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

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The equation ab = a + b has integer solutions, with one known solution being a = b = 2. To find other solutions, it is derived that b must equal a / (a - 1), meaning a must be a multiple of a - 1 for b to remain an integer. The discussion highlights that if b exceeds 2, the equation leads to no integer solutions, as demonstrated through various substitutions. The conversation emphasizes the need for a more rigorous approach to explore potential integer pairs, particularly for values of a and b less than or equal to 2. Overall, the exploration of integer solutions remains focused on the constraints imposed by the equation.
Einstein's Cat
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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?
 
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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?
 
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Einstein's Cat said:
Are there other solutions?
One more.
 
blue_leaf77 said:
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
 
Einstein's Cat said:
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!
 
Orodruin said:
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!
 
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?
 
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Einstein's Cat said:
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##?
 
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fresh_42 said:
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
 
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Einstein's Cat said:
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.
 
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