Solving ab = a + b: What are the Possible Answers?

[Einstein's Cat]

There is the equation

ab = a + b (the solutions have to be integers)

A solution to this is a= b= 2.

What are other solutions? Are there other solutions?

 

[Orodruin]

Why don't you solve for b in terms of a and think about what would be necessary for b to be an integer if a is?

 

[blue_leaf77]

Are there other solutions?

One more.

 

[Einstein's Cat]

One more.

it would be that

b= a / (a - 1)
thus b has to be a multiple of a - 1

I assume that the other solution is a=b=0

 

[Orodruin]

thus b has to be a multiple of a - 1

No, it is $a$ that needs to be a multiple of $a-1$ in order for $b$ to be an integer!

 

[Einstein's Cat]

No, it is $a$ that needs to be a multiple of $a-1$ in order for $b$ to be an integer!

what pairs of integers satisfy that? I can't for the life of me think of any!

 

[jedishrfu]

Did you try to simply insert common integers into the equation?

You got one solution of a=b=2 from a^2=(a+a) which can be converted to a quadratic expression to factor right?

 

[fresh_42]

what pairs of integers satisfy that? I can't for the life of me think of any!

So prove it.
$a$ being a multilple of $a-1$ means $a=b\cdot (a-1)$ and since $a$ is the greater number, $b$ has to be positive.
Can you show why $b$ cannot be greater than $2$?

 

[Einstein's Cat]

So prove it.
$a$ being a multilple of $a-1$ means $a=b\cdot (a-1)$ and since $a$ is the greater number, $b$ has to be positive.
Can you show why $b$ cannot be greater than $2$?

if b is greater than 2, say 3, then a= 3a - 3. no integer satisfies this from 1 to 10 and has a gets bigger the 3 is a smaller proportion of a so it can be assumed that this is the case

 

[fresh_42]

if b is greater than 2, say 3, then a= 3a - 3. no integer satisfies this from 1 to 10 and has a gets bigger the 3 is a smaller proportion of a so it can be assumed that this is the case

Well, ... yes. But the restriction to numbers until $10$ lacks a bit of rigor. $a=b(a-1) > 2(a-1) = 2a -2$ and thus $a<2$ would be more general and also leaves you with the cases $a \in \{0,1\}\,$or$\, b \in \{0,1,2\}$ which you can handle manually. Or you proceed along the lines @jedishrfu has pointed out in #7.