Generating P-adic integers from Hensels lemma, and general question about P-adic integers I know that one can use the "lifting" method in order to solve problems such as $x^2\equiv 14 \text{ mod }(625)$.  I also read that this process generates the $5$-adic integers as a sequence.  I am confused about a few things going on.  
Whenever I search $p-adic$ numbers I get articles discussing decimal representations. It seems I lack a clear understanding of what p-adic refers to.  I thought that there was a unique sequence for the $5$-adic integers, for example, such as $2,7,57,182,...$, and that we can generate this sequence from expressions like $x^2\equiv 14 \text{ mod }(625)$, but wouldn't we get a different sequence if we replaced $14$ with another integer where $x$ has solutions?  and also, what is the connection between these sequences and the all the google results I get about p-adic discussing decimal representations with infinite elements on the left side of the decimal point?
Thanks
 A: Well, first, given a $p$-adic integer $z$, there is not a unique sequence of ordinary integers that has $z$ as a limit, just as there are many sequences of integers with $p$-adic limit zero.
I want to recommend to you a way of thinking of $p$-adic integers that is becoming more and more generally used. Represent your ordinary positive integers in $p$-ary place-value notation, so that (taking $p=5$) the representation of $117_{10}$ becomes $432;$ since $117=4\cdot25+3\cdot5+2$. I like to use a semicolon to remind the reader that we’re talking $p$-adic here. Note that we put the higher powers of the radix, $5$, to the left, just as we do in elementary school with powers of ten.
Now I want you to think of a $5$-adic integer as just such an expansion in base $5$, but extending potentially infinitely to the left. So the higher and higher powers of $5$ show up farther and farther to the left. As the exponent becomes higher, the digit is less and less significant, since $5^n\to0$ as $n\to\infty$.
The utility of this notation is that addition, subtraction, and multiplication of two $p$-adic numbers proceeds precisely as it does in elementary school, assuming that you were taught to do those things. You start at the right-hand end, and proceed, using carries and borrows in exactly the familiar way. The difference is that you need only proceed to the point where you get the desired degree of accuracy. Lamentably, division as we learned it in elementary school proceeds left to right, so it has to be modified for $p$-adic computation. The method for this is eminently reasonable, but virtually impossible to explain in a static medium like the printed word.
If you ask me how to use these ideas to find the $5$-adic square root of fourteen, i.e. $\sqrt{24;}$, I’ll do it for you in an edit, but maybe you’d like to do it for yourself. I wouldn’t use the step-by-step method of getting one more power of $5$ at each repetition, but maybe something quick like Newton-Raphson.
