# The number of products with two even integers giving $n$.

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.

• Is there a question?? – Anurag A Nov 6 '19 at 18:35
• Of course: Is there a proof for the conjecture? – Sophia Antipolis Nov 6 '19 at 18:38

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).