SOLVED
When is the product of two or more disjoint blocks of consecutive integers a power? Is it true that there are only finitely many collections of disjoint intervals $I_1,\ldots,I_n$ of size $\lvert I_i\rvert \geq 4$ for $1\leq i\leq n$ such that
\[\prod_{1\leq i\leq n}\prod_{m\in I_i}m\]
is a square?
Erdős and Selfridge have proved that the product of consecutive integers is never a power. The condition $\lvert I_i\rvert \geq 4$ is necessary here, since Pomerance has observed that the product of
\[(2^{n-1}-1)2^{n-1}(2^{n-1}+1),\]
\[(2^n-1)2^n(2^n+1),\]
\[(2^{2n-1}-2)(2^{2n-1}-1)2^{2n-1},\]
and
\[(2^{2n-2}-2)(2^{2n}-1)2^{2n}\]
is always a square.
This is false: Ulas [Ul05] has proved there are infinitely many solutions when $n=4$ or $n\geq 6$ and $\lvert I_i\rvert=4$ for $1\leq i\leq n$. Bauer and Bennett [BaBe07] proved there are infinitely many solutions when $n=3$ or $n=5$ and $\lvert I_i\rvert=4$ for $1\leq i\leq n$. Furthermore, Bennett and Van Luijk [BeVL12] have found infinitely many solutions when $n\geq 5$ and $\lvert I_i\rvert=5$ for $1\leq i\leq n$.
In general, Ulas conjectures there are infinitely many solutions for any fixed size of $\lvert I_i\rvert$, provided $n$ is sufficiently large.
