= Solution
Commutativity and equality of squares give
$$
(S-T)(S+T)=S^2-T^2=0.
$$
Thus $S-T$ vanishes on $\operatorname{Ran}(S+T)$, and by <continuity> on its closure. The preceding <kernel of a sum of positive operators> identity shows that it also vanishes on $\ker(S+T)$. Since $S+T$ is <self-adjoint>, part (i) and the <Hilbert projection theorem> give the <orthogonal direct sum>
$$
X=\overline{\operatorname{Ran}(S+T)}\oplus\ker(S+T).
$$
Therefore $S-T$ vanishes on all of $X$, proving
$$
\boxed{S=T.}
$$
This proves <commuting positive operators with equal squares are equal> without assuming a square-root uniqueness theorem.
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