I am having trouble understanding how "lifting" works here. They began by finding solutions $2$ and $3$ for $x^2\equiv 14 \text{ mod}(5)$ and then insert $(2+5t)$ into $x$ for the expression $x^2\equiv 14 \text{ mod}(25)$. Why does inserting $(2+5t)$ give solutions to $25$?
and the image is from the article on "brilliant" under "Hensel's lemma".
Thanks
The basic idea behind the Hensel Lemma is the following:
Lets say you want to solve $$x^2\equiv 14 \text{ mod}(625)$$
Note that if $a \equiv b \pmod{n}$ and $d|n$ then $a \equiv b \pmod{d}$.
Therefore $$x^2 \equiv 14 \pmod{625} \Rightarrow x^2 \equiv 14 \equiv 4 \pmod{5,25,125} (*) $$
Now, you go step by step modulo $5$, then $25$ then $125$. $$x^2 \equiv 4 \pmod{5} \Rightarrow x \equiv \pm 2 \pmod{5}$$
Lets look at the first solution: $$x \equiv 2 \pmod{5}$$ means $5|(x-2)$ and hence $x-2=5t$. Now, you go to the next step in (): you know by () that $$x^2 \equiv 14 \pmod{25}$$ and you just deduced that $$x=2+5t$$ Plugging this in here you get that $x \equiv \mbox{something} \pmod{25}$ and hence $x-\mbox{something}=25s$. You next use this relation in the next step in $(*)$, and repeat.