Solve Equation II: $(1+a!)(1+b!)=(a+b)!$

[anemone]

Solve in non-negative integers the equation $(1+a!)(1+b!)=(a+b)!$.

 

[mente oscura]

Solve in non-negative integers the equation $(1+a!)(1+b!)=(a+b)!$.

Hello.

Spoiler

$$(1+a!)(1+b!)=1+a!+a!b!+b!=(a+b)!$$

$$If \ a>b \rightarrow{}b!|(a!,a!b!,(a+b)!)$$

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.

 

[anemone]

Hello.

Spoiler

$$(1+a!)(1+b!)=1+a!+a!b!+b!=(a+b)!$$

$$If \ a>b \rightarrow{}b!|(a!,a!b!,(a+b)!)$$

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.

Well done, mente oscura! Thanks for participating too!