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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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In summary, the OP has found three solutions to the equation 1/a + 1/b + 1/c = 1, a,b,c integers. xax. He has also asked a new question about solving for 7 when a, b, c are integers.

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

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- #4

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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).

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

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D H said: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.

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

try to prove if there is finite amount of solutions

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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?

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

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I thought of it as well..

Just needed a second opinion.

Thanks! :)

Just needed a second opinion.

Thanks! :)

This equation is used to solve for integer solutions when adding fractions with different denominators.

Integer solutions are values for the variables (a, b, and c) that result in whole numbers when substituted into the equation.

To solve this equation, you can use algebraic manipulation and trial and error to find values for a, b, and c that satisfy the equation.

Yes, the values of a, b, and c must all be non-zero integers. In other words, they cannot be fractions or decimals.

No, this equation can only be solved for integer solutions. If you are looking for non-integer solutions, a different equation or method would need to be used.

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