The equation \((1+a!)(1+b!) = (a+b)!\) is analyzed for ordered pairs \((a,b)\) where \(a, b \in \mathbb{N}\). The discussion reveals that the only solutions are \((0,0)\), \((1,0)\), and \((0,1)\). The factorial function \(n!\) is defined for non-negative integers, and the analysis confirms that for \(a, b \geq 2\), the equation does not hold true. Thus, the complete set of ordered pairs satisfying the equation is limited to these three combinations.
PREREQUISITESMathematicians, students studying number theory, educators teaching algebraic concepts, and anyone interested in combinatorial mathematics.
jacks said:Find all ordered pairs $(a,b)$ for which $(1+a!)(1+b!) = (a+b)!$
where $a,b\in \mathbb{N}$ and $n! = $ factorial on $n$