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 $l \equiv 1(\text{mod 3})$?

  • $\begingroup$ 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 $1$. $\endgroup$ Dec 7, 2014 at 10:00
  • $\begingroup$ Did you observe any patterns (e.g. regarding the number of roots or their location) in your computation? $\endgroup$ Dec 7, 2014 at 10:02
  • 1
    $\begingroup$ 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 $\endgroup$
    – TZE
    Dec 8, 2014 at 1:05
  • 1
    $\begingroup$ 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. $\endgroup$ Dec 8, 2014 at 10:29

1 Answer 1


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.)

  • $\begingroup$ A little edit, $x^l P_l(x^{-1})$ on the first line should be $x^{l-1} P_l(x^{-1})$. $\endgroup$
    – TZE
    Dec 10, 2014 at 2:32
  • 5
    $\begingroup$ I feel like another way to demystify this argument is to start by noticing that $P(x)=\frac{x^p+(1-x)^p-1}{p}$. :) $\endgroup$ Dec 10, 2014 at 9:30
  • 1
    $\begingroup$ @Gjergji Zaimi: Thanks for pointing this out! I did have a nagging feeling that there was something simple like this behind it... $\endgroup$ Dec 10, 2014 at 10:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.