<s>[INST] Existence of $\sqrt{-1}$ in $5$-adics, show resulting sum is convergent. I know that to prove the existence of a square root of $-1$ in $\mathbb{Z}_5$, I can just plug $x = -5$ and $a = 1/2$ into the Taylor expansion$$(1 + x)^a = \sum_{n=0}^\infty \binom{a}{n} x^n.$$However, I am confused on how I would make sure that the resulting sum is convergent. Could anyone help me with that? Thanks very much in advance.
 [/INST] The convergence should be in $\mathbb{Z}_5$, that works differently than in $\mathbb{R}$.
I try first an approximation
$$2^2 = 4 = -1 + 5$$
Say I have $a$ with $a^2 = -1$. Then 
$$\left(\frac{a}{2}\right)^2 = \frac{-1}{-1+5}= \frac{1}{1-5}$$
So take 
$$\frac{a}{2} = (1-5)^{-1/2}$$
We have the binomial formula
$$(1-t)^{-1/2} = 1+\frac{t}{2}+\frac{3 t^2}{8}+\frac{5 t^3}{16}+\frac{35 t^4}{128}+\frac{63 t^5}{256}+\frac{231 t^6}{1024}+\frac{429 t^7}{2048}+\\
+\frac{6435 t^8}{32768}+\frac{12155 t^9}{65536}+\frac{46189 t^{10}}{262144}+ \cdots$$
convergent in $\mathbb{Z}_5$ if $|t|_5<1$. Now, $t=5$ and $|5|_5=1/5 <1$. Therefore:
$$a = 2\cdot ( 1+\frac{5}{2}+\frac{3\cdot 5^2}{8}+\frac{5 \cdot5^3}{16}+\frac{35\cdot 5^4}{128}+\frac{63\cdot 5^5}{256}+\frac{231\cdot 5^6}{1024}+\frac{429\cdot 5^7}{2048}+\\
+\frac{6435\cdot 5^8}{32768}+\frac{12155\cdot 5^9}{65536}+\frac{46189 \cdot5^{10}}{262144} + \cdots)$$
with the other root $-a$.
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