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Find all the solutions to:

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

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

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1 Answer 1

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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.

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  • $\begingroup$ Thank you for your great answer! $\endgroup$
    – graffe
    May 22 at 19:27

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