In Serre's paper, On the two-dimensional modular representations of $\operatorname{Gal}(\overline{Q}/Q)$, on page 25 Section 4.3, Variants of Fermat's Theorem, Serre gives a proof that
$a^p+b^p+L^α c^p=0$
has no solutions for $a,b,c\in \mathbb Z$ and $abc \ne 0$.
$L$ is a prime number $\ne p$ belonging to the set
$S=\{3,5,7,11,13,17,19,23,29,53,59\}$
and $\alpha\ge0$.
Then he begins the proof by assuming that
$0 < α < p$.
Question 1:
I am confused by this change in the status of $\alpha$, and its affect on the values that $a$ and $b$ can take.
In particular, what happens if $\alpha$ is in fact $>p?$ Are we expected to absorb the extra $L^p$ into $c^p$? If we do, then the constraint on $\alpha$ is satisfied with a new exponent $\alpha \mod p$. BUT because of coprimality, $a$ and $b$ cannot contain any elements of $S$. Serre claims in the proof that not only are $a,b,c$ pairwise coprime, but $a^p,b^p,L^\alpha c^p$ are pairwise coprime as well. Serre uses the adjectives "obvious" and "easy" for the coprimalities and the constraint on $\alpha$. In the statement of what is to be proven, $\alpha$ is any positive number. In the proof itself, he states it is obviously bounded by $p$. It seems to me that doing that will reduce the possible values of $a$ and $b$. Is that right?
Question 2:
Explicitly, what is the calculation for $N=2L$? It appears that the $L$ somehow survived the level lowering. How did it do that?