Solve this equation :
$(2x)! = (x)! (x+2)!$
2026-04-03 07:44:04.1775202244
Solve this equation : $(2x)! = (x)! (x+2)!$
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If $x\ge 4$, $$\begin{align}(x+1)(x+2)&=\frac{(x+1)(x+2)(2x)!}{(x+2)!x!}\\&={2x\choose x}\\&>{2x\choose3}\\&=\frac{2x(2x-1)(2x-2)}{6}\\&\ge\frac x3(x+3)(x+2)\end{align}$$ quickly leads to a contradiction. Manually check all cases with $x\le 3$.