Question

43. Prove that there exists an increasing infinite sequence of tetrahedral 

numbers (i.e. numbers of the form T_n = 1/6 n(n+ 1)(n+2), n = 1,2, ... ), such 

that every two of them are relatively prime. 

Answer 1

We shall prove first that if for some positive integer m the tetra-hedral numbers $a_1<a_2<\dots<a_m$ are pairwise relatively prime, then there exists a tetrahedral number $T > a_m$ such that $(T, a_1, a_2, ... , a_m) = 1$ . In fact, let $a = a_1a_2 \dots a_n$. Put $T = T_{6a+1} = (6a+1) (3a+1) (2a+1)$; clearly $T$ is prime relatively to $a$, hence relatively to each of the numbers $a_1, \dots , a_m$, and $T>a\geq a_m$.

Thus, we can define the required increasing sequence of pairwise relatively prime tetrahedral numbers by induction: take $T_1=1$ as the first term of the sequence, and, after having defined $m$ first pairwise relatively prime tetra-hedral numbers of this sequence, define the $m+1st$ as the least tetra-hedral number exceeding all first m terms, and being relatively prime to each of them. In this manner we obtain the infinite increasing sequence of pairwise relatively prime tetrahedral numbers 

$T_1 =1, T_2 = 4, T_5 = 35, T_{17} = 969,\dots.$