[guided]This is the classical "16-term cancellation" or "octahedron" argument. The strategy is purely algebraic: pair symmetry is **not independent** of the other three symmetries — it follows automatically once skew-symmetry in $(i, j)$, skew-symmetry in $(k, \ell)$, and the first Bianchi identity all hold. Heuristically, the Riemann tensor lies in the $\mathfrak{S}_4$-subrepresentation of $V^{\otimes 4}$ cut out by exactly these three symmetries, and that subrepresentation is the irreducible piece on which pair symmetry holds automatically. We extract the identity by an explicit four-fold linear combination of the Bianchi identity. We organise the calculation around the four "anchor index" choices $(i, j, k, \ell)$, write out the cyclic Bianchi relation for each, and then take a signed sum chosen to isolate the desired diagonal $2 R_{ij, k\ell} - 2 R_{k\ell, ij}$.
Write the first Bianchi identity ($\ast$) from Step 3 in the $(0, 4)$-form: for any anchor index $a$ and remaining indices $b, c, d$,
\begin{align*}
R_{ab, cd} + R_{ac, db} + R_{ad, bc} &= 0,
\end{align*}
which cycles the last three indices while fixing the first. Apply this four times — taking each of $i, j, k, \ell$ as anchor in turn:
\begin{align*}
R_{ij, k\ell} + R_{ik, \ell j} + R_{i\ell, jk} &= 0, \tag{A} \\
R_{jk, \ell i} + R_{j\ell, ik} + R_{ji, k\ell} &= 0, \tag{B} \\
R_{k\ell, ij} + R_{ki, j\ell} + R_{kj, \ell i} &= 0, \tag{C} \\
R_{\ell i, jk} + R_{\ell j, ki} + R_{\ell k, ij} &= 0. \tag{D}
\end{align*}
We seek a linear combination of these four equations that produces $R_{ij, k\ell} - R_{k\ell, ij}$ from the diagonal terms. The diagonal terms — those involving the desired pair $(ij, k\ell)$ or $(k\ell, ij)$ — appear in (A), (B), (C), (D) respectively as $R_{ij, k\ell}$, $R_{ji, k\ell}$, $R_{k\ell, ij}$, $R_{\ell k, ij}$. By skew-symmetry in the first pair (Step 2), $R_{ji, k\ell} = -R_{ij, k\ell}$ and $R_{\ell k, ij} = -R_{k\ell, ij}$. So the combination $(A) - (B) - (C) + (D)$ contributes diagonally
\begin{align*}
R_{ij, k\ell} - R_{ji, k\ell} - R_{k\ell, ij} + R_{\ell k, ij} &= R_{ij, k\ell} + R_{ij, k\ell} - R_{k\ell, ij} - R_{k\ell, ij} \\
&= 2 R_{ij, k\ell} - 2 R_{k\ell, ij}.
\end{align*}
This is exactly what we want — provided the eight off-diagonal terms cancel.
We now verify the off-diagonal cancellation. The off-diagonal contributions to $(A) - (B) - (C) + (D)$ are:
\begin{align*}
\text{from } +(A): \quad &R_{ik, \ell j} + R_{i\ell, jk}, \\
\text{from } -(B): \quad &-R_{jk, \ell i} - R_{j\ell, ik}, \\
\text{from } -(C): \quad &-R_{ki, j\ell} - R_{kj, \ell i} = R_{ik, j\ell} + R_{jk, \ell i}, \\
\text{from } +(D): \quad &R_{\ell i, jk} + R_{\ell j, ki} = -R_{i\ell, jk} - R_{j\ell, ki},
\end{align*}
where in the last two lines we used skew-symmetry in the first pair (Step 2) to convert each term to a canonical first-pair ordering. Pairing terms with matching index patterns:
\begin{align*}
\text{from } (A), (D) \text{ (canonicalised)}: \quad &R_{i\ell, jk} - R_{i\ell, jk} = 0, \\
\text{from } (B), (C) \text{ (canonicalised)}: \quad &-R_{jk, \ell i} + R_{jk, \ell i} = 0, \\
\text{from } (A), (C): \quad &R_{ik, \ell j} + R_{ik, j\ell} = -R_{ik, j\ell} + R_{ik, j\ell} = 0, \\
\text{from } (B), (D): \quad &-R_{j\ell, ik} - R_{j\ell, ki} = R_{j\ell, ki} - R_{j\ell, ki} = 0,
\end{align*}
using skew-symmetry in the second pair (Step 1) — $R_{ik, \ell j} = -R_{ik, j\ell}$ and $R_{j\ell, ik} = -R_{j\ell, ki}$ — for the last two pairings.
All eight off-diagonal terms therefore cancel. The combination $(A) - (B) - (C) + (D)$ yields
\begin{align*}
2 R_{ij, k\ell} - 2 R_{k\ell, ij} &= 0,
\end{align*}
which gives the pair symmetry $R_{ij, k\ell} = R_{k\ell, ij}$.
A more conceptual route, which we record but do not use: define $T_{ij, k\ell} := R_{ij, k\ell} - R_{k\ell, ij}$. Then $T$ is skew in $(i, j)$ (inherited from Step 2), skew in $(k, \ell)$ (inherited from Step 1), antisymmetric under swapping the two pairs (by construction), and satisfies a cyclic identity (inherited from the first Bianchi identity applied to $R_{ij, k\ell}$ and to $R_{k\ell, ij}$). A short combinatorial argument shows that any tensor with these four properties must vanish — giving $T = 0$, hence pair symmetry. The explicit four-equation computation we just performed is in essence the unpacking of this combinatorial fact for $\mathfrak{S}_4$ acting on $V^{\otimes 4}$.[/guided]
