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Determine the area of the polygon formed by the ordered pairs (x, y) where x and y are positive integers which satisfy the equation

$\frac{1}{x} + \frac{1}{y}= \frac{1}{13}$

I got : $(x-13)(y-13)=169$

but it only has 2 solutions with positive integers, is there something I'm missing?.

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Hint: the pairs are $$(26,26)\\(14,182)\\(182,14)$$can you finish now?

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  • $\begingroup$ oops i just forgot that its symmetric .... thanks $\endgroup$ – SuperMage1 Aug 10 '18 at 10:30
  • $\begingroup$ You're welcome!! $\endgroup$ – Mostafa Ayaz Aug 11 '18 at 6:14
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It is better to isolate one variable and find the possibilities so that nothing is missed,

from $13x + 13y - xy = 0$ we can isolate for $x$ in terms of $y$,

$x = \frac{13y}{y-13} = 13 + \frac{169}{y-13}$ now you can choose y such that x is positive integer which happens when $y>13$ and 169 is divisible by $y-13$, as 169 is divisible only by 1, 13, 169 we can choose y accordingly which gives the co-ordinates given by the Mostafa's answer!

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The prime factorisation of $169$ is $13^2$. There are only $2+1$ divisors (see this answer), so there are three cases: $$(x-13)(y-13) = 1 \cdot 169$$ $$(x-13)(y-13) = 13 \cdot 13$$ $$(x-13)(y-13) = 169 \cdot 1$$

This is how we get the pairs $(14,182)$, $(26, 26)$, and $(182, 14)$.

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