Let $ n $ be a positive integer with $ n > 1 $ and let $ a, b $ be fixed coprime positive integers with $ \min\{a, b\} > 2 $. In this paper, using the Baker method, we proved that, for any $ n $, if $ a > \max\{15064b, b^{3/2}\} $, then the equation $ (an)^x+(bn)^y = ((a+b)n)^z $ has no positive integer solutions $ (x, y, z) $ with $ x > z > y $. Further, let $ A, B $ be coprime positive integers with $ \min\{A, B\} > 1 $ and $ 2|B $. Combining the above conclusion with some existing results, we deduced that, for any $ n $, if $ (a, b) = (A^2, B^2), A > \max\{123B, B^{3/2}\} $ and $ B\equiv 2\pmod 4 $, then this equation has only the positive integer solution $ (x, y, z) = (1, 1, 1). $ Thus, we proved that the conjecture proposed by Yuan and Han is true for this case.
Citation: Yongzhong Hu. An application of the Baker method to a new conjecture on exponential Diophantine equations[J]. Electronic Research Archive, 2024, 32(3): 1618-1623. doi: 10.3934/era.2024073
Let $ n $ be a positive integer with $ n > 1 $ and let $ a, b $ be fixed coprime positive integers with $ \min\{a, b\} > 2 $. In this paper, using the Baker method, we proved that, for any $ n $, if $ a > \max\{15064b, b^{3/2}\} $, then the equation $ (an)^x+(bn)^y = ((a+b)n)^z $ has no positive integer solutions $ (x, y, z) $ with $ x > z > y $. Further, let $ A, B $ be coprime positive integers with $ \min\{A, B\} > 1 $ and $ 2|B $. Combining the above conclusion with some existing results, we deduced that, for any $ n $, if $ (a, b) = (A^2, B^2), A > \max\{123B, B^{3/2}\} $ and $ B\equiv 2\pmod 4 $, then this equation has only the positive integer solution $ (x, y, z) = (1, 1, 1). $ Thus, we proved that the conjecture proposed by Yuan and Han is true for this case.
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Yongzhong Hu. An application of the Baker method to a new conjecture on exponential Diophantine equations[J]. Electronic Research Archive, 2024, 32(3): 1618-1623. doi: 10.3934/era.2024073
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