I need help on how to prove that this equation $x^2+1\equiv 0\mod {p}$ ;where p is a prime number, admits only two solutions if $p\equiv 1\pmod{4}$?
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2What do you know about the number of zeros a polynomial of degree $n$ can have in a field? – Jyrki Lahtonen Jun 13 '19 at 13:13
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Also, I recommend that you take a look at our guide for new askers. If you do, you will have a smoother ride here! – Jyrki Lahtonen Jun 13 '19 at 13:15
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a polynomial of degree n over a field has at most n roots. – Kamal Jun 13 '19 at 13:20
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3Correct! $\Bbb{Z}/p\Bbb{Z}$ is a field, so...? – Jyrki Lahtonen Jun 13 '19 at 13:50
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so for our case the polynome of degree 2 has at most 2 roots in this field. – Kamal Jun 13 '19 at 14:32
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1Correct, again. Well done! The (IMHO) harder direction is to prove that this particular quadratic has those two solutions, when $p\equiv1\pmod4$. However, that has been discussed on our site many times already. – Jyrki Lahtonen Jun 13 '19 at 15:10
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Thanks Doctor for the hints, they are indeed helpful. would you please direct me to any link in the website where this issue is discussed. Thanks again. – Kamal Jun 13 '19 at 15:24
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Recall that for every $a\neq0\pmod p$, there is a $b=1/a$ for which $ab=1\pmod p$.
Group the numbers from $2$ to $p-2$ into quartets $\{a,-a,1/a,-1/a\}$.
There are $p-3$ numbers involved, so two numbers must be left over. That is because two of the four numbers are the same. You can't have $a=-a$ since $a\ne0$, and you can't have $a=1/a$ since $a$ is neither $1$ nor $p-1$, so you must have $a=-1/a$, or $a^2+1=0$.
Empy2
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1For more on this see here and see the posts I linked in the comment (and its coments links ...) And compare to the Remark here. – Bill Dubuque Jun 13 '19 at 15:41