GoKawiilSolving Erdős Problem 123
The problem and why it resisted the usual induction
For pairwise-coprime integers a , b , c > 1 a,b,c>1 a,b,c>1, consider the numbers
a i b j c k ( i , j , k ≥ 0 ) . a^i b^j c^k \qquad (i,j,k\ge 0). a i b j c k ( i , j , k ≥ 0 ) .
The question asks whether every sufficiently large integer is a sum of distinct such numbers, with the additional requirement that no chosen summand divides another.
The divisibility condition is the real source of difficulty. Ordinary completeness arguments can use many terms from different scales, but terms from different scales tend to be comparable by divisibility. Conversely, a set chosen to be a divisibility antichain can be too arithmetically sparse to fill consecutive integers.
Earlier work had developed a powerful reduction scheme: choose a correction with the required residue modulo one base, subtract it, divide by that base, and induct. For particular triples this succeeds after a finite computer check. In general, however, it leaves a stubborn finite-seed problem: one must first represent every integer in a multiplicatively wide interval [ N , C N ] [N,CN] [N,CN]. The correction induction does not construct that interval; it only propagates it.
This explains why several attractive partial ideas did not finish the problem:
A signed identity of difference one gives two consecutive sums, but one residue representative per class necessarily has spread at least the modulus minus one. A width-one interval cannot grow under ordinary residue gluing.
Complete residue systems on a primitive level solve congruences, but say nothing about their numerical spread.
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