0

I believe we have to check all values of $k$.

$k = 0$: $a^0 = 1 = b^2$. This holds for all $a\in \mathbb{F}_{27}$, so we just have to find all $b \in \mathbb{F}_{27}$ for which $b^2 \equiv 1 \text{ (mod } 3)$. For $b \not\equiv 0 \text{ (mod } 3)$ this congruence holds (As a result of Fermat's Little Theorem). So $b$ can take $18$ values (not $0,3,6,9,12,15,18,21,24$). This means that for $k=0$, we have $27 \cdot 18$ ordered pairs.

$k = 1$: $a = 1 = b^2$. Now we only have $1 \cdot 18$ ordered pairs.

$k = 2$: $a^2 = 1 = b^2$. Same reasoning for the possible values of $a$: $a^2 \equiv 1 \operatorname{ (mod } 3)$ if $a \not\equiv 0 \operatorname{ (mod } 3)$. Now we have $18 \cdot 18$ ordered pairs.

$k = 3$: $a^3 = 1 = b^2$. Now $a^3 \equiv 1 \operatorname{ (mod } 3)$, so $a \equiv 1 \operatorname{ (mod } 3)$. There are $9 \cdot 18$ ordered pairs.

Now for $k \ge 4$: $a^k = 1 = b^2$. The calculation of the values of $a$ can be reduced to one of the prior situations, assuming that $a$ is not a multiple of 3 (and - of course - neither is $b$).

So the cases $k=1, k=2, k=3$ can be rewritten as $k \equiv 1, k\equiv 2, k \equiv 0 (k \ne 0) \operatorname{ (mod } 3)$.

Is my method correct?

MyWorld
  • 2,398
  • 1
    If the field $\Bbb{F}{27}$ we have $3=0$. This is because $\Bbb{F}{27}$ is a vector space over its subfield $\Bbb{F}_3$. – Jyrki Lahtonen Jan 29 '19 at 20:53
  • If you want a concrete model of $\Bbb{F}{27}$ you can use $\Bbb{F}{27}=\Bbb{F}3[x]/\langle x^3-x+1\rangle$. But, my recommendation is that you concentrate on using the fact that the non-zero elements $\Bbb{F}{27}^*$ is a cyclic group of order $26$. By basic properties of cyclic groups this implies that you can always replace $k$ with $\gcd(k,26)$. So you only need to consider the cases $k=0,1,2,13,26$. – Jyrki Lahtonen Jan 29 '19 at 20:57
  • For example, consider the following. In a field there are no zero divisors. So if $b^2=1$, then $0=b^2-1=(b-1)(b+1)$ implying that either $b=1$ or $b=-1=2$. You can call the latter solution $-1$ or $2$, whichever you prefer. In characteristic three they mean the same thing. – Jyrki Lahtonen Jan 29 '19 at 20:59
  • 1
    Basically, you need to internalize that $\Bbb{F}{27}$ is not the same as $\Bbb{Z}{27}$, i.e. the residue classes of integers modulo $27$. – Jyrki Lahtonen Jan 29 '19 at 21:00
  • One way of using cyclicity of $\Bbb{F}_{27}^*$ is to observe that $b^2=0$ for exactly one choice of $b$, namely $b=0$. But, in a cyclic group of order $26$ squaring is a two-to-one mapping. Meaning that it takes $13$ values twice each, and misses the remainind $13$ values completely. – Jyrki Lahtonen Jan 29 '19 at 21:04
  • But. I'm not sure how meaningful these points are to you? My advice would be that you go meet your instructor. Show this attempt to them. They can pinpoint all the misunderstandings that you have by socratically asking questions. I could write a solution to the question easily. But I am not sure how much you would learn from it due to the confusions. It may be that your confusions are easily fixed. Your instructor is better informed about your background. I can only guess. – Jyrki Lahtonen Jan 29 '19 at 21:09
  • And attempting a similar socratic dialogue here won't work here because it may take a while. A suggestion I can make at this time is that you check your course notes for an example involving a smaller field. Like $\Bbb{F}_4$ or $\Bbb{F}_9$. No $\Bbb{F}_p$ with $p$ a prime number will be helpful here (because you can treat those as residue class rings of integers). – Jyrki Lahtonen Jan 29 '19 at 21:13

0 Answers0