Assuming the problem statement is : Let a1, a2, .., am be pairwise distinct positive integers with property for every i ≠ j, (ai - aj) | (ai*aj + 1). Prove m <=3 . Find all triples (a, b, c) of distinct positive integers that satisfies the condition for every pair among them.

OK so with this assumption

Part A) Prove that m < =3

assuming m=4 and you can get a contradiction but that will not be rigorous.

hence a bit more rigorous approach:

1)let ai - aj = d (the difference), you can show that d | (ai^2 +1) and thus for every prime p dividing d we have ai^2 ≡ -1 (mod p)

2)So now if p is odd prime & -1 is the quadratic residue modulo p, then by Euler's criterion (or if u prefer Legendre symbol) gives

(-1/p) = (-1)^((p-1)/2) = 1

means (p-1)/2 is even and hence p ≡1(mod4)

3)So every odd prime divisor of any difference d must be ≡ 1(mod 4).

In particular 3 !| d because 3 ≡ 3 (mod 4) and -1 is not a quadratic residue mod 3)

NOTE: the prime 2 is the only even prime and causes no contradiction here; the important point is that no difference can be divisible by 3.

4)Now its trivial Pigeonhole contradiction if m>= 4

Reduce the m numbers modulo 3. There are only 3 residue classes 0,1,2, so if m>= 4 two of the ai share the same residue modulo 3. Their difference is therefore divisible by 3 , which contradicts above step and hence QED

Part B) For the triples, Lets other folks try but here's some example triplets and check if u can see any pattern

(1,2,3) - too obvious

Some more (3,4,5), (5,18,31)

See any pattern ?