Let $r(n)$ denote the number of integral solutions to $a^2+b^2 = n$ where $a,b,n$ are integers. Furthermore, we count the pairs with regard to order and signs. (So if $(a,b)$ is a solution, so are $(\pm a, \pm b); (\pm b, \pm a))$.
Now for a fixed $n$, let $R(n) = \sum_{k = 1}^{n} r(k)$. Why is $R(n)$ is equal to one less than the total number of lattice points in the disk $x^2+y^2 \le n$?
As franz lemmermeyer pointed out in the comment above, the reason why $R(n)$ is equal to one less than the total number of lattice points in the disk $x^2 + y^2 \leq n$ is that it doesn't include the only solution to $x^2 + y^2 = 0$ which is $(0, 0)$.