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Determine $p$ for one to have $\displaystyle\frac{1}{5}+\frac{1}{6}=\frac{1+1}{5+6}=\frac{2}{11}$, in $\mathbb{Z_p}$.

I know that $p$ must, $13, 43, 61, 101,103$.

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Try multiplying both sides by $11 \cdot 5 \cdot 6$, to get $$121 = 11 \cdot 6 + 11 \cdot 5 = 2 \cdot 5 \cdot 6 = 60$$ in $\mathbb{Z}/p\mathbb{Z}$.

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    $\begingroup$ At this opportunity, I idly wonder why we so often write "Try" this, instead of "Do". $\endgroup$ – Daniel Fischer Oct 5 '13 at 18:47
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In addition to $p=121-60=61$ we can use the extended Euclidean algorithm to obtain $5^{-1}=-12$, $6^{-1}=-10$ and $11^{-1}=-11$ in $\mathbb{Z}/61 \mathbb{Z}$. Hence $$ \frac{1}{5}+\frac{1}{6}=-12-10=-22=2\cdot (-11)=\frac{2}{11}. $$

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