I've got this equation:
$$2^{x_1}+2^{x_2}-2^{x_3}=2^y$$
Besides the trivial solutions ($x_1=x_2=x_3=y$), does it have any other natural solution for given natural y?
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$\begingroup$ $(x_1,x_2,x_2)$ works, too. $\endgroup$– Thomas AndrewsSep 24, 2022 at 18:50
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$\begingroup$ Check out our guide for new askers and fix your post. $\endgroup$– Jyrki LahtonenSep 24, 2022 at 18:51
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1$\begingroup$ I doubt it. Consider the binary representation of the LHS vs the RHS. $\endgroup$– preferred_anonSep 24, 2022 at 18:52
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$\begingroup$ in $2^a + 2^b = 2^c + 2^d$ in positive $a,b,c,d$ we demand that $b$ be the minimum. Dividing through by $2^b$ gives $2^\alpha + 1 = 2^\beta + 2^\gamma$ with $\alpha, \beta, \gamma \geq 0$ If $\alpha \neq 0$ we must have the minimum of $\beta, \gamma$ to be zero as well... $\endgroup$– Will JagySep 24, 2022 at 18:53
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$\begingroup$ By "natural solution" do you mean "solution in natural numbers"? (Non-negative integers.) $\endgroup$– Brian TungSep 24, 2022 at 18:59
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1 Answer
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Consider the binary representation of the LHS vs the RHS. Suppose $x_1,x_2,x_3$ are all different. If $x_3$ is biggest, your LHS is negative, so there are no solutions. Wherever else $x_3$ appears, subtracting $2^{x_3}$ will not decrease the number of $1$s in the binary expansion. Indeed if it's $k$ steps below $x_1$ but above $x_2$ (say), the number of $1$s increases by $k-1$ (by carrying). So you have at least two $1$s, but the RHS has one (contradiction).
So some of them are equal. By symmetry, $x_1 = x_3$ and $x_2 = x_3$ are the same case; they give a unique solution for $y$ (the non-cancelling exponent).
If $x_1=x_2$ then you have the equation $2^{x_1 + 1} - 2^{x_3} = 2^y$. By the logic above $x_3$ and $x_1+1$ must be adjacent - so $x_3 = x_1$, and you have your family of trivial solutions.
answered Sep 24, 2022 at 19:04
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$\begingroup$ So you say that there's only the trivial solution? $\endgroup$ Sep 24, 2022 at 19:22
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$\begingroup$ @MarcusRost What part of the answer are you struggling to understand? $\endgroup$ Sep 24, 2022 at 19:34
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$\begingroup$ Now I understand. Let me pose another question. If x1=x3, or x2=x3 what are the unique solutions then? $\endgroup$ Sep 24, 2022 at 19:40
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$\begingroup$ @MarcusRost If $x_1 = x_3$, then they cancel out, and the equation becomes $2^{x_2} = 2^{y}$. Can you finish from there? $\endgroup$ Sep 24, 2022 at 19:42
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$\begingroup$ Ahhhh. Now I got it. Thanks. It's simple. I just thought the wrong way around :D $\endgroup$ Sep 25, 2022 at 10:13