$x^2 = 3$ is solvable mod $3$, not solvable mod $3^n$ for any $n>1$. You should adjust the statement of the problem.

Hi @ErickWong, what do you mean by adjusting the statement of the problem? As in...plug some numbers in to try and see it more concretely? I am reading up on Newton's Method and currently looking for an approach there...thanks,

@user58865 I mean that the problem, as you have stated it, is false and therefore cannot be proven. It can be adjusted so that it is true, but we are not mind readers so it's your job to go back and check if you omitted any details from the question. One reasonable possibility is to require that $a$ is not divisible by $p$, and that $p$ be a prime $>2$. But again, not mind readers, and note that the assumptions that you're missing are almost as long as the question itself.

Hi @ErickWong, backing up just a bit, how do you know that $x^2=3$ is not solvable mod $3^n$? I tried for $n=2$, and am looking at the numbers $\frac{x^2-3}{9}$, plugging in $x = 0,1,2,3,4,5,6,7,8,9$. Based on this, I agree with your claim, but how could I think about this rigorously and not possibly be mistaken? E.g., have I checked enough cases? (Sorry, I'm brand-new to number theory / abstract algebra...so I know my questions might be a bit trivial)

@user58865 No problem, this is very easy once you focus on the right thing. In order for $x^2 - 3$ to be divisible by $3^n$ it must be divisible by $3$, but for this to happen, $x$ itself must be divisible by $3$ (this follows from uniqueness of prime factorization). However, then $x^2$ is in fact divisible by $9$, so $x^2 - 3$ is not a multiple of $9$. This proves that $x^2 - 3$ cannot be divisible by $9$ or any higher power of $3$.

I almost follow you @ErickWong, however, I'm stuck at an intermediate step: following your reasoning, then $x^2$ must be divisible by $3$. But how does this then force $x$ itself to be divisible by $3$?

Hi @ErickWong for instance if $x^2 = 3$, then $x = + / - \sqrt{3}$, which is not a multiple of $3$ ...

Hi @ErickWong, also, $p$ is prime, not necessarily $>2$: take $x^2 = 1 (mod 2). Then $x =1$ is a solution modulo 2, as well as for modulo $2^n$. Also, $a$ is not necessarily not divisible by $p$: take x^2 = 0 (mod p). Then $x=0$ is a solution for the equation (mod p) and for (mod p^n), and $0$ is divisible by $p$...

but $\sqrt 3$ is not an integer and $\sqrt 3 \mod 3$ doesn't make much sense

besides, if square roots were allowed you could just pick $x = \sqrt a$ everytime

another counterexample is $x^2=3 \pmod {2^n}$. This has a solution when $n=1$ but no solution for $n \ge 2$