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Find a solution $x \in \mathbb{Z}$ or show that no solution exists. $x^2 + x + 1 \equiv 0 \pmod 5$

So far I have been trying to the Chinese Remainder Theorem to solve this question but I still can't figure out the answer; I'm not sure whether I used the wrong theorem or I just made some calculation mistakes. Can anyone please help me with this question?

Anindya Prithvi
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sasaki085
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    You essentially only need to check the values $x = 0,1,2,3,4$. – player3236 Oct 27 '20 at 14:08
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    Why would you use the Chinese remainder theorem for a prime modulus? – J. W. Tanner Oct 27 '20 at 14:09
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    As long as you accept that square roots and fractions mean something different modulo 5 than they do in $\Bbb Z$ (well, they mean the same thing, but the concrete calculation works entirely differently), then the quadratic formula works just as well to find solutions here as it does normally. But just checking works too. – Arthur Oct 27 '20 at 14:11
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    $x^2+x+1\equiv0\implies x^3=(x^2+x+1)(x-1)\equiv0$ and $x\not\equiv1,0\implies x^9\equiv1$ and $x\not\equiv1,0\implies x\equiv1$ and $x\not\equiv1,0$, contradiction – J. W. Tanner Oct 27 '20 at 14:23
  • of course I meant $x^3\color{red}{-1}=(x^2+x+1)(x-1)$ – J. W. Tanner Oct 27 '20 at 14:29

2 Answers2

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$$\begin{align} x&=0,1,2,3,4\pmod5\\ x^2&=0,1,4,4,1\pmod5\\ x^2+x+1&=1,3,2,3,1\pmod5 \end{align}$$

So, no such $x$ exists.

Anindya Prithvi
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Here's a different way, which doesn't involve calculating five possibilities.

Assume $x^2+x+1\equiv0\bmod5$.

Then $x^3-1=(x^2+x+1)(x-1)\equiv0$ and $x\not\equiv1,0\bmod 5$.

Thus $x^9\equiv1$ and $x\not\equiv1,0\bmod 5$.

Using Fermat's little theorem, $x^4\equiv1\bmod5$, so we have $x\equiv1$ and $x\not\equiv1,0\bmod5$.

We have reached a contradiction.

J. W. Tanner
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  • Said conceptually: you deduce $,x\equiv 1,$ by raising $,x^{\large\color{#0a0}3}\equiv 1,$ to power $,\color{#0a0}{1/3}\equiv 1/(-1)\equiv 3\pmod{!\color{#c00}4},,$ since $,x\not\equiv 0\Rightarrow, x^{\large \color{#c00}4}\equiv 1,$ by Fermat, as explained here. – Bill Dubuque Oct 27 '20 at 14:33
  • That's what I did, @BillDubuque, but thanks for the reference and explanation why – J. W. Tanner Oct 27 '20 at 14:39
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    It's not at all clear what you did from what little is written - hence my comment. In particular, it's likely not clear to many readers how to generalize what you did. The prior links shows how. – Bill Dubuque Oct 27 '20 at 14:50