If $x,y,z$ are coprime, this equation has not any solution with an elementary method. I want to know a solution of this equation when $x,y,z$ are not coprime.
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Here's a way to construct a solution for each triple $(m,n,t)$ of pairwise coprime integers. Start by choosing your favorite $a,b,c$ so that $a+b=c$ and write down the prime factorizations $$a=\prod_{i=1}^k p_i^{\alpha_i} \ , \ b=\prod_{i=1}^k p_i^{\beta_i} \ , \ c=\prod_{i=1}^k p_i^{\gamma_i}.$$ Now, by the Chinese remainder theorem find large enough $\delta_i$'s so that $$\alpha_i-\delta_i\equiv 0\pmod{m}$$ $$\beta_i-\delta_i\equiv 0\pmod{n}$$ $$\gamma_i-\delta_i\equiv 0\pmod{t}$$ and denote $D=\prod_{i=1}^k p_i^{\delta_i}$. You have $$\frac{a}{D}+\frac{b}{D}=\frac{c}{D}$$ and so $x^{-m}+y^{-n}=z^{-t}$ for some integers $x,y,z$. |
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Here's another way to construct a solution for each triple $(m,n,t)$ of pairwise coprime integers. Find $u$ and $v$ such that $mnu-tv=1$ (so all that's really necessary is that $t$ be relatively prime to $mn$). Then $${1\over(2^{nu})^m}+{1\over(2^{mu})^n}={1\over(2^v)^t}$$ |
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