Does anyone know an "elementary" proof of the following theorem?
Let $k \neq 0$ be a rational integer. Then $k$ admits a square root in $\mathbb{Z}_2$ if $k = 4^a (8b+1)$ for some $a \in \mathbb{N}$, $b \in \mathbb{Z}$.
About $p$-adic numbers I don't know anything more sophisticated than Hensel lemma. Thank you!
Best Answer
You are looking for $x\in\mathbb Z_2$ such that $(1+2x)^2=8b+1$. This is equivalent to $$ x^2+x-2b=0.$$ Now mod $2$, this polynomial is $x(x+1)$ and has (at least) one simple root in $\mathbb F_2$. By Hensel's lemma, the above equation has a solution in $\mathbb Z_2$.