# Roots of a polynomial in a finite field related to Fermat's Last Theorem

In my class, we proved the following condition: define the polynomial $P_l(x)$ as

$$P_l(x) = \sum_{r=1}^{l-1}{\frac{1}{r}x^{l-1-r}}$$

Then if for all $a \in \mathbb{Z}/l\mathbb{Z}-\{0,1\},$ $P_l(x)$ does not vanish (mod $l$), then the First Case of Fermat's Last Theorem(1CFLT) is true for $\mathbb{Z}$ and exponent $l.$

It appears that the polynomials, $P_l(x)$ always have a solution in $\mathbb{Z}/l\mathbb{Z}-\{0,1\}$ whenever $l \equiv 1(\text{mod 3}).$

For instance, $P_7(x) \equiv x^5+ 4 x^4 + 5 x^3+ 2 x^2 + 3 x+6 (\text{mod 7})$ has roots $3,5.$

I have tested this on Sage for primes $l \equiv 1(\text{mod 3})$ less than $1000$. Does this hold for all primes $l \equiv 1(\text{mod 3})$?

• The condition on $P_l$ seems to be unnecessarily complicated. It is equivalent to $P_l$ not having any roots in ${\mathbb Z}/l{\mathbb Z}$ except possibly $0$ and/or $1$. Dec 7, 2014 at 10:00
• Did you observe any patterns (e.g. regarding the number of roots or their location) in your computation? Dec 7, 2014 at 10:02
• Wow! Thank you for the complete answer! Yes, I did. There were always at least two solutions excluding $0, 1.$ and they appeared in pairs $x, x^{-1}.$ Here is the Sage computation with the output: goo.gl/MyrGkx
– TZE
Dec 8, 2014 at 1:05
• You are welcome. -- I was suspecting that cube roots of unity might play a role (from the condition $l \equiv 1 \bmod 3$); looking at the first few cases confirmed that. Dec 8, 2014 at 10:29

This is true.

Let $P(x) = \sum_{n=1}^{l-1} \frac{x^n}{n} \in {\mathbb F}_l[x]$ (this is $x^{l-1} P_l(x^{-1})$). Then we have $P(1-x) = P(x)$ (check that they have the same derivative and the same value at $1/2$) and $x^l P(x^{-1}) = -P(x)$.

Now if $l \equiv 1 \bmod 3$, then there is a primitive sixth root of unity $\omega \in {\mathbb F}_l$. Note that $1-\omega = \omega^{-1}$. We then obtain $$\omega P(\omega^{-1}) = \omega^l P(\omega^{-1}) = -P(\omega)$$ and $$P(\omega^{-1}) = P(1 - \omega) = P(\omega) .$$ So $(1+\omega) P(\omega^{-1}) = 0$, which implies (since $\omega \neq -1$) that $P(\omega) = P(\omega^{-1}) = 0$. These two are actually double roots, since the derivative also vanishes.

In fact, this argument shows that $x^2 - x + 1$ divides $P(x)$ whenever $l > 3$.

Another remark: Since $$P'(x) = 1 + x + x^2 + \ldots + x^{l-2} = \frac{x^{l-1}-1}{x-1} = \prod_{a \in {\mathbb F}_l \setminus \{0,1\}}(x-a),$$ a root $a \in \overline{\mathbb F}_l$ of $P$ is a multiple root (and then of multiplicity 2) if and only if $a \in {\mathbb F}_l \setminus \{0,1\}$.

(Added later:) One can also give a "combinatorial" proof. The relations satisfied by $P$ imply that the roots of $P$ come in orbits of the $S_3$-action on ${\mathbb P}^1_{\mathbb F_l}$ generated by $a \mapsto 1-a$ and $a \mapsto a^{-1}$ (we think of $P$ has a polynomial of degree $l$ with "leading coefficient" zero, so that $\infty$ is a root). These orbits have size 6, with only three exceptions: $\{0,1,\infty\}$, $\{-1,2,1/2\}$ and $\{\omega, \omega^{-1}\}$ (this uses $l > 3$). We always have the first orbit, which leaves $l-3$ roots. Any orbit other than the third special orbit will contribute a multiple of 3 to the number of roots. So the contribution of the two-element orbit is $\equiv l \bmod 3$, and it can only be 0, 2 or 4. So for $l \equiv 1 \bmod 3$, it must be 4, implying that both are double roots and hence in ${\mathbb F}_l$, and if $l \equiv 2 \bmod 3$, the contribution must be 2, so we have simple roots and they are not in ${\mathbb F}_l$. (This actually proves that $\omega \in {\mathbb F}_l$ iff $l \equiv 1 \bmod 3$ without using that ${\mathbb F}_l^\times$ is cyclic.)

(Added Dec 11, 2014:) As Gjergji Zaimi points out in a comment below, we have that $P(x)$ is (for $l > 2$) the image of $$Q_l(x) = \frac{(x-1)^l - (x^l-1)}{l} \in \mathbb Z[x]$$ in ${\mathbb F}_l[x]$ (use that $\frac{1}{l}\binom{l}{k} = \frac{1}{k} \binom{l-1}{k-1} \equiv (-1)^{k-1}/k \bmod l$). For any prime $l > 3$, $Q_l(\omega) = 0$ (where now $\omega \in \mathbb Q(\sqrt{-3})$), which implies that $P(\bar\omega) = 0$ (where $\bar\omega$ is the image of $\omega$ in $\mathbb F_l$ or $\mathbb F_{l^2}$).

This also explains the relation to FLT. (In fact, the observation that $\omega^l + (\omega^{-1})^l = 1$ gives an $l$-adic first case solution of Fermat's equation when $l \equiv 1 \bmod 3$ gives another proof, assuming the statement on FLT in TZE's question.)

• A little edit, $x^l P_l(x^{-1})$ on the first line should be $x^{l-1} P_l(x^{-1})$.
– TZE
Dec 10, 2014 at 2:32
• I feel like another way to demystify this argument is to start by noticing that $P(x)=\frac{x^p+(1-x)^p-1}{p}$. :) Dec 10, 2014 at 9:30
• @Gjergji Zaimi: Thanks for pointing this out! I did have a nagging feeling that there was something simple like this behind it... Dec 10, 2014 at 10:15