For any smooth projective curve of <genus> $g$, a <canonical divisor> has degree $2g-2$. If this plane curve had genus two, part (ii) would imply
$$
d(d-3)=2,
$$
so $d^2-3d-2=0$. The <quadratic discriminant> is $17$, which is not a <perfect square>, so the equation has no integral solution. Thus \b[a smooth projective plane curve cannot have genus two], as recorded by <no smooth plane curve has genus two>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 24F / a / iii / Solution

If a smooth degree-$d$ <projective plane curve> had genus two, the <genus-degree formula> would give $(d-1)(d-2)=4$, equivalently $d^2-3d-2=0$. Its discriminant is $17$, so it has no integral solution for the degree $d$.

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