I am trying to find a prime p such that there exists a sequence of integers S_n where (S_n)^2 converges to 21 under the p-adic norm.
I'm familiar with the definition of p-adic distance and with the notion of convergence that applies. One idea I've had is that we're looking for primes p where 21 is a quadratic residue (mod p). I thought of this because we're looking at the squared terms in the sequence, but I'm not sure if it's helpful.
I can intuitively understand the idea of convergence to 0 in the p-adic norm (i.e. 5^x converges to 0 in 5-adic distance as x goes to infinity), but I'm having trouble extending this thinking to a non-zero convergence. Hints would be highly appreciated.
For any prime $p$ integer $x^2$, $x^2$ is close to $21$ in $p$-adic norm if $x^2 - 21$ is divisible by a large power of $p$.
Well, let's try $p=5$.
$21 - 1^2$ is divisible by $5^1$.
$21 - (5 k + 1)^2 \equiv 15 k + 20 \mod 5^2$; this is $0$ if $k = 2$, so $5k+1 = 11$.
$21 - (5^2 k + 11)^2 \equiv 75 k + 25 \mod 5^3$; this is $0$ if $k = 3$, so $5^2 k + 11 = 86$.
Can you see how to continue? Now look up Hensel lifting.