Solving $x^3\equiv2\bmod151$ using Lagrange's Theorem

Question

I'm trying to show that the congruence $$x^3\equiv2\bmod151$$ has no solution by using Lagrange's Theorem (group theory).

I know that $151$ is prime, but I don't know how can I use the theorem in this case.

Am I supposed to create cyclic groups s.t. one is subgroup of the other? How can I do that?

Thank you in advance!

Answer 1

Since $151$ is a prime, $\mathbb{Z}/151\mathbb{Z}$ is a field, and $(\mathbb{Z}/151\mathbb{Z})^{\times}$ is a multiplicative group of order $150$. Suppose $x^3\equiv 2\pmod{151}$. Notice that $2$ has multiplicative order $15$ modulo $151$ (its order has to divide $150$, and so you can check with a calculator that $151$ does not divide $2-1,2^3-1,2^5-1,2^{10}-1$, while it divides $2^{15}-1$). Since the (multiplicative) subgroup of $(\mathbb{Z}/151\mathbb{Z})^{\times}$ generated by $x$ contains the subgroup generated by $2$, we deduce that the order of the subgroup generated by $x$ is divisible by $15$ and divides $150$. So it can be $15,30,75,150$. But no one of these is admissible. For example, if $x$ has order $30$, then $$1\equiv x^{30}\equiv (x^3)^{10}\equiv 2^{10}\not\equiv 1\pmod{151},$$ contradiction.

Answer 2

Hint: $x^3 \equiv 2 \bmod 151 \implies 1 \equiv x^{150} \equiv 2^{50} \bmod 151$. But $2^{50} \not\equiv 1 \bmod 151$.