# Points on curves of genus 3

Let $$Y$$ be a smooth complex projective curve of genus two, $$X$$ a Galois cover of degree two of $$Y$$ and $$K$$ the canonical divisor of $$X$$. Let $$i$$ be the involution of $$X$$ over $$Y$$. Can one find a point $$P$$ on $$X$$ such that, if $$Q=i(P)$$, the divisor $$5P+3Q$$ is linearly equivalent to $$2K$$?

No. Note that $$P\neq Q$$ since $$i$$ is fixed-point free. Since $$i^*K=K$$, one would have $$5P+3Q\equiv 3P+5Q$$ (where $$\equiv$$ means linear equivalence), hence $$2P\equiv 2Q$$, so $$P$$ and $$Q$$ are Weierstrass points on the hyperelliptic curve $$Y$$. But then $$4P\equiv 4Q\equiv K$$, and your condition becomes $$P-Q\equiv 0$$, hence $$P=Q$$, a contradiction.