Clearly, $4$ divides $(x)(x+1)(x+2)(x+3)$
Likewise, if a prime $p^2$ divides $x$, $p^2$ does not divide $(x+p)(x+2p)\dots(x+[p-1]p)$.
Does it follow that there are an infinite number of instances $4x+1, 4x+2, 4x+3$ that are square free?
2026-03-29 10:18:12.1774779492
Are there infinite square free integers $4x+1, 4x+2, 4x+3$
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