2020
DOI: 10.1016/j.indag.2019.09.001
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Trinomials with given roots

Abstract: Riffaut [18] conjectured that a singular modulus of degree h ≥ 3 cannot be a root of a trinomial with rational coefficients. We show that this conjecture follows from the GRH, and obtain partial unconditional results.

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Cited by 7 publications
(8 citation statements)
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“…Using this, we deduce from (4.10) the inequality n 2/3 ≤ 10 15 d 5 (log(3d)) 3 . A quick calculation shows that this is incompatible with (4.1).…”
Section: Proof Of Theorem 42mentioning
confidence: 79%
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“…Using this, we deduce from (4.10) the inequality n 2/3 ≤ 10 15 d 5 (log(3d)) 3 . A quick calculation shows that this is incompatible with (4.1).…”
Section: Proof Of Theorem 42mentioning
confidence: 79%
“…In the set-up of Theorem 1.2, assume that, for a given m, the set of positive integers n with the property "the polynomial u n (x) is not identically 0 but vanishes at an mth root of unity" is not empty. Then the smallest n in this set satisfies n ≤ m(log m) 3 (X + log D).…”
mentioning
confidence: 99%
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“…We need a lower bound for |η a+1 − 1|. Corollary 4.2 in [6] shows that if we write D and h(η) for the degree and logarithmic height of η, respectively, then the inequality…”
Section: Proof Of Propositionmentioning
confidence: 99%
“…. , τ ( k/2 ) (n)} (6) there is an even number, then with m := τ ( k/2 ) (n), we have that m is even and Proposition 1 shows that P (τ (k) (n)) = P (τ (k− k/2 ) (m)) ≥ 3 k− k/2 −1 + 2 ≥ 3 k/2−1 + 2 > log(k/2) for k > 10, and therefore P (Orb τ (n, k)) ≥ P (m) > log(k/2). Thus, we may assume that all numbers in list L given at (6) are odd.…”
Section: Proof Of Propositionmentioning
confidence: 99%