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



Roots of a polynomial in a finite field related to Fermat's Last Theorem
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 $a \in {\mathbb F}\_l \setminus \{0,1\}$.




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(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.)




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(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.)


