I've got this equation:


Besides the trivial solutions ($x_1=x_2=x_3=y$), does it have any other natural solution for given natural y?

  • $\begingroup$ $(x_1,x_2,x_2)$ works, too. $\endgroup$ Sep 24, 2022 at 18:50
  • $\begingroup$ Check out our guide for new askers and fix your post. $\endgroup$ Sep 24, 2022 at 18:51
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    $\begingroup$ I doubt it. Consider the binary representation of the LHS vs the RHS. $\endgroup$ Sep 24, 2022 at 18:52
  • $\begingroup$ 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 Jagy
    Sep 24, 2022 at 18:53
  • $\begingroup$ By "natural solution" do you mean "solution in natural numbers"? (Non-negative integers.) $\endgroup$
    – Brian Tung
    Sep 24, 2022 at 18:59

1 Answer 1


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.

  • $\begingroup$ So you say that there's only the trivial solution? $\endgroup$ Sep 24, 2022 at 19:22
  • $\begingroup$ @MarcusRost What part of the answer are you struggling to understand? $\endgroup$ Sep 24, 2022 at 19:34
  • $\begingroup$ 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
  • $\begingroup$ @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
  • $\begingroup$ Ahhhh. Now I got it. Thanks. It's simple. I just thought the wrong way around :D $\endgroup$ Sep 25, 2022 at 10:13

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