1/a + 1/b + 1/c = 1

[xax]

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

[D H]

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

[xax]

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.

[CRGreathouse]

You can start by assuming, for the time being, that a\le b\le c -- you can rearrange the terms later if you need to.

If a = 1 then the sum is too large, so 2\le a\le b\le c. If a = 2, 1/b+1/c=1/2 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).

[D H]

You can start by assuming, for the time being, that a\le b\le c -- 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.

[CRGreathouse]

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.

[mhill]

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

[xax]

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?

[CRGreathouse]

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.

[xax]

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)?

[CRGreathouse]

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).