Determine all the $a \in \mathbb Z$ such the equation system is compatible in $\mathbb Z$
$$\begin{cases} 6x \equiv a^{121} \pmod {20} \\ 14x \equiv 3 \pmod {15} \end{cases}$$
So, piece of cake, I just use the Chinese remainder theorem:
$$ \begin{cases} 6x \equiv a^{121} \pmod 4 \\ 6x \equiv a^{121} \pmod 5 \end{cases}$$
$$\begin{cases} 14x \equiv 3 \pmod 5 \\ 14x \equiv 3 \pmod 3 \end{cases}$$
By the CRT, the system will have an assured solution if
$$ \begin{cases} 14x \equiv 3 \pmod 5 \\ 6x \equiv a^{121} \pmod 5 \end{cases}$$
is consistent. Using some basic properties and the fermat theorem, I finally get:
$$a \equiv 2 \pmod 5$$
So for any given $a$ the the system should work, shouldn’t it?
BUT if I take $a = 7$, it assures the condition, but $6x \equiv a^{121}\equiv3 \pmod 4$ will be impossible to solve.
What am I doing wrong?