Let a, b, c be pairwise relatively prime positive integers such that \(a^2 + b^4=c^2\) and b is odd. Then we show that the equation of the title has only one positive integer solution \((x, m, n)=(a, 4, 2)\) under some conditions.

中文翻译：

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在丢番图方程 $$\varvec{x^2+b^m=c^n}$$ x 2 + bm = cn 与 $$\varvec{a^2+b^4=c^2}$$ a 2 + b 4 = c 2

 设a ,  b ,  c为成对互质正整数，使得\(a^2 + b^4=c^2\)和b为奇数。然后我们证明题目的方程在某些条件下只有一个正整数解\((x, m, n)=(a, 4, 2)\)。