# Possible small mistake in Bilu-Hanrot-Voutier paper on primitive divisors of Lehmer sequences (?)

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.