= Solution
{c}
Pair each point $\mathbf x$ of the <square>'s boundary with $-\mathbf x$. The <vector field> obeys $\mathbf f(-\mathbf x)=\mathbf f(\mathbf x)$, whereas the outward <unit normal vector> changes sign: $\mathbf n(-\mathbf x)=-\mathbf n(\mathbf x)$. Corresponding <arc length> elements are equal. Thus the two contributions to the <flux> cancel, proving without evaluating any boundary <integral> that
$$
\boxed{\int_C\mathbf f\cdot\mathbf n\,ds=0.}
$$
The same cancellation appears in the <divergence>, which is an <odd function> of $x$ over a region symmetric under $x\mapsto-x$. To justify using the planar <divergence theorem> despite the lack of <differentiability> at the origin, remove a disk of radius $\varepsilon<1$. The <vector field> is continuously differentiable on the remaining region, its <divergence> has zero <double integral> by this symmetry, and the magnitude of the inner <flux> is at most $2\pi\varepsilon^2$, since $\|\mathbf f\|\leq\varepsilon$ there. Letting $\varepsilon\to0$ recovers the result. This is <flux around a removable point singularity>.
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