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.
 A: We can solve each of the following separately and find the common solutions:
$x^{103} \equiv 2 \pmod{11}$
$x^{103} \equiv 2 \pmod{13}$
They simplify by Fermat's little theorem to:
$x^3 \equiv 2 \pmod{11}$
$x^7 \equiv 2 \pmod{13}$
It is now easy to check all small cases to get the solutions:
$x \equiv 7 \pmod{11}$
$x \equiv 11 \pmod{13}$
When solved these give:
$x \equiv 128 \pmod{143}$
A: If you want to, you can use Euler to say (mod $143$) that $x^{120}\equiv 1$ so that $x^{-17} \equiv 2$
Now note that $103=6\times 17 +1$ so that $$2\equiv x^{103}\equiv (x^{17})^6x$$ whence , on dividing by $(x^{17})^6$, and using $x^{-17}\equiv 2$ $$x\equiv 2\times (x^{-17})^6\equiv 2^7\equiv 128$$
A: 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 
$$x\equiv128 \pmod {143}$$
