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I am working on the following exercise: For which values of $a \in \mathbb{Z}$ is the following system of congruences

$$ \begin{alignat}{2} x^{27} & \equiv x^2 && \mod 144 \\ 10x & \equiv a && \mod 25 \\ 2^x - 1 & \equiv 4 && \mod 11 \end{alignat} $$

resolvable? Find the solutions.

From the last equation, I understand that $2^x \equiv 5 \mod 11 \Rightarrow x = 10n + 4, n \in \mathbb{Z}_{\ge 0}$ by Fermat’s little theorem. What to do with the other equations?

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    By the second equation, $a$ must be a multiple of $5$, and then $10x = 10(10n+4) = 100n + 40 \equiv 15 \mod 25$ always happens, right? – 2017-01-13
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    Thank you, but how do I know if the first equation is fulfilled for $x = 10n + 4$? Sorry, I don’t see the trick... – 2017-01-13
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    The same $x=10n+4$ (or a subset of this) has to work for all three equations for the whole system to be resolvable, right? So that's why I just substituted, and found that $n$ can be anything, but the value of $a$ will be $15$ always. – 2017-01-13
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    True, but what can you say about the _first_ equation? – 2017-01-13
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    I'm sorry I don't know how to solve that equation. – 2017-01-14

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