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Find all positive integers $m$ and $n$, such that: $$\frac 1m + \frac 1n - \frac 1{mn}=\frac 25$$ Actually, I have already solved this problem using inequality. The solutions I have found are: $$\{(3,10),\;(4,5),\;(10,3),\;(5,4)\}$$ But I want to solve it in a different way.

I have tried this: $$\frac {m+n-1}{mn}=\frac 25\implies m+n-1=2k \;\;\&\;\; mn=5k$$ where $k$ is a positive integer. Then $m+n=2k+1$ and $mn=5k$. I wonder if I can do anything from here. I have tried $x^2 -(2k+1) + 5k=0$ where the solutions of the equations are $m,n$ where $m≠n$, because $m+n$ is odd. But it doesn't seem that it is leading me to the solution.

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It is probably a good idea to multiply both sides my $10mn$. This gets you $$10n+10m-10=4mn$$

Now there are just integers. Also, the number in front of $mn$ is a square. This can be rewritten as $4mn-10n-10m+25=15$, and hence $(2m-5)(2n-5)=15$. That means that $2m-5$ a divisior is of 15. This gives you that $m \in \{3,4,5,10\}$, and you get the values for $n$ form there.

This is roughly the same approach as peter.petrov, however, multiplying by $10mn$ gives us only integers to work with, which means we can use divisibility properties.

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The LHS is equal to: $1-(1-\frac{1}{m}) \cdot (1-\frac{1}{n})$, maybe that could somewhat help you.

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Since $$ \left(1-\frac1m\right)\left(1-\frac1n\right)=\frac35 $$ we get that $m,n\ge3$; if $m=2$ or $n=2$, the left side is less than or equal to $\frac12$.

Furthermore, $$ n=\frac{5m-5}{2m-5} $$ $m=3\implies n=10$
$m=4\implies n=5$
$m=5\implies n=4$
$m=10\implies n=3$

For $m\gt10$, we have $n\lt3$, so these are all the solutions.

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