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Need to find all the posible solutions (a,b,c) for 1/a + 1/b + 1/c = 1, a,b,c integers.

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- Thread starter xax
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- #1

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Need to find all the posible solutions (a,b,c) for 1/a + 1/b + 1/c = 1, a,b,c integers.

- #2

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xax, please show your work, and post your homework questions in the correct forum.

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- #4

CRGreathouse

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If a = 1 then the sum is too large, so [itex]2\le a\le b\le c.[/itex] If a = 2, [itex]1/b+1/c=1/2[/itex] so clearly b needs to be no more than 4 (to get 1/4 + 1/4 = 1/2). Check the cases of b = 3 and b = 4.

If a = 3 then b and c need to be small to make the sum 1, so check cases similarly.

Finally, rearrange the solutions you have as needed (so a need not be the smallest).

- #5

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That is true only if a, b, c are restricted to the positive integers. The OP just said integers.You can start by assuming, for the time being, that [itex]a\le b\le c[/itex] -- you can rearrange the terms later if you need to.

If a = 1 then the sum is too large

- #6

CRGreathouse

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That is true only if a, b, c are restricted to the positive integers. The OP just said integers.

True, I missed that. There is one family of solutions which includes negative numbers.

- #7

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for example if a=-b an c is arbitrary big or similar we have an approximate solution in integers to your equation.

or simply a=-b and c=1 b=-c and a=1 a=-c and b=1

or simply a=-b and c=1 b=-c and a=1 a=-c and b=1

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- #8

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Thanks alot to all of you and expecialy to CRGreathouse. I've proven that a<=3, b<=4 and found all the solutions(there are 3 in total).

Edit: If one of them is negative the solution is (t,1,-t). Do you think there are more?

Edit: If one of them is negative the solution is (t,1,-t). Do you think there are more?

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- #9

CRGreathouse

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Do you think there are more?

If two are negative you have 1 - 1/a - 1/b < 1. If three are negative you have -1/a - 1/b - 1/c < 0.

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- #11

CRGreathouse

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- #12

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2, 3, 6

2, 6, 3

3, 2, 6

3, 6, 2

6, 2, 3

6, 3, 2

6 solutions..

1/2 + 1/3 + 1/6 = 1.

Then i got a new question.

1/a + 1/b + 1/c = 7.

If a, b, c, are integers, how many solutions are there?

- #14

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Zero! Because 1/a, 1/b and 1/c are all <=1. Therefore their sum is <=3.

- #15

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I thought of it as well..

Just needed a second opinion.

Thanks! :)

Just needed a second opinion.

Thanks! :)

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