Research article

On the conjecture of Je$ \acute{\textbf{s}} $manowicz

  • Received: 28 December 2022 Revised: 29 March 2023 Accepted: 03 April 2023 Published: 17 April 2023
  • MSC : 11D61, 11D75

  • Let $ k, l, m_1 $ and $ m_2 $ be positive integers and let both $ p $ and $ q $ be odd primes such that $ p^k = 2^{m_1}-a^{m_2} $ and $ q^l = 2^{m_1}+a^{m_2} $ where $ a $ is a positive integer with $ a\equiv {\pm 3}\pmod 8 $. In this paper, using only the elementary methods of factorization, congruence methods and the quadratic reciprocity law, we show that Je$ \acute{s} $manowicz' a conjecture holds for the following set of primitive Pythagorean numbers:

    $ \frac{q^{2l}-p^{2k}}{2}, p^kq^l, \frac{q^{2l}+p^{2k}}{2}. $

    We also prove that Je$ \acute{s} $manowicz' conjecture holds for non-primitive Pythagorean numbers:

    $ n\frac{q^{2l}-p^{2k}}{2}, np^kq^l, n\frac{q^{2l}+p^{2k}}{2}, $

    for any positive integer $ n $ if for $ a = a_1a_2 $ with $ a_1\equiv 1 \pmod 8 $ not a square and $ \gcd(a_1, a_2) = 1 $, then there exists a prime divisor $ P $ of $ a_2 $ such that $ \left(\frac{a_1}{P}\right) = -1 $ and $ 2|m_1, a\equiv 5 \pmod 8 $ or $ 2\not|m_2, a\equiv 3\pmod 8 $.

    Citation: Nan Fan, Jiagui Luo. On the conjecture of Je$ \acute{\textbf{s}} $manowicz[J]. AIMS Mathematics, 2023, 8(6): 14232-14252. doi: 10.3934/math.2023728

    Related Papers:

  • Let $ k, l, m_1 $ and $ m_2 $ be positive integers and let both $ p $ and $ q $ be odd primes such that $ p^k = 2^{m_1}-a^{m_2} $ and $ q^l = 2^{m_1}+a^{m_2} $ where $ a $ is a positive integer with $ a\equiv {\pm 3}\pmod 8 $. In this paper, using only the elementary methods of factorization, congruence methods and the quadratic reciprocity law, we show that Je$ \acute{s} $manowicz' a conjecture holds for the following set of primitive Pythagorean numbers:

    $ \frac{q^{2l}-p^{2k}}{2}, p^kq^l, \frac{q^{2l}+p^{2k}}{2}. $

    We also prove that Je$ \acute{s} $manowicz' conjecture holds for non-primitive Pythagorean numbers:

    $ n\frac{q^{2l}-p^{2k}}{2}, np^kq^l, n\frac{q^{2l}+p^{2k}}{2}, $

    for any positive integer $ n $ if for $ a = a_1a_2 $ with $ a_1\equiv 1 \pmod 8 $ not a square and $ \gcd(a_1, a_2) = 1 $, then there exists a prime divisor $ P $ of $ a_2 $ such that $ \left(\frac{a_1}{P}\right) = -1 $ and $ 2|m_1, a\equiv 5 \pmod 8 $ or $ 2\not|m_2, a\equiv 3\pmod 8 $.



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