gpt pro 5.2 attempt on Math-990

1) FORMAL RESTATEMENT

Let $A=\{n_1<n_2<\cdots\}$ be the increasing sequence of powerful (squarefull) numbers, i.e. positive integers $m$ such that for every prime $p\mid m$ we have $p^2\mid m$.

Question (main): Are there only finitely many indices $k$ such that the three consecutive terms $n_k,n_{k+1},n_{k+2}$ form a 3-term arithmetic progression, i.e. \[n_{k+1}-n_k = n_{k+2}-n_{k+1}?\]

Related conjecture (stated): There is no triple of powerful numbers of the form $n,n+1,n+2$.

2) QUICK LITERATURE/CONTEXT CHECK

The problem statement reports only Erdős's conjectural remark about $n,n+1,n+2$. I do not use any external results.

3) ATTACK PLAN

- Use a structural description of powerful numbers to (i) count them and (ii) generate them for computation. - Compute the initial segment of $A$ and check for arithmetic-progression triples among consecutive terms.

4) WORK

Fast reality check (explicit computation). Using the standard parametrisation of powerful numbers and enumerating all powerful numbers $\le 10^6$: - There are 2027 powerful numbers $\le 10^6$. - There are 3 instances where $n_k,n_{k+1},n_{k+2}$ are in arithmetic progression: \[(1728,1764,1800)\ (\text{common difference }36),\] \[(6912,7056,7200)\ (\text{common difference }144),\] \[(729000,729316,729632)\ (\text{common difference }316).\] - There are no triples $n,n+1,n+2$ of powerful numbers with $n+2\le 10^6$.

Lemma 938.1 (Powerful numbers are $a^2 b^3$ with $b$ squarefree). A positive integer $m$ is powerful if and only if it can be written in the form \[m=a^2 b^3\] for some integers $a\ge 1$ and squarefree $b\ge 1$.

Proof. ($\Rightarrow$) Let $m=\prod_p p^{e_p}$ with all $e_p\ge 2$. For each prime $p$, write $e_p$ as $e_p=2u_p+3v_p$ where $v_p\in\{0,1\}$: - If $e_p$ is even, take $v_p=0$ and $u_p=e_p/2$. - If $e_p$ is odd, then $e_p\ge 3$; take $v_p=1$ and $u_p=(e_p-3)/2$. Define \[a:=\prod_p p^{u_p},\qquad b:=\prod_{p:\,v_p=1} p. \] Then $b$ is squarefree by construction (each such prime occurs to exponent 1), and \[a^2 b^3 = \prod_p p^{2u_p} \prod_{p:\,v_p=1} p^{3} = \prod_p p^{2u_p+3v_p} = \prod_p p^{e_p}=m. \]

($\Leftarrow$) Conversely, if $m=a^2 b^3$ with $b$ squarefree, then for any prime $p\mid m$, either $p\mid a$ (so $p^2\mid a^2\mid m$) or $p\mid b$ (so $p^3\mid b^3\mid m$, hence again $p^2\mid m$). Thus every prime divisor of $m$ divides it to exponent at least 2, i.e. $m$ is powerful. QED.

Lemma 938.2 (Crude counting bound). Let $N(X)$ be the number of powerful integers $m\le X$. Then $N(X)=O(X^{1/2})$.

Proof. By Lemma 938.1, every powerful $m\le X$ can be written as $m=a^2 b^3$ with $b$ squarefree and $b^3\le X$, hence $b\le X^{1/3}$. For fixed $b$, the condition $a^2 b^3\le X$ implies $a\le \sqrt{X/b^3}=X^{1/2} b^{-3/2}$. Therefore \[N(X) \le \sum_{\substack{b\le X^{1/3}\\ b\ \mathrm{squarefree}}} \Big\lfloor X^{1/2} b^{-3/2}\Big\rfloor \le X^{1/2} \sum_{\substack{b\ge 1\\ b\ \mathrm{squarefree}}} b^{-3/2}. \] The series $\sum_{b\ge 1} b^{-3/2}$ converges (p-series with exponent $3/2>1$), and the restriction to squarefree $b$ only decreases the sum. Hence the sum is bounded by a constant independent of $X$, giving $N(X)\ll X^{1/2}$. QED.

5) VERIFICATION

- Lemma 938.1: For odd $e_p\ge2$, we used $e_p\ge3$ to make $(e_p-3)/2\ge0$, which holds. - Computation: the three displayed triples were extracted by scanning the sorted list of powerful numbers $\le 10^6$ and checking consecutive triples. This validates existence of such triples at least up to this bound.

6) FINAL

UNRESOLVED (i) Strongest proved partial result: Powerful numbers admit the $a^2b^3$ description (Lemma 938.1) and satisfy the crude count $N(X)=O(X^{1/2})$ (Lemma 938.2). Computationally, there are at least three arithmetic-progression triples among consecutive powerful numbers up to $10^6$, including $(1728,1764,1800)$. (ii) First gap: Prove either that the set $\{k: n_k,n_{k+1},n_{k+2}\ \text{is an AP}\}$ is finite, or construct an infinite family of such indices while controlling the "consecutive in the powerful sequence" condition. (iii) Top 3 next moves: 1. Search for parametrised families of AP triples of powerful numbers and test whether the "no other powerful numbers in between" property persists. 2. Develop local density/gap estimates for powerful numbers in intervals $[x,x+\Delta]$ to bound when a given AP triple can be consecutive. 3. Extend brute-force computation to larger bounds and record the gap patterns around candidate AP triples. (iv) Minimal counterexample structure: If there are infinitely many such AP triples of consecutive powerful numbers, one expects a structured family $(a_k, a_k+d_k, a_k+2d_k)$ of powerful numbers with unusually large gaps to exclude intermediate powerful values; understanding how powerful numbers cluster in residue classes modulo large squares/cubes may be essential.