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For what values of $p$ prime does the equation $a+a^{-1}=1$ hold true for some $a\in \mathbb F_{p}$ such that $a \ne0$? I tried counting field elements that are inverses of each other and playing with modular arithmetic, but I can't seem to work it out.

Matt R
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    How many solutions has $a^2+1=a$ in $\mathbb{F}_p$ ? Certainly not more than $2$ ?! – Dietrich Burde Dec 06 '13 at 21:07
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    Very nice hint, @Dietrich. Make it an answer? – Cameron Buie Dec 06 '13 at 21:19
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    I don't really understand the question. Isn't it obvious that for $a=1$ you have $1+1^{-1}=2\neq 1$? This is never true, in any (unital) ring, unless you allow $1=0$. – tomasz Dec 06 '13 at 21:45
  • In some fields it works. For example, in $\mathbb F_{7}$, $5=3^{-1}$ and $5+3=8=1$ – Matt R Dec 06 '13 at 21:49
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    @MattR He is right, it won't work for $a = 1$. It can work for particular $a$s, but you said for all $a$ there. If it were true for some $a\neq 0$, we can conclude that, since $a^3 + 1 = (a + 1) (a^2 - a + 1) = 0$, then $a^3 = -1$ and $a^6 = 1$ hence (check why) $6|(p-1)$, as $a^{p-1}=1$. For the converse... think that $a+a^{-1}=1$ if and only if $a^3 = -1$ and $a\neq -1$. – Pablo Rotondo Dec 06 '13 at 21:53
  • @tomasz You are right. I edited the question: for some a – Matt R Dec 06 '13 at 21:54
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    @Serkan in other words $6|(p-1)$. – Pablo Rotondo Dec 06 '13 at 22:04

1 Answers1

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We note that $a + a^{-1} = 1$ if and only if $a^2 - a + 1 = 0$.

Now note that $a^3 + 1 = (a+1)(a^2 - a + 1) = 0$ thus $a^3 = -1$, but $a\neq -1$ if $p\neq 3$, as otherwise $-2 = (-1) + (-1)^{-1} = 1$ and so $p=3$.

Conversly, since $a^3 + 1 = (a+1)(a^2 - a + 1)$, we get that if $a^3 = -1$ with $a\neq -1$, then $a^2 - a + 1 = 0$.

Observation 1: If $p\neq 3$: $a + a^{-1} = 1$ occurs if and only if $a^3 = -1$ and $a\neq -1$.

I will give you indications for the rest, when $p\neq 3$:

Question: What is the order of $a$ ? What does that imply by recalling Fermat's Little Theorem?

For the converse : Remember that ${\mathbb F}_p^\times$ is cyclic (the multiplicative group), use this to show for the $p$ as you found, there is at least one $a$ satisfying $a+a^{-1}=1$.

Observation 2 This proof actually works finite fields $F$ with a minor change. Indeed, change $p\neq 3$ above for $\text{char}(F) \neq 3$, since for characteristic $3$ we can take $a=-1$.

Edit A more routinary way would probably be the following: $a^2 - a + 1 = 0$ if and only if $(2a-1)^2 = -3$, so $-3$ must be a nonzero quadratic residue if $p\neq 3$. Next what is $\left(\tfrac{-1}{p}\right)$? and use quadratic reciprocity to find $\left(\tfrac{3}{p}\right)$, don't forget $\left(\tfrac{p}{3}\right)\equiv p (\bmod. 3)$ too. You will get a congruence for $p(\bmod. 3)$, and turn it into one $\bmod.6$ by observing that $p$ can't be even ;)

Pablo Rotondo
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