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Can anyone give a method for solving the congruence: $10^{n+1} - 9n - 10 \equiv 0$ (mod 7), where $n$ is a natural number? I am told that you have to perform the Euclidean algorithm twice on $n$ before attempting to use Fermat's Little Theorem, but why is this necessary? Why not just one application of the algorithm?

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well, because you have $n$ the unknown in the base and in power. lets start solving it first $\phi(7)=6$ where $\phi(n)$ is the Euler phi function, and because $n$ is in the power and in the base we will have a $7*6=42$ cycle or period over $n$ since its $6$ cycle for the $n$ in the power and $7$ in the base,by brute force we get that the values for $n = \{0,22,26,31,39,41\}$ will work and so for the general solution to $n$ is :

$$ n = 42 k +\{0,22,26,31,39,41\} $$ where $k$ is a non negative integer.

hope its what you are looking for.

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  • $\begingroup$ Yes, that's right. But why is a six cycle for the power? I'm familiar with the phi function but not for working out cycles. Also, by cycle do you mean cycling the numbers 0,1,2,3,4,5,6,7? How did you incorporate the 42 when finding the answers by brute force? $\endgroup$ – wrb98 Mar 15 '17 at 10:48
  • $\begingroup$ @Will why six ? because of Fermat's little theorem which says that $a^{\phi(n)} = 1 \mod n$ which means that every $\phi(n)$(sometimes its one of $\phi(n)$ divisors) which means at most after $\phi(n)$ for any $n$ we cycle again (return the same sequence as before ) for example : {1,2,4,1,2,4,1,2,4,..} is a 3-cycle sequence and because $\phi(7)=6$ we will need 6-cycle to return to the same sequence mod 7 $\endgroup$ – Ahmad Mar 15 '17 at 10:56

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