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Claim. Fix $l\geq 2$ and a prime $q$, and put $f(x)=(x+1)\cdots(x+l)$. Theorem 2 of M. Skałba, Products of disjoint blocks of consecutive integers which are powers, Colloq. Math. 98 (2003), no. 1, 1--3, concerns the equation $\prod_{i=1}^k f(x_i)=y^q$ when the number $k$ of blocks may vary. It has solutions in nonnegative integers with $x_{j+1}\geq x_j+l$, the least with $x_k+l<e^{cql}$ for some $c<1.25506$, and the number $N(x)$ of solutions with $x_k+l\leq x$ satisfies $N(x)\geq 2^{(x/l)(1+o(1))}$. With $q=2$ and $l=4$ this gives infinitely many collections of disjoint blocks of four positive integers whose product is a square. So the question of Problem 363, read with the number of intervals free, has the answer no. The paper says that it refutes Erdős and Graham's suggestion only because the number of blocks varies. The journal gives the issue's year and no day, so the page is dated by the year's first day.

Argument, in outline. The proof applies the Davenport constant of the group $C_q^{\pi(x)}$ and Olson's count of zero-sum subsequences to the blocks that partition $[1,x]$.

Acceptance. The result appeared in a refereed journal, Colloquium Mathematicum, in 2003: the refereed evidence. The site's commentary does not credit the paper, so there is no reviewed evidence. The disproofs with a fixed number of blocks are Ulas's, Bauer and Bennett's and Bennett and Van Luijk's.