# Is every integer $\ge 312$ the sum of two integers with triangular divisors?

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.

• I think this question is quite hard. If one writes $N$ as the set of numbers with triangular divisors, then you are asking whether $N$ is an asymptotic additive basis of order 2. One knows that the analogous problem is unknown for primes (where necessarily only even numbers can be represented) which have density $X/\log X$ and it is known for sums-of-two-squares which have density $X/(\log X)^{1/2}$. $N$ is even thicker, but it has far less structure than either of these sets. In particular, it does not seem to be multiplicatively closed (continued...) – Stanley Yao Xiao Jun 7 '20 at 13:30
• Therefore it is not clear to me what techniques would work. Problems relating additive and multiplicative properties of integers is notoriously difficult, as the Goldbach problem shows. In general we don't have a good handle on how to handle sums of multiplicative functions, which are probably the nicest class of arithmetical functions. Thus the answer to the question is "your conjecture is probably true, but we don't know how to prove it". – Stanley Yao Xiao Jun 7 '20 at 13:32
• The second conjecture is false since $7$ cannot be written as the sum of two integers both of which has triangular divisors. It seems that you meant something like "Every integer $\ge 8$ except ...". – mathlove Jun 7 '20 at 13:44
• @mathlove Yes, I had missed the $\ge 8$. Added it. Thanks! – Nilotpal Kanti Sinha Jun 7 '20 at 14:11
• @GerhardPaseman . More is true; every powerful number (in the sense every prime factor is raised to at least the 2nd power) has triangular divisors. Heath-Brown proved that every sufficiently large number is the sum of 3 powerful numbers. So that gets us down to 3 if we allow sufficiently large. I don't know if anyone has made sufficiently large explicit in Heath-Brown's work. – JoshuaZ Jun 9 '20 at 15:54

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.

• Another elementary observation: k*4^n has triangular divisors where k is less than 4^n (oops, less than 2^(n+1). This means every number m is within about sqrt(m) (again oops, more like m to the two thirds) of such a number, and gives a weak logarithmic bound to the number of terms such a sum. One might be able to tweak this to get better bounds. Gerhard "Solve Sum Bit By Bit" Paseman, 2020.06.09. – Gerhard Paseman Jun 9 '20 at 16:09