GPT 6 Astra Ultra attempt on Math-5605

Erdős Problem #364: three consecutive powerful integers

1. Statement and scope

A positive integer is powerful, or $2$-full, when every prime exponent in it is either zero or at least two. Are there any three consecutive positive powerful integers? We prove structural restrictions, reconstruct the infinite family of consecutive pairs, and give the precise conditional consequence of $abc$. An exact finite search is included as a bounded check, not as an answer to the infinite question, which remains unresolved.

2. Unique parametrization and an exhaustive search

Every powerful integer has a unique expression \[ n=a^2b^3,\qquad a,b\ge1,\quad b\text{ squarefree}. \] For each odd positive prime exponent $e$, place that prime once in $b$ and $(e-3)/2$ times in $a$; for an even exponent use $e/2$ copies in $a$. This construction is forced by parity and proves uniqueness. There is no requirement that $a$ and $b$ be coprime. In particular the number of powerful integers at most $X$ is at most \[ \sum_{b\le X^{1/3}}\left\lfloor\sqrt{X/b^3}\right\rfloor \le \sqrt X\sum_{b\ge1}b^{-3/2}. \] Enumerating precisely the squarefree $b\le X^{1/3}$ and $1\le a\le\lfloor\sqrt{X/b^3}\rfloor$ therefore gives an exhaustive finite algorithm, with no factorization heuristic. For $X=10^{10}$ the exact enumeration contains $214122$ integers. Checking membership of $n,n+1,n+2$ finds no triple all of whose members lie in $[1,X]$. This computation makes no claim about larger integers and is far smaller than published computational searches.

3. Local obstructions and an infinite family of pairs

Four consecutive integers cannot all be powerful, because one has residue $2$ modulo $4$. If $m-1,m,m+1$ are powerful, this also shows that the center satisfies $m\equiv0\pmod4$. Residues $3,6$ modulo $9$ are impossible for a powerful integer, so examining the three consecutive residues gives \[ m\equiv0,1,8\pmod9. \] The two odd neighbors cannot both be squares: two positive squares differing by $2$ would give $(v-u)(v+u)=2$, which is impossible for positive integers $u<v$. Thus at least one neighbor has a nontrivial squarefree cube factor in the parametrization above.

These restrictions do not prevent consecutive pairs. Starting at $(x_1,y_1)=(3,1)$, define \[ x_{j+1}=3x_j+8y_j,\qquad y_{j+1}=x_j+3y_j. \] Direct expansion preserves $x_j^2-8y_j^2=1$, and both coordinates increase. Hence $8y_j^2$ and $8y_j^2+1=x_j^2$ give infinitely many distinct powerful pairs. Every prime exponent in the first number is even except possibly that of $2$, which is $3+2v_2(y_j)$ and therefore at least three.

4. What the $abc$ conjecture would actually imply

Assume the usual $abc$ conjecture: for every $\varepsilon>0$ there is $C_\varepsilon$ such that coprime positive $a+b=c$ satisfy $c\le C_\varepsilon\operatorname{rad}(abc)^{1+\varepsilon}$. If $m-1,m,m+1$ are powerful, each obeys $\operatorname{rad}(u)\le\sqrt u$. Apply the conjecture to \[ (m^2-1)+1=m^2. \] The summands are coprime, and \[ \operatorname{rad}\big((m^2-1)m^2\big) =\operatorname{rad}\big((m-1)m(m+1)\big) \le\sqrt{(m-1)m(m+1)}<m^{3/2}. \] It follows that $m^2\le C_\varepsilon m^{3(1+\varepsilon)/2}$. Taking $\varepsilon<1/3$ bounds $m$. Thus $abc$ implies only finitely many such triples, not that there are none. No use of this conjecture is made in the unconditional results or the finite search.

5. Assessment and references

The remaining obligation is an unconditional existence proof or an obstruction valid for every candidate center. The congruences, parametrization, and conditional finiteness do not provide it. Source: https://www.erdosproblems.com/364, checked 2026-09-25, which records Mahler's Pell observation and the conditional $abc$ consequence. The prior GPT Pro 5.2 argument was checked; its parametrization and elementary obstruction are reconstructed here. No novelty or formal Lean verification is claimed.

COMPLETION ESTIMATE

COMPLETION: 35%