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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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    $\begingroup$ $$\frac1{2^{1638}}+\frac1{2^{1638}}=\frac1{2^{1637}}$$ $\endgroup$ Aug 1, 2010 at 10:33
  • $\begingroup$ @wadim thank you for example, buti need complate solution $\endgroup$ Aug 1, 2010 at 11:22
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    $\begingroup$ I need an explanation: Why 1638? Isn't $\frac{1}{2^{42}}+\frac{1}{2^{42}}=\frac{1}{2^{41}}$ more canonical? $\endgroup$
    – Helge
    Aug 1, 2010 at 12:03
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    $\begingroup$ @Helge, that a hint: mathoverflow.net/questions/33265 . But your solution is worth canonisation as well. :-)$$\ $$ @Hashem, yes I have the feeling that you wish more. ;-) $\endgroup$ Aug 1, 2010 at 13:12
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    $\begingroup$ Wadim, Helge is correct. The Hitchhiker's Guide to the Galaxy. In the first novel and radio series, a group of hyper-intelligent pan-dimensional beings demand to learn the Ultimate Answer to the Ultimate Question of Life, The Universe, and Everything from the supercomputer, Deep Thought, specially built for this purpose. It takes Deep Thought 7½ million years to compute and check the answer, which turns out to be 42. $\endgroup$
    – Will Jagy
    Aug 2, 2010 at 3:32

2 Answers 2

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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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  • $\begingroup$ that's what you get from my construction above, if you start with $1+1=2$ :) $\endgroup$ Aug 1, 2010 at 23:18
  • $\begingroup$ True. I guess instead of calling it "another way" I should have called it the simplest instance of your way. $\endgroup$ Aug 2, 2010 at 1:20

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