GPT 6 Astra Ultra attempt on Math-5606
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Erdős Problem #365: consecutive powerful numbers and their count
1. The two questions must be separated
The precise first question is whether, whenever $n,n+1$ are powerful, at least one must be a square. Its answer is known to be no. The second asks whether \[ Q(X)=\#\{n\le X:n,n+1\text{ are powerful}\} \le(\log X)^{O(1)}. \] We prove the first answer, an explicit infinite family of nonsquare counterexamples, and an $O(\log X)$ upper bound after fixing the two squarefree kernels. Uniformly controlling all kernels remains unresolved, so this is not a full solution to the two-part problem.2. Golomb's counterexample and an explicit infinite continuation
The factorizations \[ 12167=23^3,\qquad12168=8\cdot39^2=2^3\,3^2\,13^2 \] show that both integers are powerful and neither is a square. We can continue this same example indefinitely. Start with $x_0=1,y_0=39$, and set \[ x_{j+1}=24335x_j+624y_j,\qquad y_{j+1}=949026x_j+24335y_j. \tag{1} \] To verify the constants, let $\lambda=24335+3588\sqrt{46}$. Exact calculation gives $24335^2-46\cdot3588^2=1$, and multiplication yields \[ (4y_j+23x_j\sqrt{46})\lambda =4y_{j+1}+23x_{j+1}\sqrt{46}. \] The norm identity therefore preserves \[ 8y_j^2-23^3x_j^2=1. \] The positive integer coordinates strictly increase in $1$. Hence \[ n_j=23^3x_j^2,\qquad n_j+1=8y_j^2 \] are infinitely many consecutive powerful pairs. Their respective $23$-adic and $2$-adic exponents are $3+2v_{23}(x_j)$ and $3+2v_2(y_j)$, both odd, so neither number can be a square. The explicit multiplication formula also gives $n_j\asymp\lambda^{2j}$, and consequently $Q(X)\gg\log X$. This lower bound is compatible with the proposed polylogarithmic upper bound.3. A logarithmic upper bound for each fixed pair of kernels
Every powerful number has a unique expression $b^3u^2$ with $b$ squarefree. Indeed, the primes occurring to odd exponent form $b$, and the remaining exponents are even and nonnegative. Fix squarefree $b,c\ge1$ and consider solutions \[ c^3v^2-b^3u^2=1,\qquad u,v\ge1. \tag{2} \] This is exactly the restriction to consecutive powerful numbers with these two kernels. For a solution, put $\theta=\sqrt{b^3/c^3}$. Then \[ \frac vu=\sqrt{\theta^2+\frac1{c^3u^2}}>\theta, \qquad 0<\frac vu-\theta <\frac1{2\sqrt{b^3c^3}\,u^2}. \tag{3} \] If $u_1<u_2$ are two solution denominators, their ratios $v_i/u_i$ strictly decrease as $u_i$ increases. The positive integer $v_1u_2-v_2u_1$ is at least $1$. Therefore $3$ gives \[ \frac1{u_1u_2} \le\frac{v_1}{u_1}-\frac{v_2}{u_2} <\frac1{2\sqrt{b^3c^3}\,u_1^2}, \quad\text{so}\quad u_2>2(bc)^{3/2}u_1. \tag{4} \] The denominators thus grow geometrically. Since $b^3u^2\le X$ implies $u\le\sqrt X$, equation $2$ has $O_{b,c}(\log X)$ solutions below $X$. For $b=c=1$ there are no positive consecutive squares anyway. No structural theorem about Pell equations is needed for this bound.4. Assessment and provenance
The unresolved step is to sum $4$ over kernels that grow with $X$. There can be up to $X^{1/3}$ choices for each squarefree kernel before the equation is imposed; a bound for each fixed equation does not produce the required uniform polylogarithmic total. The first subquestion is completely disproved, but the counting question remains unresolved in this attempt.Source: https://www.erdosproblems.com/365, checked 2026-09-25, which attributes the displayed seed to Golomb and the existence of infinitely many nonsquare pairs to Walker. The previous GPT Pro 5.2 and 5.4 attempts were checked; the latter correctly separated the counting question. Formula $1$ is a fully verified Pell-unit continuation of Golomb's seed, offered as a reconstruction rather than a novelty claim. No formal verification or independent proof of the proposed counting bound is claimed.