# Find all solutions to a sum of fractions [closed]

Find all the solutions to:

$$\frac{1}{x}+\frac{2}{y}-\frac{3}{z}=1$$

where $$x, y, z$$ are positive integers.

Solutions

$$(x,y,z)$$ must be one of the following $$(1, 2k, 3k)$$ for any positive integer $$k$$, $$(k, 2, 3k)$$ for any positive integer k, $$(2,1,2)$$ or $$(2,3,18)$$

Reasoning

If $$x>1$$ and $$y>3$$ then the sum of the first two terms on the left hand side is at most $$1$$ and there is obviously no solution.
To analyse the other cases, let us first multiply across by $$xyz$$ (necessarily positive) to get $$yz + 2xz - 3xy = xyz$$ If $$x=1$$ then we have $$2z=3y$$. This means that $$z$$ must be divisible by $$3$$ and can be written as $$3k$$ for some positive integer $$k$$ and thus $$y=2k$$.

If $$y=1$$ then the equation reduces to $$xz - 3x + z= 0$$ or $$(x+1)(z-3) = -3$$.
The factors on the left hand side of this equation are integers and must either be $$-1$$ and $$3$$ or $$1$$ and $$-3$$ in some order. Since $$x$$ and $$z$$ are positive, it must be that $$x+1=3$$ and $$z-3 = -1$$ giving us $$x=2, z=2$$ as the only solution in this branch.

If $$y=2$$ then we have $$z = 3x$$ and this gives a solution for any integer $$x$$.

Finally, if $$y=3$$, then we have $$xz +9x - 3z = 0$$ or $$(x-3)(z+9) = -27$$.
Again, since both factors are positive, it must be that $$z+9 > 9$$ and so must be $$27$$ which means $$x-3=-1$$ and so $$x=2, z=18$$ which is the only solution in this branch.

• Thank you for your great answer! May 22 at 19:27