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I have to solve $3^{\phi(n)}\equiv 4 \pmod{11}$, where $\phi(n)$ is Euler's function. I have no idea after this observation: $\phi(n)\equiv 4\pmod{5}$.

Thanks.

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    Are you supposed to find all positive integers $n$ that solve this? – 2017-02-09
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    Use Discrete Logarithm – 2017-02-09
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    Of course, $n = 5$ works, as would $n = 10$. – 2017-02-09
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    Yes I want to find all positive integers. – 2017-02-09
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    Works also $n=8$ and $n=12$; but I'm looking for a general idea, maybe elementary, because I dont't know the discrete logarithm. – 2017-02-09
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    Are you sure that you did not make an error when copying the question from your source to here? – 2017-02-09
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    No why? However this is an interesting problem. – 2017-02-09
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    Are you sure that you did not make an error when copying the question from your source ? Maybe the 11 in the modulus was an $n$ instead? You correctly used discrete logarithms when you reduced your original question to $\phi(n)\equiv 4\pmod{5}$. But using the standard formula (https://en.wikipedia.org/wiki/Euler%27s_totient_function) for $\phi(n)$ there is nothing that would indicate that the set of such $n$ is any easier to describe than the formula itself. Obviously, $25 \nmid n$ and $p \nmid n$ for any prime $p \equiv 1 \pmod 5$, but every other prime seems possible as prime factor for $n$. – 2017-02-09
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    No error. The text ask for $0 \le n\le10$, but this is trivially. I want to discover some proprieties, if they exists. – 2017-02-09

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