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Induct on the Seifert genus $g_s(K)$. A genus-zero knot is the unknot. If $K$ is prime there is nothing to prove. Otherwise write $K=K_1\#K_2$ with both summands nontrivial. Part c givesso each summand has strictly smaller genus than $K$. Apply the induction hypothesis to both. Since the genus drops at every nontrivial split, the process terminates after finitely many steps and expresses $K$ as a finite connected sum of prime knots.

Solved by gpt-5.6-sol high.

If $K=K_1\#K_2$ with both summands nontrivial, the sphere separating the two punctured three-balls in the connected-sum construction is a splitting sphere of a knot. If it were trivial, the corresponding one-string tangle would be boundary-parallel and one of $K_1,K_2$ would be the unknot. The sphere is therefore nontrivial.

Conversely, a splitting sphere $S$ divides $S^3$ into three-balls $B_1,B_2$, and $K\cap B_i$ is a properly embedded arc. Join its endpoints by a fixed arc on $S$ and push that joining arc slightly into $B_i$; this closes the two tangles to knots $K_1,K_2$. Reversing the construction showsIf either $K_i$ were unknotted, an innermost-disk argument for a spanning disk of $K_i$ would make the corresponding tangle boundary-parallel, which is exactly the stated triviality condition for $S$. A nontrivial splitting sphere therefore makes both summands nontrivial, so $K$ is composite.

Solved by gpt-5.6-sol high.