TITLE: Is every integer $\ge 312$ the sum of two integers with triangular divisors?
QUESTION [5 upvotes]: We say that a natural number $n$ has triangular divisors if it has at least one triplet of divisors $n = d_1d_2d_3$, $1 \le d_1 \le d_2 \le d_3$, such that $d_1,d_2$ and $d_3$ form the sides of a triangle (non degenerate)
E.g.: $60$ has triangular divisors because $60 = 3.4.5$ and $3,4,5$ form a triangle. Note that another triplet of divisors of $60 = 1.4.15$ does not form a triangle but because of the triplet $3,4,5$ the number $60$ qualifies a number with triangular divisors. On the other the number $10$ does not have any triplet of triangular divisors.
I found the following conjectures experimentally. Can they be proved or disproved?

Weak conjecture: Every integer $\ge 8$ which has triangular divisors can be written as the sum of two integers both of which has triangular divisors.
Strong conjecture: Every integer $\ge 8$ except $11, 14, 15,23, 38, 47, 55, 71, 103, 113$ and $311$ can be written as the sum of two integers both of which has triangular divisors.

Note: This question was posted in MSE 3 month ago year. It got some upvotes but to answers. Hence posting in MO.
Related question: How many numbers $\le x$ can be factorized into three numbers which form the sides of a triangle?

REPLY [4 votes]: Not a complete answer, but it made sense I think to summarize the comment thread above:
Theorem: Every sufficiently large positive integer is expressible as a sum of three numbers which have triangular divisors.
To see this, note that every perfect square, $n^2$, has triangular divisors $(1,n,n)$. 
Also, if $m$ has triangular divisors, so does $mk^3$ for any $k$. This is because we can just take our three divisors $d_1$, $d_2$, and $d_3$ and scale them up to be $kd_1$, $kd_2$, $kd_3$. 
So any number of the form $n^2k^3$ has triangular divisors. But these are precisely the powerful numbers, numbers in which all prime factors are raised to at least the second power. (Note these are sometimes called squarefull numbers.) 
To prove the theorem we then use the theorem of Heath-Brown that every sufficiently large positive integer is expressible as the sum of three powerful numbers.