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Question Number 94786 by MJS last updated on 21/May/20

$$x,y\in \mathbb{N}\mathrm{\setminus }\{0,1\}\wedge x⩽y$$$$\mathrm{find}z\in \mathbb{N}\mathrm{with}z!=x!y!$$

Commented by hknkrc46 last updated on 21/May/20

$$\left(\mathrm{a}\right)\mathrm{x}!=\mathrm{y}+1\Rightarrow \mathrm{z}!=(\mathrm{y}+1)!\Rightarrow \mathrm{z}=\mathrm{y}+1$$$$(\mathrm{a}.1)\mathrm{x}=3\Rightarrow \mathrm{y}=5\Rightarrow \mathrm{z}=6$$$$(\mathrm{a}.2)\mathrm{x}=4\Rightarrow \mathrm{y}=23\Rightarrow \mathrm{z}=24$$$$(\mathrm{a}.3)\mathrm{x}=5\Rightarrow \mathrm{y}=119\Rightarrow \mathrm{z}=120$$$$\cdot \cdot \cdot \cdot$$$$\{\mathrm{x},\mathrm{y},\mathrm{z}:\mathrm{x}!=\mathrm{y}+1=\mathrm{z};\mathrm{x}⩾3\}$$$$\left(\mathrm{b}\right)\mathrm{y}!=\mathrm{x}+1\Rightarrow \mathrm{z}!=(\mathrm{x}+1)!\Rightarrow \mathrm{z}=\mathrm{x}+1$$$$(\mathrm{b}.1)\mathrm{y}=3\Rightarrow \mathrm{x}=5\Rightarrow \mathrm{z}=6(\mathrm{x}\nleqq \mathrm{y})$$$$(\mathrm{b}.2)\mathrm{y}=4\Rightarrow \mathrm{x}=23\Rightarrow \mathrm{z}=24(\mathrm{x}\nleqq \mathrm{y})$$$$....$$

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