+(\log_{2024}d_4+\log_{2024}d_{11})+(\log_{2024}d_5+\log_{2024}d_{10})+(\log_{2024}d_6+\log_{2024}d_9)+(\log_{2024}d_7+\log_{2024}d_8)=7\times 1=7\in\mathbb{N}$
Example 2 CiP $x=perfect\; square$
\begin{array}{c|c|}2025&d_0=1&d_1=3&d_2=5&d_3=9&d_4=15&d_5=25&d_6=27&d_7=45\\\hline &d_{14}=2025&d_{13}=675&d_{12}=405&d_{11}=225&d_{10}=135&d_9=81&d_8=75&d_7=45\\\end{array}
$2025=3\times 673=5\times 405=9\times 225=15\times 135=25 \times 81=27\times 75=\color{Red}{45\times 45}$
$$\sum_{i=1}^{13}\frac{1}{\log_{d_i}2025}=\sum_{i=1}^{13}\log_{2025}d_i=(\log_{2025}d_1+\log{2025}d_{13})+(\log_{2025}d_2+\log_{2025}d_{12})+(\log_{2025}d_3+\log_{2025}d_{11})+$$
$+(\log_{2025}d_4+\log_{2025}d_{10})+(\log_{2025}d_5+\log_{2025}d_9)+(\log_{2025}d_6+\log_{2025}d_8)+\color{Red}{\log_{2025}d_7}=6\times 1+\color{Red}{\log_{2025}d_7}=6+\frac{1}{2}$
So we have to take the sum twice, to get :
$$2\sum_{i=1}^{13}\frac{1}{\log_{d_i}2025}=12\times 1+(\log_{2025}d_7+\log_{2025}d_7)=12+1=13\in\mathbb{N}$$
.
Solution CiP
Let us note that in the sum (E) only the proper divisors of the number $x$ appear, excluding $1$ and $x$ , in number of $\tau (x)-2$. Therefore, in (E) we have the divisors :
$(1<)\;d_1<d_2<\dots<d_{n-2}<d_{n-1}\;(<x) \tag{1}$
But, we observe that if $ d\mid x$ then $\frac{x}{d}\mid x$ , so the same list (1) coincides with :
$(1<)\;\frac{x}{d_{n-1}}<\frac{x}{d_{n-2}}<\dots<\frac{x}{d_2}<\frac{x}{d_1}\;(<x) \tag{2}$
As long as $x$ is not a perfect square (when $\tau(x)$ is an even number), the complete list (1) or (2) is given by :
$(1<)\;d_1<d_2<\dots<d_k<\frac{x}{d_k}<\dots<\frac{x}{d_2}<\frac{x}{d_1}\;(<x)\;\;\;,\;\;k=\frac{\tau(x)-2}{2} \tag{3}$
In the sum
$$S:=\sum_{i=1}^{n-1}\frac{1}{\log_{d_i}x}$$
there appear two by two terms of the form
$\frac{1}{\log_{d_i}x}+\frac{1}{\log_{\frac{x}{d_i}}x}=\log_xd_i+\log_x(\frac{x}{d_i})=1$
and then $S=1\times \frac{\tau{x}-2}{2}=\frac{\tau(x)}{2}-1.$
But if $x$ is a perfect square (when $\tau(x)$ is an odd number), $x=d_k^2$ then the complete list of divisors is
$(1<)\;d_1<d_2<\dots<d_{k-1}<d_k=\frac{x}{d_k}<\frac{x}{d_{k-1}}<\dots<\frac{x}{d_1}\;(<x)\;\;\;,\;k=\frac{\tau(x)-1}{2} \tag{4}$
and $S$ is written
$$S=\sum_{i=1}^{k-1}\left (\frac{1}{\log_{d_i}x}+\frac{1}{\log_{\frac{x}{d_i}}x} \right )+\frac{1}{\log_{d_k}x}$$
and, because $\frac{1}{\log_{d_k}x}=\log_x{d_k}=\log_x\frac{x}{d_k}=1-\log_x{d_k}=1-\frac{1}{\log_{d_k}x} \Rightarrow \frac{1}{\log_{d_k}x}=\frac{1}{2}$
we finally get $S=\left (\frac{\tau(x)-1}{2}-1\right )\times 1+\frac{1}{2}=\frac{\tau(x)}{2}-1.$
$\blacksquare\;\;\;\blacksquare$
