= Solution
To form the <connected sum of knots> $K_1\#K_2$, remove a short unknotted arc from each oriented <knot> in two disjoint balls and join the remaining endpoints by a pair of parallel arcs respecting the orientations. Equivalently, a <sphere> meeting a <knot> in two points cuts it into two one-string <tangles>; close each by an arc in the <sphere> to recover its summands. Sliding the small joining balls along the <knots> shows that the <isotopy> type does not depend on these choices. The <unknot> is the identity for this operation. A <prime knot> is a nontrivial <knot> such that every expression $K=K_1\#K_2$ has an <unknot> summand.
The <Seifert algorithm> supplies a <Seifert surface> for every oriented <knot>: smooth the crossings in their oriented directions, span the resulting <Seifert circles> by disks at different heights, and restore each crossing with a half-twisted band. If needed, join components by tubes away from the boundary. Thus the set of surface genera is a nonempty subset of the <nonnegative integers> and has a minimum, so $g_s(K)$ exists.
We first prove <additivity of Seifert genus>, which also supplies a terminating measure for decomposition. Joining minimal <Seifert surfaces> along their boundary arcs gives $g_s(K_1\#K_2)\leq g_s(K_1)+g_s(K_2)$. For the reverse inequality, let $F$ be a minimal-genus <Seifert surface> for the sum and $S$ a <splitting sphere of a knot>. Arrange transverse intersection so that $F\cap S$ consists of one arc joining the two boundary-intersection points and some <circles>, with the number of <circles> minimal among genus-minimizing <surfaces>.
Each <circle> separates the <sphere> into two disks; since it is disjoint from the intersection arc, one disk contains neither marked endpoint. Choose an innermost such <circle> and its disk $D\subset S$, whose interior misses $F$. If its boundary were essential in $F$, compressing $F$ along $D$ would lower the genus of its component with boundary K: for a nonseparating curve genus drops by one, and for a separating essential curve the removed boundaryless side has positive genus. This contradicts minimal genus. Its boundary therefore bounds a disk in $F$. Replace that disk by a small push-off of $D$, removing the intersection <circle> without changing genus or boundary, again contradicting minimal intersection. Thus no <circles> remain.
Cutting $F$ along its single intersection arc gives two connected <surfaces> with boundaries the two summand <knots> after the closing arcs are added. <Euler characteristic> shows that their genera add to $g(F)$. Each genus is at least the minimum genus of its summand, so
$$
\boxed{g_s(K_1\#K_2)=g_s(K_1)+g_s(K_2).}
$$
Genus zero means that the <knot> bounds a disk and is the <unknot>. Every nontrivial <knot> therefore has positive genus. Induct on this nonnegative integer: a nontrivial nonprime <knot> splits into two nontrivial summands, each with strictly smaller genus by the equality above, and induction decomposes both into primes. This proves \b[every <knot> is a finite connected sum of <prime knots>], with the <unknot> the empty sum.
The same induction does not follow from the three suggested replacements. For <crossing number of a knot>, joining minimal diagrams gives only subadditivity; a reverse inequality giving strict descent for every nontrivial summand is not provided by this argument. The <Jones polynomial of a connected sum> does make its <breadth of a Laurent polynomial> additive, but the needed implication that zero breadth forces the <unknot> is the <Jones unknot detection problem>. For the <Alexander polynomial>, even this implication is false: nontrivial untwisted <Whitehead doubles> have <Alexander polynomial> one and hence breadth zero. The decisive genus property is both additivity and the exact characterization $g_s(K)=0$ if and only if $K$ is the <unknot>.
For uniqueness use complete <decomposing sphere systems for a knot>. Cut each system and cap the marked boundary <spheres> with straight arcs; the resulting oriented factors are <prime knots>. Compare two systems in transverse position, and use the <prime-ball exchange lemma> to simplify their intersection. Choose an innermost intersection <circle> bounding a disk in a sphere of one system and lying in a prime-factor ball of the other. If the disk misses the knotted arc it cuts off an empty ball. If it meets the arc once it cuts the prime factor into two <knots>, so one is an <unknot>. Replace the old ball by the side carrying the same prime factor, perturb its boundary slightly, and remove that intersection circle. This preserves the unordered factors: the discarded empty or trivial ball merely changes the joining arc, and any intervening capped boundaries still carry their original factors. If the chosen disk contains both marked points, take another innermost disk without marked points, or the complementary disk when the intersection circle is the only one.
Iteration yields an innermost prime-factor ball disjoint from both systems. Its arc has no nontrivial further decomposition, so it represents one factor in each. The boundary can be included in both systems using the same ball exchange; replace its arc by a straight boundary arc and remove this common factor. Induction on the number of factors matches the remaining factors. When the systems already have no intersections, start directly with an innermost ball and the same removal step. All replacements preserve the arc <orientation>, so the match is of oriented <knots>, not only unoriented types. Hence the <prime decomposition of knots> is unique up to reordering and oriented <link isotopy>.
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