anemone said:Solve in non-negative integers the equation $(1+a!)(1+b!)=(a+b)!$.
mente oscura said:Hello.
[tex](1+a!)(1+b!)=1+a!+a!b!+b!=(a+b)![/tex]
[tex]If \ a>b \rightarrow{}b!|(a!,a!b!,(a+b)!)[/tex]
Therefore:
[tex]b!|1 \rightarrow{}b!=1 \rightarrow{}b=0 \ or \ b=1[/tex]
[tex]If \ b=1 \rightarrow{}(1+a!)2=(a+1)![/tex]
[tex]2+2a!=(a+1)a! \rightarrow{}a=2[/tex]
[tex]If \ b=0 \rightarrow{}(1+a!)2=a![/tex]
[tex]a!=-2 \ absurdity[/tex]
The end:
[tex]If \ a=b \rightarrow{}(1+a!)(1+a!)=1+2a!+a!a!=(2a)![/tex]
[tex]a!|1 \rightarrow{}a=0 \ or \ a=1[/tex]
[tex]If \ a=0 \rightarrow{}2*2=1 \ absurdity[/tex]
[tex]If \ a=1 \rightarrow{}2*2=2 \ absurdity[/tex]
I will conclude by:
[tex]If \ a>b \rightarrow{}a=2 \ and \ b=1[/tex]
Same b>a ...
Regards.
To solve this equation, you need to simplify the terms and isolate the variables on one side of the equation. Then, you can use algebraic manipulation and properties of factorials to solve for the values of a and b that make the equation true.
The values of a and b can be any non-negative integers, including 0. However, there may be restrictions on the values depending on the context of the equation.
Yes, this equation can have multiple solutions. In fact, there are infinite solutions since the values of a and b can vary as long as they satisfy the equation.
Yes, there are specific methods and strategies for solving equations with factorials. These include expanding factorials, using properties of factorials, and substitution.
This equation can be used in various fields such as probability, combinatorics, and number theory. It can also be applied in solving problems related to permutations and combinations.