Calculating $x^{103} \equiv 2 \pmod{143}$ I need to find x, given that:
$0\leq x \leq 143$ and $x^{103}\equiv 2 \pmod{143}$.
I tried to use Euler's theorem $p(143)=120$, but it didn't help.

Consider 
$$x^{103}\equiv2 \pmod {13}$$
$$x^{103}\equiv2 \pmod {11}$$
en
$$x^7\equiv2 \pmod {13}$$
$$x^3\equiv2 \pmod {11}$$
$7$ is the common  primitive root $\bmod {13}$ and $\bmod {11}$ 
$7^{11}\equiv2 \pmod {13}$, $7^3\equiv2 \pmod {11}$, and
if $(7^t)^7\equiv2 \pmod {13}$, $7t\equiv 11 \pmod {12}$, then $t=5\pmod {12}$
if $(7^s)^3\equiv2 \pmod {13}$, $3s\equiv 3 \pmod {10}$, then $s=1\pmod {10}$
so, the root $x$ is decided by 
$$x\equiv 7^5\equiv 11 \pmod {13}$$
$$x\equiv7 \pmod {11}$$
then 
$$x\equiv128 \pmod {143}$$
