Proving that $\mathbb{Q}_p$ is not formally real. I've been looking for a concrete proof without results. The only hint that I have found says:
1) $\mathbb{Q}_2$ contains a square root of $-7$.
2) $\mathbb{Q}_p$ ($p>2$) contains a square root of $1-p$.
Question 1: How to prove 1 and 2? if it is very long to write as an answer, is there any book or paper showing the proof?
Question 2: Is there any better way to prove that $\mathbb{Q}_p$ is not formally real?
 A: Here is a sketch of the three approaches suggested in the comments.


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*You want to know whether the equation $x^2 - a$ has solutions in $\mathbb{Q}_p$, for $a = 1 - 8$ when $p = 2$ and for $a = 1 - p$ when $p \ge 3$. You can first show that these equations have solutions $\bmod 8$ and $\bmod p$ respectively, then apply Hensel's lemma repeatedly to show that these solutions lift to solutions in $\mathbb{Z}_2$ and $\mathbb{Z}_p$ respectively.

*The binomial theorem with exponent $\frac{1}{2}$ gives $(1 + x)^{1/2} = \sum_{n \ge 0} {1/2 \choose n} x^n$, and this equation continues to make sense over $\mathbb{Q}_p$ provided that the RHS converges $p$-adically, which it does when $x = -8$ and $p = 2$ (this requires a little work to bound the $2$-adic valuation of ${1/2 \choose n}$) and when $x = -p$ and $p \ge 3$ (this is easy because ${1/2 \choose n} \in \mathbb{Z}_p$ in this case). 

*Lubin's suggestion, which directly addresses the formal realness question, is to use the Teichmuller character $\lim_{n \to \infty} a^{p^n}$ to show that $\mathbb{Q}_p$ contains all $(p-1)$th roots of unity. When $p \ge 5$ we get roots of unity which don't exist in any formally real field. 
A: The sketch by @QiaochuYuan of my suggestion about taking successive $p$-th powers is good enough for me.
But my method of seeing that $\sqrt{-7}\in\Bbb Q_2$ is this. Start with $f(X)=X^2+7$, form $F(X+1)=X^2+2X+8$, and notice that the Newton polygon plots the points $(0,3)$, $(1,1)$, and $(2,0)$ so that there are two segments of width $1$, thus two $\Bbb Q_2$-rational factors of degree one.
If you aren’t comfortable with Newton, just form $F(2X+1)=4X^2+4X+8=4(X^2+X+2)$, which has a root by any version of Hensel that you like.
