I am looking for some insights on how to prove that for every integer $l$, there exists an integer $x$ such that:
$$x^2-1 = l(l+1)(l+2)(l+3).$$
So far, what I did was:
$$x^2-1=(x-1)(x+1).$$
2026-03-29 14:58:50.1774796330
Insights for proving $(x^2-1) = l(l+1)(l+2)(l+3)$
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$$l(l+1)(l+2)(l+3)+1=l^4+6l^3+11l^2+6l+1=(l^2+3l+1)^2$$