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I need to solve: $2^k\equiv7(\text{mod }30)$. Also there exists a homomorphism $\varphi: U_{30}\rightarrow U_{30}$ such that $\text{Ker}(\varphi)={\{\bar{1},\bar{11}\}} $

our teacher chalenged us also to solve: $x^k\equiv7(\text{mod }30)$, i.e.: find a method to find an x that for whom there exists a k that solved this congruence.

eagleye
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    Well...that would be your challenge, then. Not ours. Just trying all the possible values of $k$ would be one approach. I note that $\phi(30)=8$ so there aren't even very many values to check. – lulu Oct 03 '18 at 19:56
  • I tried with trial and error and came to these numbers: 7^5, 7^9, 13^3, 13^7. I wonder if there is a more general way (involves primes maybe?) to get there.. – eagleye Oct 03 '18 at 20:25

2 Answers2

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Your first congruence has no solutions. Note that for any $k\in\mathbb{N}$ we get that $2^k$ is even and hence the remainder after dividing it by $30$ will always be an even number, so it will never be $7$. Try to use that idea for the second part where you have to find $x$.

Mark
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$$2^k\equiv7(\text{mod }30)$$ $$2^k-7\equiv0(\text{mod }30)$$

it means that $$2^k-7=30m\hspace{15pt} m\in Z$$

where right side is even and left side is odd means it is not possible

Deepesh Meena
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