Show that if some equation is solvable modulo $p$, then it is also solvable modulo $p^n$. 
Show that if $x^2=a$ is solvable modulo $p$, then it is also solvable modulo $p^n$ for all positive integers $n.$  (Note that $p$ is prime, not necessarily >2, and $a$ is not required to be divisible by $p$.)

My work, so far:
By assumption, $x^2 =a$ is solvable modulo $p$ - call the solution $x = k$.
Then $\frac {x^2 - a}{p} \in \mathbb{Z}$,which implies 
$$k^2 - a = mp$$
for $m, k \in \mathbb{Z}$
Now I'm trying to show the next implication, namely that there exists some $x$ such that 
$$\frac{x^2-a}{p^2} \in \mathbb{Z}$$
Any comments or suggestions are welcome.  
Thanks,
 A: I am proceeding with the assumption that $ p $ is prime. The following statement is the simplest form of Hensel's lemma:
Theorem (Hensel's lemma). Let $ f(X) \in \mathbf Z[X] $ be a polynomial with a root $ \alpha $ in $ \mathbf Z/p\mathbf Z $ such that $ f(\alpha) = 0 $ and $ f'(\alpha) \neq 0 $ ($f' $ is the formal derivative of $ f $). Then, $ f(X) $ has a unique root modulo every $ p^n $ for $ n \geq 1 $ which is congruent to $ \alpha $ modulo $ p $.
The proof is simple and proceeds by induction, you may find an argument in this wikipedia page.
To see why this resolves our problem, let $ p $ be an odd prime. Then, the formal derivative of $ X^2 - a $ is $ 2X $, which does not vanish unless $ X = 0 $. Thus, Hensel's lemma kicks in and gives us a root in $ \mathbf Z_p $, that is, a root modulo every $ p^n $. However, we have a problem if $ p = 2 $: the formal derivative is then zero, since $ 2 = 0 $ in characteristic 2. Indeed, for example, $ 3 $ is a perfect square modulo $ 2 $, but it is not a perfect square modulo $ 4 $.
This has to do with the structure of the group $ \mathbf Z_2^{\times} $ - in short, an odd integer is a perfect square modulo every $ 2^n $ if and only if it is a perfect square modulo $ 8 $. In comparison, the result for odd primes is simpler, as we have seen above.
A: Yes, $p$ denotes a prime number.  To say that $x^2 = a$ is solvable modulo $p$ means that the congruence $x^2 \equiv a \mod p$ has a solution.
A: $p$ does not generally have to be prime, although often it is chosen as the letter to represent the modulus to emphasize that it is prime.  If it is required to be prime, the author should say so.  If I want generic naturals, I start with $n,m$, but if I need more then $p,q$ are next even if primality is not required. In this example, $3^2=9 \equiv 3\pmod {6}$. $6$ is not prime, but we can say $x^2=3$ is solvable $\bmod 6$. This example shows that the statement requires something else (like $p$ being prime) as $x^2=3$ has no solution $\bmod {36}$.  Just try all the cases.
