Long runs of non-squarefree numbers

Idea. The Chinese remainder theorem lets us prescribe divisibility conditions independently, as long as the moduli are pairwise coprime.

Proof. Take distinct primes $p_1,\dots,p_n$. The moduli $p_1^2,\dots,p_n^2$ are pairwise coprime, so by the Chinese remainder theorem there is a positive integer $x$ with
$$x\equiv-i\pmod{p_i^2}\qquad(i=1,\dots,n).$$
Then $p_i^2\mid x+i$ for each $i$, so $x+1,x+2,\dots,x+n$ are $n$ consecutive integers each divisible by a square greater than $1$.

Example. $n=3$ with primes $2,3,5$: $x\equiv-1\pmod4$, $x\equiv-2\pmod9$, $x\equiv-3\pmod{25}$ gives $x=547$: $548=4\cdot137$, $549=9\cdot61$, $550=25\cdot22$ ✓.

Remark. So the gaps between consecutive squarefree numbers are unbounded.

Lesson. To find consecutive integers with prescribed divisibility properties, use the Chinese remainder theorem.