- #1

Einstein's Cat

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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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In summary, the conversation discusses the equation ab = a + b and its solutions for integers a and b. One solution is a = b = 2, and the conversation explores other possible solutions and the conditions necessary for b to be an integer if a is. The conversation also mentions the quadratic expression of the equation and how it can be used to find solutions. It concludes with a discussion on the restrictions for a and b and how they can be handled manually.

- #1

Einstein's Cat

- 182

- 2

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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- #2

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- #3

blue_leaf77

Science Advisor

Homework Helper

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One more.Einstein's Cat said:Are there other solutions?

- #4

Einstein's Cat

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it would be thatblue_leaf77 said:One more.

b= a / (a - 1)

thus b has to be a multiple of a - 1

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

- #5

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No, it is ##a## that needs to be a multiple of ##a-1## in order for ##b## to be an integer!Einstein's Cat said:thus b has to be a multiple of a - 1

- #6

Einstein's Cat

- 182

- 2

what pairs of integers satisfy that? I can't for the life of me think of any!Orodruin said:No, it is ##a## that needs to be a multiple of ##a-1## in order for ##b## to be an integer!

- #7

jedishrfu

Mentor

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

- #8

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So prove it.Einstein's Cat said:what pairs of integers satisfy that? I can't for the life of me think of any!

##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##?

- #9

Einstein's Cat

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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 casefresh_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##?

- #10

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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.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

The main purpose of solving ab = a + b is to find the possible values of a and b that satisfy the equation. This can help in solving various mathematical problems and equations involving variables.

No, there are infinite solutions for the equation ab = a + b. This is because any value of a and b that satisfy the equation can be considered a solution.

Yes, the equation ab = a + b can be solved for all real numbers. This is because the equation does not have any restrictions on the values of a and b.

Yes, there are specific values of a and b that satisfy the equation ab = a + b. Some examples include a = 1 and b = 2, a = 0 and b = 0, and a = 2 and b = 4.

Solving the equation ab = a + b can be useful in various real-life scenarios, such as calculating the total cost of items when a discount is applied, finding the dimensions of a rectangle with a given perimeter, and solving for the growth rate of a population. It is a fundamental equation in algebra and can be applied in many different contexts.

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