I'm trying to read the proof of
COROLLARY. The only solutions of
$x^2+7=2^m$ ($x,m \in \mathbb{Z}$) (6.15)
have $m=3,4,5,7,15$.
I don't see why there could be a + in $y\pm \alpha$ (6.17) or in $y \pm \beta$ (6.18):
PROOF. Clearly $x$ is odd, say $x=2y-1$ ($y \in \mathbb{Z}$) and then
$y^2-y+2=2^{m-2}$ (6.16)
The ring $\mathbb{Z}[\alpha]$, where $\alpha^2-\alpha+2=0$, has a Euclidean algorithm and so is a UFD. On considering factorization of both sides of (6.16), we get
$y \pm \alpha = \pm \alpha^{m-2}$ (6.17)
(for some choice of signs). Then
$y\pm \beta = \pm \beta^{m-2}$ (6.18)
for the conjugate root $\beta$.
My understanding of the situation so far:
By equating coefficients in $(y-\alpha)(y-\beta)=y^2-y+2$, we get $\alpha\beta=2$.
$\alpha$ and $\beta$ have norm 2, so are irreducible, so $2^{m-2}=\alpha^{m-2}\beta^{m-2}$ is the factorisation into irreducibles of $2^{m-2}$.
I can show that $\beta$ doesn't divide into $y- \alpha$ and $\alpha$ doesn't divide into $y - \beta$. (Edit: it turns out that I made a mistake here. This is only true for even $y$, but you can show $y$ is even for $m\geq 6$.) Then using the fact we're in a UFD gives
$y-\alpha=\epsilon \alpha^{m-2}$
$y-\beta=\epsilon^{-1} \beta^{m-2}$
for some unit $\epsilon\in\mathbb{Z}[\alpha]^\times$.
Now I want to find the units. For $a, b \in \mathbb{Z}$, the norm of $a+b\alpha$ is
$N(a+b\alpha)=\frac{1}{2}(a+b)^2+\frac{1}{2}a^2+\frac{3}{2}b^2$
so $N(a+b\alpha)=1$ only if $b=0$ and $a=\pm1$. This gives
$y-\alpha=\pm \alpha^{m-2}$ and $y-\beta=\pm \beta^{m-2}$.
Is this right and was there an easier way to see all this?