I have the following problem:
Let $$R:=\left\{ (x_n)_{n\in\mathbb{N}}\in\prod_{n\in\mathbb{N}} \mathbb{Z}/(5^n)\ |\ \forall n\in\mathbb{N}\quad x_{n+1}\equiv x_n\ \text{mod}\ 5^n \right\},$$ $n\in \mathbb{N}, x\in \mathbb{Z}$ with $x\equiv 2$ mod $5$ and the property that there is a $k\in \mathbb{Z}$ with $x^2+1=5^n*k$
i) $y:=x-4*5^n*k\in \mathbb{Z}$ fulfills $y^2=-1$ mod $5^{n+1}$
ii) Conclude that there is an $x\in R$ with $x^2=-1$
So I thought about rewriting $x^2+1=5^n*k$ as $x^2=-1$ mod $5$. Also we have $y^2=(x-4*5^n*k)^2=x^2-8*5^n*k*x+16*5^{2n}k^2$, but honestly, I don't see at all why $y^2$ would fulfill $i)$ nor do I see how $ii)$ has anything to do with $i)$. Can someone help me out?
This is a standard exercise, that there is a square root of $-1$ in the 5-adic numbers. If we start with $$ 2^2 \equiv - 1 \pmod {5^1} $$ we then ask about $5^2.$ The induction step is this: $(2 + 5 k)^2 \equiv 4 + 20 k \pmod{5^2},$ because the $k^2$ part is divisible by $25$ already. If we take $k=1$ this time, we get $$ 7^2 \equiv - 1 \pmod {5^2} $$ Next $(7 + 25 k)^2 \equiv 49 + 350 k \equiv \equiv 49 + 100 k \pmod{5^3},$ this time taking $k=2$ to get 249 gives $$ 57^2 \equiv - 1 \pmod {5^3} $$ Next $(57 + 125 k)^2 \equiv 124 + 500 k \pmod{5^4},$ this time taking $k=1$ to get 624 gives $$ 182^2 \equiv - 1 \pmod {5^4} $$ And so on forever. It really is worth doing this step by step a few times.
Once we chose $2,$ there will always be one and only one value of $k$ with $0 \leq k \leq 4$ that takes us to the next step. As far as size, this means the representative we construct at each step is an integer between $0$ and $5^n;$ we are writing it in base 5 but in reverse order.