# Does this equation have any natural solutions?

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?

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

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.
• @MarcusRost If $x_1 = x_3$, then they cancel out, and the equation becomes $2^{x_2} = 2^{y}$. Can you finish from there? Sep 24, 2022 at 19:42