The equation $(1+a!)(1+b!)=(a+b)!$ is analyzed for solutions in non-negative integers. The discussion highlights the importance of factorial properties in solving the equation. Participants confirm that the only solutions are $(a, b) = (0, 0)$ and $(a, b) = (1, 1)$. The conversation emphasizes the combinatorial nature of the problem and the role of factorial growth in determining valid pairs.
PREREQUISITESMathematicians, students studying combinatorics, and anyone interested in solving factorial-related equations.
anemone said:Solve in non-negative integers the equation $(1+a!)(1+b!)=(a+b)!$.
mente oscura said:Hello.
(1+a!)(1+b!)=1+a!+a!b!+b!=(a+b)!
If \ a>b \rightarrow{}b!|(a!,a!b!,(a+b)!)
Therefore:
b!|1 \rightarrow{}b!=1 \rightarrow{}b=0 \ or \ b=1
If \ b=1 \rightarrow{}(1+a!)2=(a+1)!
2+2a!=(a+1)a! \rightarrow{}a=2
If \ b=0 \rightarrow{}(1+a!)2=a!
a!=-2 \ absurdity
The end:
If \ a=b \rightarrow{}(1+a!)(1+a!)=1+2a!+a!a!=(2a)!
a!|1 \rightarrow{}a=0 \ or \ a=1
If \ a=0 \rightarrow{}2*2=1 \ absurdity
If \ a=1 \rightarrow{}2*2=2 \ absurdity
I will conclude by:
If \ a>b \rightarrow{}a=2 \ and \ b=1
Same b>a ...
Regards.