250 Problems in Elementary Number Theory - Sierpinski (1970)

36 250 PROBLEMS IN NUMBER THEORYIt follows that if we have a finite increasing sequence of pairwise relativelyprime triangular numbers, then we can always find a triangular numberexceeding all of them and pairwise relatively prime to them. Taking alwaysthe least such number we form the infinite sequencet 1 =1, t2=3, t 4 =10, t 13 =91, t22=253,of pairwise relatively prime triangular numbers.43. We shall prove first that if for some positive integer m the tetrahedralnumbers al < a2 < ... < am are pairwise relatively prime, then thereexists a tetrahedral number T > am such that (T, al, a2, ... , am) = 1. In fact,let a = ala2 ... an- Put T = T 6a + 1 = (6a+l) (3a+l) (2a+l); clearly T isprime relatively to a, hence relatively to each of the numbers at, ... , am,and T> a ~ am.Thus, we can define the required increasing sequence of pairwise relativelyprime tetrahedral numbers by induction: take Tl = 1 as the first term of thesequence, and, after having defined m first pairwise relatively prime tetrahedralnumbers of this sequence, define the m + 1 st as the least tetrahedralnumber exceeding all first m terms, and being relatively prime toeach of them. In this manner we obtain the infinite increasing sequenceof pairwise relatively prime tetrahedral numbersTl =1, T2 = 4, Ts = 35, T17 = 969,44. Let a ,and b be two different integers. Assume for instance a < b,and let n = (b-a)k+ I-a. For k sufficiently large, n will be positive integer.We have a+n = (b-a)k+l, b+n = (b-a) (k+l)+l, hence a+n and b+nwill be positive integers. If we had dla+n and dlb+n, we would havedla-b, and, in view of dla+n, also dll, which implies that d = 1. Thus,(a+n, b+n) = 1.45*. If the integers a, b, c are distinct, then the numberh = (a-b) (a-c) (b-c)is different from zero. In case h :F ± 1, let Ql, ... , qs denote all prime> 3divisors of h.If two or more amo~g numbers a, b, c are even, put r = 1, otherwise putr = O. Clearly, at least two of the numbers a+r, b+r, c+r will be odd. If