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Addendum. I realized that the last paragraph of [2, Proof of Proposition 1.9] is flawed, so that the unimodular row $(21 + 4x, 12, x^2 + 20)$ is not guaranteed to be non-stable in $\mathbb{Z}[X]^3$. Indeed, it is assumed there that $SK_1(\mathcal{O}, 2\mathcal{O}) = \mathbb{Z}/2\mathbb{Z}$ for $\mathcal{O} = \mathbb{Z} + \mathbb{Z} \sqrt{-5}$, whereas $SK_1(\mathcal{O}, 2\mathcal{O})= 1$ by the Bass-Milnor-Serre Theorem (In other words, $f = 2$ does not have the property $(*)$ defined at the top of page 191). This glitch in the final step of the proof is harmless and can be fixed easily by considering $f = 4$, i.e., $B = \mathbb{Z} + 4 \mathbb{Z} \sqrt{-5} \subset \mathcal{O}$ instead and a suitable quadratic residue symbol. My answer below is yet another way to address this problem.

Addendum. I realized that the last paragraph of [2, Proof of Proposition 1.9] is flawed, so that the unimodular row $(21 + 4x, 12, x^2 + 20)$ is not guaranteed to be non-stable in $\mathbb{Z}[X]^3$. Indeed, it is assumed there that $SK_1(\mathcal{O}, 2\mathcal{O}) = \mathbb{Z}/2\mathbb{Z}$ for $\mathcal{O} = \mathbf{Z} + \mathbf{Z} \sqrt{-5}$, whereas $SK_1(\mathcal{O}, 2\mathcal{O})= 1$ by the Bass-Milnor-Serre Theorem (In other words, $f = 2$ does not have the property $(*)$ defined at the top of page 191). This glitch in the final step of the proof is harmless in the sense that changing sligthly the conductor $f$ yields a valid, but different, non-reducible unimodular row. Setting for instance $f = 4$, i.e., $B = \mathbf{Z} + 4 \mathbf{Z} \sqrt{-5} \subset \mathcal{O}$, and considering a suitable quadratic residue symbol, will do the job. My answer below is yet another way to address this problem.

The Bass-Milnor-Serre theorem [3, Theorem 11.33] is key to understand step 1: let $F$ be a totally imaginary number field and $m$ is the number of elements of finite order in $F^{\ast}$. For each ideal $I$ in the ring of integers $R$ of $F$, the relative special Whitehead group $SK_1(R, I)$ is a cyclic group of finite order $r$. For each prime $p$, $\text{ord}_p(r)$ is the nearest integer in the interval $\lbrack 0, \text{ord}_p(m) \rbrack$ to $$\min_{\mathfrak{p}} \left\lfloor \frac{\nu_{\mathfrak{p}}(I)}{\nu_{\mathfrak{p}}(pR)} - \frac{1}{p - 1} \right\rfloor$$ where $\lfloor x \rfloor$ denotes the greatest integer $\le x$ and $\mathfrak{p}$ ranges over the prime ideals of $R$ containing $p$.

The Bass-Milnor-Serre theorem [3, Theorem 11.33] is key to understand step (1): let $F$ be a totally imaginary number field and $m$ is the number of elements of finite order in $F^{\ast}$. For each ideal $I$ in the ring of integers $R$ of $F$, the relative special Whitehead group $SK_1(R, I)$ is a cyclic group of finite order $r$. For each prime $p$, $\text{ord}_p(r)$ is the nearest integer in the interval $\lbrack 0, \text{ord}_p(m) \rbrack$ to $$\min_{\mathfrak{p}} \left\lfloor \frac{\nu_{\mathfrak{p}}(I)}{\nu_{\mathfrak{p}}(pR)} - \frac{1}{p - 1} \right\rfloor$$ where $\lfloor x \rfloor$ denotes the greatest integer $\le x$ and $\mathfrak{p}$ ranges over the prime ideals of $R$ containing $p$.

Addendum. I realized that the last paragraph of [2, Proof of Proposition 1.9] is flawed, so that the unimodular row $(21 + 4x, 12, x^2 + 20)$ is not guaranteed to be non-stable in $\mathbb{Z}[X]^3$. Indeed, it is assumed there that $SK_1(\mathcal{O}, 2\mathcal{O}) = \mathbb{Z}/2\mathbb{Z}$ for $\mathcal{O} = \mathbb{Z} + 2 \mathbb{Z} \sqrt{-5}$, whereas $SK_1(\mathcal{O}, 2\mathcal{O})= 1$ by the Bass-Milnor-Serre Theorem (In other words, $f = 2$ does not have the property $(*)$ defined at the top of page 191). This glitch in the final step of the proof is harmless and can be fixed easily by considering $\mathcal{O'} = \mathbb{Z} + 4 \mathbb{Z} \sqrt{-5}$ instead and a suitable quadratic residue symbol. My answer below is yet another way to address this problem.

Addendum. I realized that the last paragraph of [2, Proof of Proposition 1.9] is flawed, so that the unimodular row $(21 + 4x, 12, x^2 + 20)$ is not guaranteed to be non-stable in $\mathbb{Z}[X]^3$. Indeed, it is assumed there that $SK_1(\mathcal{O}, 2\mathcal{O}) = \mathbb{Z}/2\mathbb{Z}$ for $\mathcal{O} = \mathbb{Z} + \mathbb{Z} \sqrt{-5}$, whereas $SK_1(\mathcal{O}, 2\mathcal{O})= 1$ by the Bass-Milnor-Serre Theorem (In other words, $f = 2$ does not have the property $(*)$ defined at the top of page 191). This glitch in the final step of the proof is harmless and can be fixed easily by considering $f = 4$, i.e., $B = \mathbb{Z} + 4 \mathbb{Z} \sqrt{-5} \subset \mathcal{O}$ instead and a suitable quadratic residue symbol. My answer below is yet another way to address this problem.