= Solution
Boundary-connected-summing minimal <Seifert surface>[Seifert surfaces] for $K$ and $K'$ gives
$$
g_s(K\mathbin{\#}K')\leq g_s(K)+g_s(K').
$$
For the reverse inequality, let $F$ be a minimal-genus Seifert surface for the <connected sum of knots> and let $S$ be its standard splitting sphere. A minimal-genus Seifert surface is incompressible in the knot exterior: a compression either lowers its genus or separates off a closed component that can be discarded. Put $F$ and $S$ in transverse position and minimize the number of intersection circles. An innermost circle on $S$ either gives a compression of $F$ or bounds a disk on $F$ across which it can be removed. Both alternatives contradict minimality, so $F\cap S$ consists only of the single arc joining the two points of $S\cap\partial F$.
Cutting $F$ along this arc gives Seifert surfaces $F_1,F_2$ for $K,K'$. Their Euler characteristics satisfy
$$
\chi(F)=\chi(F_1)+\chi(F_2)-1,
$$
which, since all three surfaces have one boundary component, is equivalent to
$$
g(F)=g(F_1)+g(F_2)\geq g_s(K)+g_s(K').
$$
This proves additivity.
The <torus knot> $T_{2,3}$ bounds a once-punctured torus, and its degree-two <Alexander polynomial of a knot> forces every Seifert surface to have genus at least one. Thus $g_s(T_{2,3})=1$. If it were a <composite knot>, both nontrivial summands would have positive Seifert genus, and additivity would give genus at least two. Hence $T_{2,3}$ is a <prime knot>.
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