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Claim. There are infinitely many pairs of consecutive powerful numbers $n$, $n+1$ with neither member a square, so the first question of Problem 365, read as the site reads it, has answer no infinitely often. D. T. Walker, Consecutive integer pairs of powerful numbers and related Diophantine equations, Fibonacci Quart. 14 (1976), no. 2, 111--116, describes every consecutive powerful pair with neither member a square through the solutions of $mX^2-nY^2=\pm1$ for which $mX^2$ and $nY^2$ are both powerful (the paper's property $Q$); by his Theorems 3.2 and 3.5 these are the odd powers of the least such solution. His example $7X^2-3Y^2=1$, whose least solution is $(X,Y)=(2,3)$, gives infinitely many solutions of $7^3x^2=3^3y^2+1$ and so infinitely many pairs $(3^3y^2,\,7^3x^2)$ with neither member a square. The least such pair is $48{,}689{,}748{,}233{,}307=3\cdot4{,}028{,}637^2=3^3\cdot1{,}342{,}879^2$ and $48{,}689{,}748{,}233{,}308=7\cdot2{,}637{,}362^2=7^3\cdot376{,}766^2$. The same family answers the question Guy's B16 asks, whether infinitely many pairs do not come from Pell equations $x^2-dy^2=\pm1$, which the library's card Guy 2004 records; the card Walker 1976 records the Pell parametrization.

Covers. The first question only (the part pell), refuted with an infinite family; the single counterexample is Golomb's. The second question, the $(\log x)^{O(1)}$ bound on the count, is untouched.

Depends on. Nothing in this wiki; the claim rests on the cited paper.

Acceptance. Refereed: the paper appeared in the Fibonacci Quarterly, a refereed journal, in April 1976, the month this page is dated to. The site's commentary credits Walker with the infinitely many counterexamples, but the site labels the problem OPEN (page last edited 31 October 2025), so that commentary is not acceptance of the problem and the page lists no reviewed evidence. The Pell identity and the least pair are checked above; the proof is not reviewed in this corpus.