Solve 1/a + 1/b + 1/c = 1: Find All Integer Solutions

  • 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.
  • #1

xax

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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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  • #2
xax, please show your work, and post your homework questions in the correct forum.
 
  • #3
D H, I thought I was in the right forum. What I did so far: b+c = (a-1)*t and b*c = a*t and (c-1)*(b-1) = t+1. I'm stuck because I get too many posibilities.
 
  • #4
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, 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
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
That is true only if a, b, c are restricted to the positive integers. The OP just said integers.
 
  • #6
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.
 
  • #7
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
 
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  • #8
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?
 
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  • #9
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.
 
  • #10
Yes CRGreathouse, that's why I said that only one can be negative. My question was are there other solutions when one number is negative besides (t,1,-t)?
 
  • #11
If exactly one value is negative and another value is 1, the values are (1, v, -v). Otherwise, the sum of the reciprocals of the positive values is at most 1, so not a solution (since the negative will lower the result).
 
  • #12
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
 
  • #13
This is easy:
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
Zero! Because 1/a, 1/b and 1/c are all <=1. Therefore their sum is <=3.
 
  • #15
I thought of it as well..
Just needed a second opinion.
Thanks! :)
 

1. What is the equation 1/a + 1/b + 1/c = 1 used for?

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

2. What are integer solutions?

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

3. How do I solve this 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.

4. Are there any restrictions on the values of a, b, and c?

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

5. Can this equation be solved for non-integer solutions?

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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