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I was wondering if there exists a proof showing that there exists no Pythagorean Triple such that when its bases are swapped with the exponents the left and right hand side of the Pythagorean Triple are still equal to one another.

In other words, is there a proof showing $$2^x + 2^y = 2^z$$ for a Pythagorean Triple $$x^2 + y^2 = z^ 2?$$ I have searched everywhere for such a proof but I can't seem to find one and I didn't want to start writing my own proof in case there already exists one.

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Hint: The only way to satisfy $2^x+2^y=2^z$ with positive integers is for $x=y$.

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  • $\begingroup$ 2^2 + 2^2 = 8 = 2^3 $\endgroup$ – user25406 Apr 12 '15 at 14:30
  • $\begingroup$ But $2^2 + 2^2 \neq 3^2$, so $(2, 2, 3)$ is not a Pythagorean triple, hence there does not exist.... $\endgroup$ – Jordan Glen Apr 12 '15 at 14:32
  • $\begingroup$ If so, I was wondering if there exists such a formula that would give us the error that occurs for any Pythagorean Triple with such a property. More clearly, does there exist a formula that would give us the error for $$2^x + 2^y= 2^z$$ . $\endgroup$ – Reinhild Van Rosenú Apr 12 '15 at 14:41
  • $\begingroup$ How do you want to measure the error? $\endgroup$ – paw88789 Apr 12 '15 at 14:50
  • $\begingroup$ I want to measure the error as in the error that prevents the Pythagorean Triple from still being equal when its exponents and bases are swapped. I'm certain, logically, the formula is $$2^z - (2^x + 2^y)$$ . However, I was wondering if the errors with each succeeding Pythagorean Triple follows a particular sequence. $\endgroup$ – Reinhild Van Rosenú Apr 12 '15 at 14:52

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