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Let $p$ be an odd prime. Show that the congruence $x^4$$\equiv -1\text{ (mod }p\text{)}\ $ has a solution if and only if $p$ is of the form $8k+1$.

Here is what I did

Suppose that $x^4$$\equiv -1\text{ (mod }p\text{)}\ $ and let $y$=$ind_rx$. Then $-x$ is aolso a solution and $ind_r(-x)$ $\equiv ind_r(-1) +ind_r(x)\text{}$ $\equiv (p-1)/2 +y(mod p-1)$ Without loss of generality, we may take $0<y<(p-1)/2$ or $0<4y<2(p-1)$. Taking indices of both sides of the congruence yields $4y$$\equiv ind_r(-1) \text{ }\ $ $\equiv ind_r(p-1)/2(mod p-1) \text{ }\ $. So $4y = (p-1)/2$ + $m$$(p-1)$ for some $m$. But $4y<2(p-1)$, so $4y=(p-1)/2$ and so $p=8y+1$ or $4y=3(p-1)/2$. In this case, 3 must divide $y$, so we have $p=8(y/3)+1$. In either case, $p$ is of the form. Conversely, suppose $p=8k+1$ and let $r$ be a primitive root of $p$. Take $x$ $=$ $r^k$. Then $x^4$$\equiv r^{4k} \text{ }\ $ $\equiv r^{(p-1)/2} \text{ }\ $ $\equiv -1 (mod p) \text{ }\ $. This $x$ is a solution.

We just learned some of this topic. My TA said do not worry about it yet. Is this correct?

Is there another way to do this? If so, please show me.

9959
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    You've asked a handful of questions, but haven't accepted any. After you ask a question here, if you get a helpful answer, you should "accept" it (one answer per question), by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. If you don't do this, people are less likely to answer your later questions. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. – amWhy May 01 '13 at 04:16
  • @9959 how can that be true though? $x^4\equiv-1 mod 9$ doesn't have an integer solution. – rurouniwallace May 01 '13 at 04:20
  • @amWhy Sorry, I am still new here. I did not know that. I thought they get points by just answering the question or me saying thanks. – 9959 May 01 '13 at 04:23
  • @amWhy Thank for answering some of my questions though. You were helpful. – 9959 May 01 '13 at 04:23
  • @amWhy I also just learned about these points. I did not know what they were. – 9959 May 01 '13 at 04:25
  • Don't worry, @9959. I thought I'd post the comment with links in case you didn't know about accepting, that's all! It takes time getting used to a new site, and all the "rules" and "recommendations." It's certainly not too late to accept answers! BTW: I am very pleased that you posted the effort you put into thinking about this question. – amWhy May 01 '13 at 04:28
  • That goes for everyone who has helped me. Thanks. – 9959 May 01 '13 at 04:28
  • @amWhy Thanks. I am still learning LATEX too. ha. Still thank you and everyone else too. – 9959 May 01 '13 at 04:29
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    @ZettaSuro, is $9$ prime? – lab bhattacharjee May 01 '13 at 04:36
  • @labbhattacharjee That is what I was thinking. 9 is not a prime? What are your thoughts to this question? – 9959 May 01 '13 at 04:41
  • Everyone feel free to post a solution or your thoughts to this. I would like to see it. Is my solution correct though? I just learned some of this topic. – 9959 May 01 '13 at 04:43

1 Answers1

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There is a simpler way:

If $x^4\equiv-1\pmod p, x^8=(x^4)^2\equiv(-1)^2\pmod p\equiv1$

So,$ord_px$ must divide $8$

But it can not divide $4$ as $x^4\equiv-1\pmod p\implies ord_px=8$

Using this, $ 8$ divides $\phi(p)=p-1$

Conversely seems to be fine in your approach

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