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Given an integer $n$, let $P(n)$ denote the set of odd prime divisors of $n$. Let $\Delta$ be the simplicial complex over the set of sets of odd prime numbers which consists of the simplices $S$ such that for every triple $\{A, B, C\} \subseteq S$ there are integers $a$, $b$ and $c$ which satisfy $P(a) = A$, $P(b) = B$, $P(c) = C$ and $a + b = c$.

Question 1: Which is the largest dimension of a simplex in $\Delta$?

Question 2: Does $\Delta$ have nontrivial automorphisms — and if so, which?

Example: It is easy to check that $\{\emptyset, \{3\}, \{5\}, \{7\}, \{11\}, \{13\}\} \in \Delta$. — The largest triples of integers to be looked at for this are $7 + 13^2 = 2^4 \cdot 11$, $3^2 + 2^4 \cdot 7 = 11^2$, $2^4 \cdot 3 + 11^2 = 13^2$ and $5^3 + 2^2 \cdot 11 = 13^2$.

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    $\begingroup$ All of your examples have a common factor $2$; is this intentional? It doesn't show up in the formulation of the problem. In addition, do you have examples of triples that provably do not satisfy the condition? $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jan 23, 2023 at 12:42
  • $\begingroup$ @R.vanDobbendeBruyn Well -- it is clear that not all numbers can be odd, as the sum of two odd numbers is even. On the other hand, not all numbers need to be even. E.g. we have $\{\emptyset, \{3\}, \{5\}\} \in \Delta$ since $2 + 3 = 5$. $\endgroup$
    – Stefan Kohl ♦
    Commented Jan 23, 2023 at 16:01
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    $\begingroup$ @R.vanDobbendeBruyn To prove that a given triple does not to satisfy the condition, one can use the algorithm described in: B. M. M. de Weger, Solving exponential Diophantine equations using lattice basis reduction algorithms, J. Number Theory 26 (1987), no. 3, 325–367. An example is $\{\{5\},\{13\},\{17\}\}$. $\endgroup$
    – Stefan Kohl ♦
    Commented Jan 23, 2023 at 17:15
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    $\begingroup$ Ah, I'll have a look! (My comment about the factors of 2 is just that without loss of generality, you can assume $a$, $b$, and $c$ have no common factors, so in particular two of them are odd and the last one even. I think it's customary to get rid of common factors when solving Diophantine equations, but it's a free world!) $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jan 23, 2023 at 20:18
  • $\begingroup$ @R.vanDobbendeBruyn Thanks! -- I edited out the redundant factor of 2 from the example. $\endgroup$
    – Stefan Kohl ♦
    Commented Jan 24, 2023 at 8:26

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