pavel25 wrote:

$d_k = {n\over d_1}$
$d_{k-1} = {n\over d_2}$
$d_1 = {n\over d_k}$
So we can write it like this:
$d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ = ${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}$
The Maximum that can be is:
${{n^2}\over 1*2}+{{n^2}\over 2*3}+\,\ldots\,+{{n^2}\over (n-1)n}$
We can see that the sum starting with the form:
${m\over {m+1}}{n^2}$
We check what happens if we add the next term:
${m\over {m+1}}{n^2} + {1\over {(m+1)(m+2)}}{n^2} = {{(m+1)^2}\over {(m+1)(m+2)}}{n^2} = {{m+1}\over {m+2}}{n^2}$
From here is easy to define that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ = ${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}<{n^2}$

$d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ = ${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}$ is a divisor of ${n^2}$ when $n$ is a prime number:

$d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ = ${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k} = n$

if $n$ isn't a prime we have at least 3 divisors of $n$
We proved that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ = ${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}<{n^2}$

So if it's a divisor of ${n^2}$ the maximum that can be: ${{n^2}\over d_2}$

$d_1 = 1$

$d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ = ${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k} = {{n^2}\over d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k} > {{n^2}\over d_2}$
We see from here that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ is a divisor of ${n^2}$ only if $n$ is a prime!

I have asked this question as another link sorry for the same question <http://www.mathlinks.ro/Forum/viewtopic.php?p=531646#531646>