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Prove that $3^{28}$ is a multiplicative inverse of $9^{34}$ modulo $17$, i.e. show that $3^{28}9^{34}\equiv 1\pmod {17}$.

I really have no idea how to approach this example other than applying Fermat's theorem.

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Hint $\,\ 3^{28}(3^2)^{34} = (\color{#c00}{3^{16}})^6\,$ and $\,\color{#c00}{3^{16}\equiv 1}\pmod{17}\,$ by little Fermat

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