A proof on Hilbert symbol for 2-adic numbers. 
I am stuck at several places in this proof related to some properties of Hilbert symbol for 2-adic numbers. This is from the book Number theory by Coppel. First of all I think there is a typo in the third line: $\alpha$ should be replaced by $-\alpha$. Corollary VI.14 as mentioned in the proof says that any p-adic number can be written as a Laurent series and in this case for $x$ the first term in the representation of $x$ should be $2^{-\alpha}$. Given that correction everything below makes sense untill the (corrected) statement: $1=2^{-2\alpha}(a_0+b_0y_0^2+8z')$ implies that $\alpha=0,y_0^2\equiv 4 (mod\, 8), a_0=5$. This is something I can't prove concretely. I am sure it's not brute force. Somehow I need to show $\alpha=0$ first and then everything else should follow. I feel that somehow dominance principle ($|c|_2<|d|_2$ implies $|c+d|_2=|d|_2$) should be used to establish something about the absolute value of $a_0+b_0y_0^2+8z'$ which would force $\alpha =0$. Proposition VI.16 as mentioned in the proof says that an element $c\in \mathbb{Q}_2$ with $|c|_2=1$ is a square if and only if $|c-1|_2\leq 2^{-3}$ (this follows from Hensel's lemma). I hope I am not missing something obvious, but I am badly stuck at this seemingly easy looking modular arithmetic problem.
Thanks for any help on explanation of this proof.
 A: I think you're right and we have to make little corrections in the proof, but the statement of the proposition remains true. So we need to get
$$1 =2^{-2\alpha} (a_0+b_0y_0^2 +8z')$$
with the restrictions $a_0,b_0 \in \{3,5,7 \}, y_0^2\equiv 0,1,4$ mod $8$ and $z' \in \mathbb Z_2$. Well, as you say yourself, if $\alpha=0$ then checking $(a_0+b_0y_0^2 +8z')$ modulo $8$ with these restrictions gives what is claimed.
So let's assume $\alpha \ge 1$. Then $\lvert a_0+b_0y_0^2 +8z'\rvert_2 \le \frac14$ so that, necessarily, $a_0+b_0y_0^2$ has to be divisible by $4$, which under the given restrictions forces (this is a quick case-by-case check, not too brute a force) $y_0^2=1$ mod $8$ and then further $\{a_0, b_0\}=\{3,5\}$ (first case) or $=\{5,7\} $ (second case). In the second case, we would get the nonsensical $1= 2^{-2\alpha}(12 +8z')=2^{2-2\alpha}(3+2z')$; but the first case can actually happen, as in $\alpha=2, a_0=3, b_0=b=5, z'=a'=1$ (i.e. $x=y=\frac14, a=11, b=5$). Note however that either way, either $a_0$ or $b_0$ would be $\equiv 5$ mod $8$ and hence either $a-4$ or $b-4$ would be a square in $\mathbb Z_2$. So we have to amend the possibility of that case in the proof, but the assertion that $(a,b)_2=1$ entails that at least one of $a,b, a-4,b-4$ is a square in $\mathbb Q_2$ still holds true.
