I have the following congruence equation: $x^{3} \equiv 3 \bmod(357035)$ which I'm having troubles solving.
The prime factorisation of $35705$ is $5\cdot 7 \cdot 101^{2}$. I thought I would solve the congruences \begin{align}x^{3} & \equiv 3 \bmod(5) \\ x^{3} & \equiv 3 \bmod(7) \\ x^{3} & \equiv 3 \bmod(101^{2})\end{align} using the Chinese Remainder Theorem, but I somehow ended up with the solution $x^{3} \equiv 3 \bmod(357035)$ which doesn't help me. Some hints or a detailed answer would be much appreciated.
Best Answer
Not my strong suit, but just by testing, it appears the second condition is impossible. If $x\equiv0\pmod7, x^3\equiv0\pmod7$. Otherwise, by Fermat's Little Theorem
$$x^6\equiv1\pmod7$$ $$(x^3+1)(x^3-1)\equiv0\pmod7$$ $$x^3\equiv-1,1\pmod7$$
Therefore, there are no solutions.