The square knot is a nontrivial ribbon knot

Apply [F1] with $K=T$. Its explicit ribbon immersion has oriented boundary $T\mathrm{\#}{T}^{†}$, so the square knot is ribbon by the A-page definition. Apply [F6] to that immersion to obtain a smooth properly embedded disk in $B^4$ with the same boundary, proving smooth sliceness as well. Applying [F7] to the same immersion also gives a no-maximum disk movie with its required radial product boundary collar.

In the two-strand braid chart with parameter increasing from bottom to top, let $S$ be the color transfer for the positive generator whose overstrand runs from lower left to upper right. By [F3], $S(a,b)=(2a-b,a)$; the inverse crossing has transfer $S^{-1}$. Direct calculation gives $S^2(a,b)=(3a-2b,2a-b)$ and $S^3(a,b)=(4a-3b,3a-2b)=(a,b)$ in ${\mathbb{F}}_3^2$. Hence $S^{-3}$ is also the identity. Starting each trefoil tangle with the ordered pair $(0,1)$ therefore satisfies the braid closure conditions for both the trefoil and its mirror. The first arc in each closure has color $0$; choose those arcs for the connected-sum datum. The colorings glue because the shared arc colors agree, and the color-$1$ arcs remain, giving a nonconstant Fox 3-coloring of the square knot.

A crossing-free diagram of the unknot has one arc and exactly three Fox 3-colorings, one for each residue. The square-knot diagram has those three constant colorings and the nonconstant coloring from step 1.2, so its coloring count is at least four. If it were isotopic to the unknot, [F5] would give a finite Reidemeister sequence between their diagrams, and [F4] would preserve the coloring count, a contradiction. Thus the square knot is nontrivial.