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Dergiler / Turkish Journal of Mathematics / 2022 / Cilt: 46 - Sayı: 4

A parametric family of ternary purely exponential Diophantine equation $A^x + B^y = C^z$

Yasutsugu FUJITA
Maohua LE
Sayfa
1224–1232
Yayın Tarihi
2022-11-01
DOI
—
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⚡ TL;DR(1 Cümlede Hızlı Çıkarım)

In this paper, using the Baker method, we prove that if m > $max{10^8 , c^2 }$, then the equation $(am^2 + 1)x + (bm^2 − 1)^y = (cm)^z$ has only one positive integer solution (x, y, z) = (1, 1, 2).

📌 3 Maddede Bu Makale

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Amaç & Hipotez:Let a, b, c be fixed positive integers such that $a + b = c^2 , 2 ∤ c and (b/p) ̸= 1$ for every prime divisor p of c , where (b/p) is the Legendre symbol.
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Temel Bulgular & Veri:Further let m be a positive integer with m > 1.
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Sonuç & Katkı:In this paper, using the Baker method, we prove that if m > $max{10^8 , c^2 }$, then the equation $(am^2 + 1)x + (bm^2 − 1)^y = (cm)^z$ has only one positive integer solution (x, y, z) = (1, 1, 2).
🔬Tür:Araştırma Makalesi
📍Kapsam:Genişletilmiş Akademik Örneklem
⚙️Değişkenler:Alan Parametreleri

Abstract

Let a, b, c be fixed positive integers such that $a + b = c^2 , 2 ∤ c and (b/p) ̸= 1$ for every prime divisor p of c , where (b/p) is the Legendre symbol. Further let m be a positive integer with m > 1. In this paper, using the Baker method, we prove that if m > $max{10^8 , c^2 }$, then the equation $(am^2 + 1)x + (bm^2 − 1)^y = (cm)^z$ has only one positive integer solution (x, y, z) = (1, 1, 2).

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Bu Makaleye Atıf Yap

FUJITA, Y., LE, M. (2022). A parametric family of ternary purely exponential Diophantine equation $A^x + B^y = C^z$. Turkish Journal of Mathematics, Cilt: 46 - Sayı: 4, 1224–1232. https://ajindex.com/articles/604276

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Kaynakça (14)

  1. Alan M. On the exponential Diophantine equation $(18m^2 + 1)^x + (7m^2 − 1)^y = (5m)z$. Turkish Journal ofMathematics 2018
  2. 42 (4): 1990-1999. doi: 10.3906/mat-1801-76 [2] Bertók C. The complete solution of the Diophantine equation $(4m^2 + 1)^x + (5m^2 − 1)^y = (3m)^z$. Periodica Mathematica Hungararica 2016
  3. 72 (1): 37-42. doi: 10.1007/s10998-016-0111-x [3] Bugeaud Y. Linear forms in p-adic logarithms and the Diophantine equation $(x^n − 1)/(x − 1) = y^q$. Mathematical Proceedings of the Cambridge Philosophical Society 1999
  4. 127 (3): 373-381. doi: 10.1017/S0305004199003692 [4] Deng N-J, Wu D-Y, Yuan P-Z. The exponential Diophantine equation $(3am^2 − 1)^x + (a(a − 3)m^2 + 1)^y = (am)^z$. Turkish Journal of Mathematics 2019
  5. 43 (5): 2561-2567. doi: 10.3906/mat-1905-20 [5] Deng N-J, Yuan P-Z. Diophantine equation $((c + 1)m^2 + 1)^x + (cm^2 − 1)^y = (am)z$. Mathematics in Practice and Theory 2020
  6. 50 (20): 220-226 (in Chinese). [6] Fu R-Q, Yang H. On the exponential Diophantine equation $(am^2 + 1)^x + (bm^2 −1)^y = (cm)z$ with c | m. Periodica Mathematica Hungararica 2017
  7. 75 (2): 143-149. doi: 10.1007/s10998-016-0170-z [7] Kizildere E, Le M-H, Soydan G. A note on the ternary purely exponential Diophantine equation $A^x+B^y = C^z withA+B = C2$. Studia Scientiarum Mathematicarum Hungarica 2020
  8. 57 (2): 200-206. doi: 10.1556/012.2020.57.2.1457 [8] Kizildere E, Miyazaki T, Soydan G. On the Diophantine equation $((c + 1)m^2 + 1)^x + (cm^2 − 1)^y = (am)z$. Turkish Journal of Mathematics 2018
  9. 42 (4): 2690-2698. doi: 10.3906/mat-1803-14 [9] Kizildere E, Soydan G. On the Diophantine equation $(5pm^2−1)^x+(p(p−5)m^2+1)^y = (pm) z$. Honam Mathematical Journal 2020
  10. 42 (1): 139-150. doi: 10.5831/HMJ.2020.42.1.139 [10] Laurent M. Linear forms in two logarithms and interpolation determinants II. Acta Arithmetica 2008
  11. 133 (4): 325–348. doi: 10.4064/aa133-4-3 [11] Le M-H, Scott R, Styer R. A survey on the ternary purely exponential Diophantine equation $a^x + b^y = c^z$ Surveys in Mathematics and its Applications 2019
  12. Miyazaki T, Terai N. On the exponential Diophantine equation $(m^2 + 1)^x + (cm^2 − 1)^y$ = (am) z. Bulletin of the Australian Mathematical Society 2014
  13. 90 (1): 9-19. doi: 10.1017/S0004972713000956 [13] Pan X-W. A note on the exponential Diophantine equation $(am^2 + 1)^x + (bm^2 − 1)^y = (cm)^z$. Colloquium Mathematicum 2017
  14. 149 (2): 265-273. doi: 10.4064/cm6878-10-2016 [14] Scott R, Styer R. Number of solutions to ax + by = cz