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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


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?

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I believe this is correct. If you feel your question has been answered, you could post an answer and accept it (see…). – M Turgeon Sep 26 '12 at 16:34

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