Star Fleet Math

The problem and why it resisted the usual induction

For pairwise-coprime integers $a,b,c>1$

$$a^i b^j c^k \qquad (i,j,k\ge 0).$$

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 $[N,CN]$

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.
• Van der Waerden and Hales–Jewett arguments produce arbitrarily long arithmetic progressions of primitive sums, but initially with an uncontrolled common difference.
• Even after fixing the common difference, a progression $B_0+r d$

The important lesson was that large additive width is not enough. The lower endpoint has to remain under quantitative control.

The homogeneous-level coordinate system

The first structural simplification is to work on one homogeneous exponent level

$$i+j+k=D.$$

For pairwise-coprime bases greater than one, divisibility of monomials is coordinatewise comparison of their exponents. Therefore two distinct monomials on the same level can never divide one another. Every subset of a homogeneous level is automatically primitive.

This turns the problem into an additive question about subset sums while making primitiveness essentially free—as long as all pieces of the construction can be placed on the same exact degree.

An edge-code construction supplies $c^n$

Turning one AP into a large lattice interval

Order the bases as

$$1$$

Choose

$$H=\operatorname🔥(c),\qquad u=H+2,$$

and then choose $v>0$

$$2b a^v\le c^v.$$

Define two coprime homogeneous translation weights

$$A=a^🔥,\qquad B=b^u c^v.$$

Copies of one AP digit family are translated by the weights

$$A^🔥B^r.$$

The choice $u>H+1$

A bounded homogeneous-radix lemma proves that coefficient sums

$$\sum_💬^{M} s_r A^{M-r}B^r, \qquad 0\le s_r<4AB,$$

contain a full interval of width at least $2AB^{M+1}$

Filling residues with face corrections

The next ingredient constructs, on every sufficiently high exact degree, a primitive correction for each residue modulo any prescribed modulus. The corrections are supported on the three coordinate faces and, in the ordered case, have total size bounded by

$$C_{\mathrm{corr}}c^D.$$

Apply this with modulus $abc\,d$

$$\frac{B}{c^{u+v}}=\left(\frac bc\right)^u>1,$$

so exponential domination gives

$$C_{\mathrm{corr}}c^D=o(B^M).$$

Thus the correction spread is eventually smaller than the radix width. Residue gluing converts the lattice interval into an ordinary consecutive interval $[L_M,U_M]$

$$abc\,B^M\le U_M-L_M, \qquad L_M\le K B^{M+1}$$

for a fixed constant $K$.

At this stage there is a genuine interval, but its multiplicative width is still only bounded by a constant. This is exactly where the earlier baseline problem remained.

The key breakthrough: an optional interior shell

The decisive idea was to exploit monomials that had not yet been used, on the same exact homogeneous level.

For each of linearly many indices $s$, fix a $b$-exponent just beyond every AP band. Among the remaining $a,c$

$$b^s a^{R-k}c^k$$

below a target of size $c^{vM}$

$$z\le X, \qquad aX\le cz, \qquad X=abc\,B^M.$$

Every optional term is therefore no larger than the already available interval width. Adding such a term optionally—either use it or do not—extends a consecutive interval without changing its lower endpoint. Since all optional terms remain on the same exact level and lie beyond the AP exponent bands, primitiveness and disjointness are preserved.

Their combined contribution is

$$\Omega(MB^M),$$

while the lower endpoint remains $O(B^M)$

$$[N,RN].$$

This interior-shell amplification is what removes the finite-seed obstruction. The successful coordinate change was not merely “work on a homogeneous level,” but “place the main growth along an interior homogeneous ray, then use the unused transverse strip as optional mass.”

Completing the induction

The residue-reduction argument was strengthened to a flexible finite-seed gate: there are constants $N_0$

Applying the arbitrary-width construction with $R=C$

Verification

The proof is formalized in Lean 4 with Mathlib. The final theorem is

Erdos123.erdos_123 : Erdos123.IntendedStatement

where IntendedStatement quantifies over all pairwise-coprime natural bases greater than one and asserts an explicit eventual threshold for primitive representations.

The complete project builds successfully, and a source scan finds no proof placeholders. Lean’s axiom report for the final theorem is exactly

[propext, Classical.choice, Quot.sound]

with no sorryAx. The webpage literally writes $a,b,c\ge1$