Gonitzoggoβ
This is a proof based problem added for learning purposes and does not accept submissions.
You can view the solution by clicking on the solution tab.
You can view the solution by clicking on the solution tab.
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$x! + 2^y = (x+1)! $
$ \implies 2^y = (x+1)\cdot x! - x!$
$ \implies 2^y = x\cdot x!$
Now, if $x>2$, $3 \mid x\cdot x!$. But $3\nmid 2^y$. $\therefore x \leq 2$. $x=0$ doesn't work because $2^y = 0$ has no solutions.
By putting $x=1,2$, we obtain that $(x,y)=(1,0),(2,2)$ and we are done.