GPT 6 Astra Ultra attempt on Math-5443
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FORMAL RESTATEMENT
Erdős #123; reconstruction, 2026-09-23. Let $a,b,c>1$ be pairwise coprime. Must every sufficiently large integer be a sum of distinct numbers $a^ib^jc^k$, with nonnegative exponents, such that no selected number divides another?QUICK LITERATURE AND CONTEXT CHECK
The current page records a recent affirmative resolution. We reconstruct the level-set argument in the primary exposition hosted by Thomas Bloom, expanding its residue-correction and interval steps. Van der Waerden's theorem is an external classical input. This is an exposition of that known proof, not a new solution or a report of running its formalization.WORK
Relabel the bases so that $a<c<b$, and write $L_D=\{a^ib^jc^k:i+j+k=D\}$. Pairwise coprimality implies divisibility is coordinatewise in $(i,j,k)$; thus every subset of one $L_D$ is an antichain for divisibility. We seek long intervals of subset sums on one level.Progressions with a bounded $b$ exponent
Put $\phi=\varphi(c)$ and $H=\phi(c-2)$. For $E\ge H$ and $0\le r<c$, take \[ P_E(r)=\{a^{E-H+\phi t}b^{\phi(c-2-t)}:0\le t<r\}\subseteq L_E. \] Its sum is $r a^{E-H}$ modulo $c$, by Euler's theorem. Given a word $(r_0,\ldots,r_{h-1})\in\{0,\ldots,c-1\}^h$, form \[ F=\bigcup_{i=0}^{h-1}c^iP_{H+h-1-i}(r_i)\subseteq L_{H+h-1}. \] These unions have no repeated terms since their $c$ exponents differ. If two words first differ at index $i$, the difference of their sums has $c$-adic order exactly $i$: after division by $c^i$, the difference is $(r_i-r_i')a^{h-1-i}$ modulo $c$, which is nonzero. Hence the sums represent all residues modulo $c^h$, exactly once. All $b$ exponents are at most $H$. Since $a<c$, summing the resulting geometric series bounds every sum by $Kc^h$, with $K$ independent of $h$. Color each residue in $[0,c^h)$ by the quotient of its sum by $c^h$. The number of colors is bounded. Van der Waerden's theorem therefore gives, for any prescribed $L>1$, sets $E_0,\ldots,E_{L-1}\subseteq L_\delta$ whose sums are $s,s+d,\ldots,s+(L-1)d$ with $d>0$, and whose $b$ exponents are at most $H$.Two auxiliary interval and residue facts
For coprime integers $1<A<B$, put $L=4AB$. The digit sums \[ T_N=\left\{\sum_{e=0}^N r_e A^{N-e}B^e:0\le r_e<L\right\} \] contain integer intervals $[U_N,V_N]$ satisfying \[ V_N-U_N\ge2AB^{N+1},\qquad 0\le U_N\le V_N\le K_1B^{N+1}. \tag{1} \] For completeness start with $U_0=0,V_0=L-1$. Given an interval at stage $N-1$, use $T_N=AT_{N-1}+\{0,\ldots,L-1\}B^N$. For each residue modulo $A$, the allowed digits form a progression of step $A$. Their translated intervals overlap in that residue, since $V_{N-1}-U_{N-1}\ge B^N-1$. All residues therefore fill the interval with endpoints \[ U_N=AU_{N-1}+(A-1)B^N,\quad V_N=AV_{N-1}+(L-A)B^N. \] Its width is at least $2A^2B^N+(4AB-2A+1)B^N\ge2AB^{N+1}$. The endpoint upper bound follows by summing a geometric series with ratio $A/B<1$.Fix a modulus $\Delta$. For all large $D$, every residue modulo $\Delta$ is a subset sum of face monomials in $L_D$, of total size at most $K_2c^D$. Here a face means that at least one exponent is zero. To verify this sometimes useful detail, choose a common eventual period $T$ and threshold for the three sequences $a^j,b^j,c^j$ modulo $\Delta$. Choose a fixed $R$ larger than that threshold plus $\Delta T$. For large $D$, the three numbers \[ b^Rc^{D-R},\quad a^Rc^{D-R},\quad a^{D-R}b^R \] have collective gcd $1$, so their residues generate $\mathbb Z/\Delta$. Any desired residue is a combination with coefficients in $\{0,\ldots,\Delta-1\}$. Replace a coefficient by that many distinct copies obtained by moving $0,T,2T,\ldots$ units of exponent along the same face. Eventual periodicity preserves every residue. The three families lie in different face interiors and so are disjoint. Their $b$ exponents are bounded independently of $D$, and $a<c$, proving the claimed $K_2c^D$ sum bound.
Amplification to every large integer
Choose $u>H+1$ and then $v\ge2$ so large that $ba^v<c^v$. Set \[ A=a^{u+v},\quad B=b^uc^v,\quad q=abc, \] so $A<B$ and $(A,B)=1$. Obtain the progression of sets $E_r$ above with length $L=4AB$, and put $\Delta=qd$. At stage $N$, multiply $E_{r_e}$ by $qA^{N-e}B^e$ for $0\le e\le N$ and take their union. All terms lie in \[ L_{D_N},\qquad D_N=\delta+(u+v)N+3. \] Their $b$ exponents lie in the disjoint ranges $[ue+1,ue+H+1]$, so there are no repeated terms. Their sums range over \[ O_N+\Delta T_N,\qquad O_N=qs\sum_{e=0}^NA^{N-e}B^e. \] Every such term lies in the interior of the exponent simplex. We can therefore adjoin the face corrections without collisions. By $1$, all integers in \[ J_N=[\lceil O_N+\Delta U_N+K_2c^{D_N}\rceil, O_N+\Delta V_N] \] are represented: choose a face correction of the needed residue modulo $\Delta$ and subtract it. Since $c^{D_N}/B^N=c^{\delta+3}(c/b)^{uN}\to0$, the interval $J_N$ has length at least $X=qB^N$ for large $N$, and its lower endpoint is at most $KB^{N+1}$ for a fixed $K$.For each $H+2\le s\le N$, consider the interior monomials with $b$ exponent $uN+s+1$. Along this line successive values have ratio $c/a$. The smallest value, with $c$ exponent $1$, divided by $X$ is \[ a^{\delta-s}(ba^v/c^v)^N b^{s-N}\le a^\delta(ba^v/c^v)^N\longrightarrow0. \] Equivalently its exact ratio is $a^{\delta-s}b^s(a/c)^{vN}$; the displayed bound uses $s\le N$ and $a^{-s}\le1$. The largest value, with $a$ exponent $1$, has ratio $c^\delta(b/c)^s>1$. The line therefore contains a term $z_s$ with $(a/c)X<z_s\le X$. These terms are distinct, lie above every earlier block in $b$ exponent, and lie off the faces. They may be adjoined independently. Adding a new available summand no larger than the length of an interval of represented integers extends that interval by the summand. Thus we obtain an interval $I_N$ of subset sums on $L_{D_N}$ with \[ \inf I_N\le KB^{N+1},\qquad |I_N|-1\ge\kappa N B^N \] for fixed $\kappa>0$ and large $N$. Choose an integer $t$ such that $KB^{1-t}\le1$. For large $M$ take $N=M-t$. Then the lower endpoint is at most $B^M$, while the interval length is at least $B^{M+1}$. Hence $[B^M,B^{M+1})\subseteq I_N$. These intervals cover every sufficiently large integer. Each representation belongs to one level and is therefore primitive. $\square$