Goormaghtigh equation

TheoremGoormaghtigh solutions with \(\gcd(m - 1, n - 1) > 1\) and \(m \le 50\)#●ready to prove

Let \(x, y, m, n\) be natural numbers with \(2 \le x < y\), \(n > 2\), \(m \le 50\) and \(\gcd(m - 1, n - 1) > 1\), and suppose

\[N = \sum_{i=0}^{m-1} x^i = \sum_{i=0}^{n-1} y^i,\]

that is, \(N = (x^m - 1)/(x - 1) = (y^n - 1)/(y - 1)\). Then \(N = 31\) or \(N = 8191\).

The paper prints the side conditions of its Equation (1) as \(y > x \ge 2\) and \(n > m > 2\). Since \(y > x\), equal repunits force \(m > n\) (the paper's own solution \(31 = 1 + 2 + 4 + 8 + 16 = 1 + 5 + 25\) has \(m = 5 > n = 3\)), so the printed \(n > m\) is read as \(m > n\); that inequality follows from the other hypotheses and is not assumed. The paper quotes this result as part of Theorem 4 of Bennett, Garbuz and Marten (2020).

public theorem_wanted goormaghtigh_eq_of_gcd_sub_one_gt_one_of_le_fifty
    (x y m n : ℕ) (hx : 2 ≤ x) (hxy : x < y) (hn : 2 < n)
    (hgcd : 1 < Nat.gcd (m - 1) (n - 1)) (hm : m ≤ 50)
    (heq : ∑ i ∈ Finset.range m, x ^ i = ∑ i ∈ Finset.range n, y ^ i) :
    ∑ i ∈ Finset.range m, x ^ i = 31 ∨ ∑ i ∈ Finset.range m, x ^ i = 8191

Sources

• Jon Grantham, No New Goormaghtigh Primes up to 10^700, INTEGERS 24 (2024), Article A98, Theorem 1.

TheoremNo Goormaghtigh primes below \(10^{700}\) besides \(31\) and \(8191\)#●ready to prove

Let \(p\) be a prime with \(p < 10^{700}\), and let \(x, y, m, n\) be natural numbers with \(2 \le x < y\) and \(n \ge 3\) such that

\[p = \sum_{i=0}^{m-1} x^i = \sum_{i=0}^{n-1} y^i,\]

that is, \(p\) is a repunit of length \(m\) in base \(x\) and of length \(n\) in base \(y\). Then \(p = 31\) or \(p = 8191\).

public theorem_wanted goormaghtigh_prime_lt_ten_pow_seven_hundred
    (p x y m n : ℕ) (hp : p.Prime) (hplt : p < 10 ^ 700) (hx : 2 ≤ x) (hxy : x < y)
    (hn : 3 ≤ n) (hpx : p = ∑ i ∈ Finset.range m, x ^ i)
    (hpy : p = ∑ i ∈ Finset.range n, y ^ i) :
    p = 31 ∨ p = 8191

Sources

• Jon Grantham, No New Goormaghtigh Primes up to 10^700, INTEGERS 24 (2024), Article A98, Section 2, discussion following Theorem 1.

TheoremUnder abc, finitely many Goormaghtigh numbers have both lengths at least five#●ready to prove

Assume the abc conjecture in the following form: for every real \(\varepsilon > 0\), there are only finitely many triples \((a, b, c)\) of positive integers with \(a + b = c\), \(\gcd(a, b) = 1\) and \(\operatorname{rad}(abc)^{1 + \varepsilon} \le c\), where \(\operatorname{rad}(k)\) is the product of the distinct primes dividing \(k\). Then there are only finitely many natural numbers \(N\) for which there exist natural numbers \(x, y, m, n\) with \(2 \le x < y\) and \(n \ge 5\) such that

\[N = \sum_{i=0}^{m-1} x^i = \sum_{i=0}^{n-1} y^i.\]

Such an \(N\) is a repunit of length \(m\) in base \(x\) and of length \(n \ge 5\) in base \(y\); since \(x < y\), equality forces \(m > n\), so both lengths are at least five. The paper states the abc conjecture for relatively prime integers; the positive-triple form used here is the standard equivalent one.

public theorem_wanted goormaghtigh_numbers_length_ge_five_finite_of_abc
    (habc : ∀ ε : ℝ, 0 < ε →
      {t : ℕ × ℕ × ℕ | 0 < t.1 ∧ 0 < t.2.1 ∧ t.1 + t.2.1 = t.2.2 ∧ Nat.Coprime t.1 t.2.1 ∧
        ((∏ q ∈ (t.1 * t.2.1 * t.2.2).primeFactors, (q : ℝ)) ^ (1 + ε) ≤ (t.2.2 : ℝ))}.Finite) :
    {N : ℕ | ∃ x y m n : ℕ, 2 ≤ x ∧ x < y ∧ 5 ≤ n ∧
      N = ∑ i ∈ Finset.range m, x ^ i ∧ N = ∑ i ∈ Finset.range n, y ^ i}.Finite

Sources

• Jon Grantham, No New Goormaghtigh Primes up to 10^700, INTEGERS 24 (2024), Article A98, Theorem 2.

TheoremUnder abc, there are only finitely many Goormaghtigh primes#●ready to prove

Assume the abc conjecture in the following form: for every real \(\varepsilon > 0\), there are only finitely many triples \((a, b, c)\) of positive integers with \(a + b = c\), \(\gcd(a, b) = 1\) and \(\operatorname{rad}(abc)^{1 + \varepsilon} \le c\), where \(\operatorname{rad}(k)\) is the product of the distinct primes dividing \(k\). Then there are only finitely many primes \(p\) for which there exist natural numbers \(x, y, m, n\) with \(2 \le x < y\) and \(n \ge 3\) such that

\[p = \sum_{i=0}^{m-1} x^i = \sum_{i=0}^{n-1} y^i.\]

public theorem_wanted goormaghtigh_primes_finite_of_abc
    (habc : ∀ ε : ℝ, 0 < ε →
      {t : ℕ × ℕ × ℕ | 0 < t.1 ∧ 0 < t.2.1 ∧ t.1 + t.2.1 = t.2.2 ∧ Nat.Coprime t.1 t.2.1 ∧
        ((∏ q ∈ (t.1 * t.2.1 * t.2.2).primeFactors, (q : ℝ)) ^ (1 + ε) ≤ (t.2.2 : ℝ))}.Finite) :
    {p : ℕ | p.Prime ∧ ∃ x y m n : ℕ, 2 ≤ x ∧ x < y ∧ 3 ≤ n ∧
      p = ∑ i ∈ Finset.range m, x ^ i ∧ p = ∑ i ∈ Finset.range n, y ^ i}.Finite

Sources

• Jon Grantham, No New Goormaghtigh Primes up to 10^700, INTEGERS 24 (2024), Article A98, Corollary 1.

4 ready to prove

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