Show $|x^2 + y^2 - 7|_7 \geq \frac{1}{7}$ in p-adics Working in the $7$-adic numbers, I want to show that for any $x, y \in \mathbb{Q}$, we can guarantee that $|x^2 + y^2 - 7|_7 \geq c$ for some $c$.

By some amount of trial and error, I've guessed that $c = 1/7$. I showed this must be true for any $|x^2 - 7|_7$, but I'm having trouble extending the insight for one square to the sum of two squares.
To simplify notation a bit, put $p = 7$. Let $\nu : \mathbb{Q} \to \mathbb{Z}$ denote the $p$-adic valuation, so that $|x|_p = p^{-\nu(x)}$. Then $\nu(x + y) = \min\{\nu(x), \nu(y)\}$ for $\nu(x)\not= \nu(y)$. Since $(-1/p) = -1$, we have $\nu(x^2 + y^2)\not = 1$, and thus $|x^2 + y^2 - 7|_p \geq |7|_p = \frac{1}{7}$.
There are a number of cases to consider, depending on $|x|_7$ and $|y|_7$. 
I'll just do the most interesting one.  Suppose $|x|_7 = |y|_7 = 1$.
Write $x = x_1 + 7 x_2 + \ldots$ and $y = y_1 + 7 y_2 + \ldots$, with $x_1, y_1 \in \{1,\ldots,6\}$.  Since the nonzero squares mod $7$ are $1$, $2$, and $4$, $x_1^2+y_1^2$ can't be divisible by $7$, so 
$|x^2 + y^2 - 7|_7 = |x_1^2 + y_1^2|_7 = 1$.