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Show that there are no solutions to $x^3 \equiv 2(mod151)$

I really want to show some way how to solve this, but I have no idea what to do when I have something other than $x$. How can I approach $x^3$?

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Assume there is such an $x$. Fermat's little theorem says that $x^{150}\equiv 1\pmod{151}$. But $x^{150} = (x^3)^{50} = 2^{50}\equiv 32\pmod {151}$, which is a contradiction.

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    Thank you very much Arthur! – 2017-01-08
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    How did you get to remainder $32$? (with a calc it works ofcurse, but is there a fast way on a paper)? – 2017-01-08
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    use that $$2^5 \equiv 32 \mod 151$$ – 2017-01-08
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    @Dr.SonnhardGraubner How do I get rid of the $10$ now? because it is $2^{{5}^{10}} = 2^{50}$ – 2017-01-08
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    @Ilan $5^{10}\neq 50$. Rather, $(2^5)^{10}=2^{50}$. But it's easiest to calculate by so-called [repeated squaring](https://en.m.wikipedia.org/wiki/Exponentiation_by_squaring). You calculate $2^{50}$ by starting with $2$ and then square, double, square, square, square, double, square. It's not too much work if you remember to reduce modulo $151$ each step. – 2017-01-08
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Calculate the multiplicative order of 2 mod 151, i.e. the least n such that $2^n \cong 1$ mod 151. Then think of the possible multiplicative order of $x^3$ mod 151. Remember $\mathbb{Z}/\mathbb{nZ}^{*} \cong \mathbb{Z}/\phi(p) \mathbb{Z}$, which in this case means $\mathbb{Z}/\mathbb{151Z}^{*} \cong \mathbb{Z}/150 \mathbb{Z}$, since 151 is prime.