= Solution
For a nonsingular projective integral curve, apply the <Riemann-Roch theorem> to the <divisor on an algebraic curve> $mP$. It gives $\ell(mP)\geq m\deg P+1-g$, where $g$ is the <geometric genus>. For sufficiently large $m$, this exceeds the dimension of the space of constant functions; over the usual geometrically integral convention that dimension is one. Thus $L(mP)$ contains a nonconstant <rational function>.
By the definition of $L(mP)$, the <principal divisor on an algebraic curve> of that function plus $mP$ is effective, so it has no pole away from $P$. Therefore \b[there is a nonconstant <rational function> regular on $X\setminus\{P\}$]. If $P$ is not a rational point, using its closed-point degree in the same inequality handles that case. A positive pole order at $P$ is unavoidable, since <regular functions> on a projective integral curve are constant in its constant field.
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