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

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

 

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).

 

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.

 

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

 

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.

 

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).

 

try to prove if there is finite amount of solutions

 

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?

 

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

 

I thought of it as well..
Just needed a second opinion.
Thanks! :)