For which prime numbers $p$ does the congruence $x^2 + x + 1 \equiv 0 \pmod{p}$ have solutions?

Question

For which prime numbers p does the congruence $x^2 + x + 1 \equiv 0 \pmod{p}$ have solutions?

We've recently learnt about quadratic reciprocity in class, however I am not sure how to tackle this problem. I have tried starting with the $b^2-4ac$ (discriminant) but that hasn't really helped.

Welcome to MathSE. I am glad you found the help you needed on this question. There is information about writing mathematics on this site here: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference. Also, the quantity $b^2 - 4ac$ is called the discriminant. — N. F. Taussig, Jan 31 '15 at 17:57

score 5 · Accepted Answer · answered Jan 31 '15 at 17:14

5

Hint: Consider $4(x^2 +x+1) = (2x+1)^2 + 3$. Therefore, if it has a solution modulo $p$, $-3$ must be a quadratic residue. Use quadratic reciprocity.

answered Jan 31 '15 at 17:14

Nacho Darago

1,004

So following from that I should find p for which (-3/p)=1? – The Problem Jan 31 '15 at 17:21
Yes. Remember that $(-3|p) = (-1|p)(3|p)$. Then, you either need that both -1 and 3 are quadratic residues or both are quadratic non-residues. $(-1|p) \equiv (-1)^{(p-1)/2} \pmod{p}$, and for 3 use quadratic reciprocity – Nacho Darago Jan 31 '15 at 17:28
Okay, so I get (-3/p)=(-1)^(p-1)(p/3) which is dependent on pmod3 then it works for p=1 and p=2, does this mean it holds for all p equivalent to 1 or 2 mod 3? Also what about p=3? – The Problem Jan 31 '15 at 17:31
For $p=3$ you must handle it separately. You can see that $x=1$ is a solution. And yes, if $a\equiv b\pmod{p}$ then $(a|p)=(b|p)$. Also, be careful, $2$ is not a quadratic residue modulo $3$. – Nacho Darago Jan 31 '15 at 17:36
So if it holds for p=3, does this imply it holds for all p as we already have it working for p=1 or 2 mod 3? – The Problem Jan 31 '15 at 17:42
It doesn't work for $p\equiv 2\pmod{3}$ because, as you've seen $(-3|p) = (p|3)$ and $2$ is not a quadratic residue modulo $3$. – Nacho Darago Jan 31 '15 at 17:44
I thought (-3|p)=(-1)^(p-1) (p|3) so 2 would work? – The Problem Jan 31 '15 at 17:47
You can check that $p=2$ doesn't work by hand, and then, if $p$ is a prime, it must be odd. So $p-1$ is even and then $(-1)^{p-1}=1$. That is, $(-3|p) = (p|3)$. – Nacho Darago Jan 31 '15 at 17:50
Ah I see, and then you have it holds for p as long as p is equivalent to 1 or 3 mod3? – The Problem Jan 31 '15 at 17:53
Yep, that's it :) – Nacho Darago Jan 31 '15 at 17:53
Cool, thanks for the help :) – The Problem Jan 31 '15 at 17:54

score 3 · Answer 2 · edited Apr 13 '17 at 12:20

3

Simpler than reciprocity: $\,x^2+x+1 \equiv 0 \,\overset{\large {\cdot\, (x-1)}}\Rightarrow\,x^3\equiv 1,\ $ so by basic results on cyclic groups $\ldots$

edited Apr 13 '17 at 12:20

Community

1

answered Jan 31 '15 at 17:59

Bill Dubuque

272,048

For which prime numbers $p$ does the congruence $x^2 + x + 1 \equiv 0 \pmod{p}$ have solutions?

2 Answers2

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