Solutions for $2^x+2^y=2^z$ for integers x, y and z

Question

I'm trying to prove/disprove that the equation

$2^x+2^y=2^z$

where $x$, $y$ and $z$ are positive integers, only have trivial solutions. The obvious case where this is true would be for $x=y$ but I'm not certain if any solutions exits for $x\ne y$

It feels like there are no other solutions but I'm not sure how one would formally prove it. Anyone have any tips on how to do this?

Sorry if this is a trivial question / known problem.

Possible duplicate of Can $2^k + 2^j$ be expressed as $2^n$? and also Solving $2^{x-z} + 2^{y-z} = 1$ — Sil, Apr 28 '19 at 20:14
Basically if $x\neq y$ there is simple argument: $2^x+2^y$ has two $1$'s in its binary representation, while $2^z$ only has one $1$, so they cannot be equal. — Sil, Apr 28 '19 at 20:32

score 6 · Accepted Answer · answered Apr 27 '19 at 21:22

If $x\neq y,$ then we may assume that $x<y.$ Of course, since $2^y>0$, we have that $z>x.$ Then, we can cancel the power $2^x$ in the equation and get $1+2^{y-x}=2^{z-x}.$ This is a contradiction, because the inequalities $y>x$ and $z>x$ ensure that $2^{y-x}$ and $2^{z-x}$ are even, which in turn implies that the left hand side of the new equation is odd, while the right hand side is even.

score 1 · Answer 2 · answered Apr 27 '19 at 21:23

1

Since the equation is symmetrix in $x,y$ I can suppose that $x\geq y$. So I can do this : $$2^y(2^{x-y}+1)=2^z$$ This means that $(2^{x-y}+1)$ has to be a power of $2$ and this clearly happens only if $x=y$ :)

answered Apr 27 '19 at 21:23

Kandinskij

3,709

score 0 · Answer 3 · answered Apr 27 '19 at 21:24

Hint: Assume that $x\neq y$. Without loss of generality we assume that $x>y$ then we can express $x = y + \bar{x}$ in which $\bar{x}>0$.

Then we obtain:

$$2^{y+\bar{x}}+2^{y}=2^z$$ $$2^{\bar{x}}+1=2^{z-y}$$

The right-hand side is even the left-hand side is odd. Hence we have a contradiction.

Solutions for $2^x+2^y=2^z$ for integers x, y and z

3 Answers3