I think that I might have spotted I small mistake (a missing $5$-defective Lehmer pair) in the classification of terms of Lehmer sequences without primitive divisors given in:

*1 Bilu, Hanrot, and Voutier, Existence of primitive divisors of Lucas and Lehmer numbers, Journal für die reine und angewandte Mathematik (Crelles Journal), 2001.* (available on the web)

Since I might very much be wrong myself, I would like to know if my reasoning is correct or not. I recall below the main definitions of 1.

A *Lehmer pair* is a pair of complex numbers $(\alpha, \beta)$ such that $(\alpha + \beta)^2$ and $\alpha\beta$ are non-zero coprime integers and $\alpha / \beta$ is not a root of unity. Two Lehmer pairs $(\alpha_1, \beta_1)$ and $(\alpha_2, \beta_2)$ are said to be *equivalent* if $\alpha_1 / \alpha_2 = \beta_1 / \beta_2 \in \{-1,+1,\sqrt{-1},-\sqrt{-1}\}$. Given a Lehmer pair $(\alpha, \beta)$, the associated Lehmer sequence is
$$\widetilde{u}_n(\alpha, \beta) := \begin{cases} (\alpha^n - \beta^n) / (\alpha - \beta) & \text{ if $n$ is odd} \\
(\alpha^n - \beta^n) / (\alpha^2 - \beta^2) & \text{ if $n$ is even}\end{cases}$$
for every positive integer $n$ (it is an integer sequence).

A prime number $p$ is a *primitive divisor* of $\widetilde{u}_n(\alpha, \beta)$ if $p$ divides $\widetilde{u}_n(\alpha, \beta)$ but does not divide $(\alpha^2 - \beta^2)^2 \widetilde{u}_1(\alpha, \beta) \cdots \widetilde{u}_{n-1}(\alpha, \beta)$. If $\widetilde{u}_n(\alpha, \beta)$ has no primitive divisor then the Lehmer pair $(\alpha, \beta)$ is *$n$-defective.*

One of the claims of Theorem 1.3 of 1 is that, up to equivalence, all $5$-defective Lehmer pairs are of the form $((\sqrt{a} - \sqrt{b})/2, (\sqrt{a} + \sqrt{b})/2)$ with

$$(1) \qquad (a, b) = (\phi_{k-2\varepsilon}, \phi_{k-2\varepsilon} - 4\phi_k) \quad (k \geq 3)$$

or

$$(2) \qquad (a, b) = (\psi_{k-2\varepsilon}, \psi_{k-2\varepsilon} - 4\psi_k) \quad (k \neq 1) ,$$

where $k$ is a nonnegative integer, $\varepsilon \in \{-1, +1\}$, $(\phi_n)$ is the sequence of Fibonacci numbers, and $(\psi_n)$ is the sequence of Lucas numbers.

**Claim 1:** *The Lehmer pair $(\alpha_0, \beta_0) := ((1 - \sqrt{5}) / 2, (1 + \sqrt{5}) / 2)$ is $5$-defective.*

First, note that $(\alpha_0 + \beta_0)^2 = 1$ and $\alpha_0\beta_0 = -1$ are non-zero coprime integers and $\alpha_0 / \beta_0 = (\sqrt{5}-3) / 2$ is not a root of unity, so that $(\alpha_0, \beta_0)$ is indeed a Lehmer pair. Second, for the associated Lehmer sequence we have $\widetilde{u}_5 = 5$ and $(\alpha_0^2 - \beta_0^2)^2 = 5$, thus $\widetilde{u}_5$ has no primitive divisor and $(\alpha_0, \beta_0)$ is $5$-defective.

**Claim 2:** *The Lehmer pair $(\alpha_0, \beta_0)$ is not equivalent to a pair of the form $(\alpha, \beta) = ((\sqrt{a} - \sqrt{b})/2, (\sqrt{a} + \sqrt{b})/2)$ with $(a, b)$ as in (1) or (2).*

For the sake of contradiction suppose $(\alpha_0, \beta_0)$ is equivalent to a pair of the form $(\alpha, \beta) = ((\sqrt{a} - \sqrt{b})/2, (\sqrt{a} + \sqrt{b})/2)$ with $(a, b)$ as in (1) or (2). Then $(a - b) / 4 = \alpha\beta = \pm \alpha_0 \beta_0 = \pm 1$ so that $a - b = \pm 4$. In case (1), we have $a - b = 4\phi_k \geq 8$, because $k \geq 3$. In case (2), we have $a - b = 4\phi_k \geq 8$, because $k \neq 1$. Absurd.

**Possible source of the error:** I think that the missing $5$-defective pair is lost in the last paragraph of case $n = 5$ in section "Small $n$" of 1. It is said that:

*"By (28), we have $k \geq 3$ in the case (34), and $k \neq 1$ in the case (35)."*

But, in case (35), $k = 1$ (and $\varepsilon = 1$) are not in contradiction with (28).

In other words, (2) should allow $k = 1$ (and $\varepsilon = 1$). This would lead to the $5$-defective pair $((\sqrt{-1} + \sqrt{-5}) / 2, (\sqrt{-1} - \sqrt{-5}) / 2)$, which is equivalent to $(\alpha_0, \beta_0)$.

Thank in advance to anyone who takes the time to check.