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Premises: Let $c(n)$ denote the number of products of the form $$ n = a \cdot b $$ where $a$ and $b$ are positive even integers and $a < b$.

Conjecture: If $4$ divides $n$ then $$ c(n) = \Bigl\lfloor \frac{\tau(n/4)}{2} \Bigr\rfloor , $$ where $\tau(n)$ is the number of divisors of $n$.

Example: Let $n=24$, then the products with two parts (disregarding the order) which are divisors of $n$ are $$ 1 \cdot 24,\ 2 \cdot 12,\ 3 \cdot 8,\ 4 \cdot 6. $$ Those with even integers only are $\{2, 12\}$ and $\{4, 6\}.$ Also $$ \lfloor \tau(24/4)/2 \rfloor = \tau(6)/2 = 4/2 = 2, $$ as predicted.

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  • $\begingroup$ Is there a question?? $\endgroup$ – Anurag A Nov 6 '19 at 18:35
  • $\begingroup$ Of course: Is there a proof for the conjecture? $\endgroup$ – Sophia Antipolis Nov 6 '19 at 18:38
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Each decomposition $n = ab$ (where $a< b$ are both even) corresponds with a unique decomposition $n/4 = (a/2)(b/2)$ (where $a/2 < b/2$) of $n/4$.

The number of such decompositions of $n/4$ situation is $\lfloor \tau(n/4)/2\rfloor$ (specifically it is $\tau(n/4)/2$ if $n/4$ is not a perfect square, and $(\tau(n/4)-1)/2$ if $n/4$ is a perfect square).

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