Find $x$ and $y$ in $2^{x-y} + 1 = 2^x,$ where $x,y$ are integers

Question

I have no idea what to do now.

Is there any way to find the integers $x$ and $y$ by factoring?

Thank you.

Answer 1

Dividing by $2^x$ gives $$ 2^{-x}+2^{-y}=1 $$ Since $2^{x}\gt0$ for any $x$, we must have both $2^{-x}\lt1$ and $2^{-y}\lt1$; that is, $x\gt0$ and $y\gt0$. Since $x$ and $y$ are integers, the biggest that $2^{-x}$ or $2^{-y}$ can be is $\frac12$ and since their sum is $1$, they must both be $\frac12$. That means $x=y=1$.

Answer 2

Hint: If they are integers, Which powers of two are 1 apart?

$2^x>1$ so $x > 0$

if $x<y$ then $2^{x-y}\leq\frac{1}{2}$ so $1<2^x \leq \frac32$ and this is impossible.

So $x \geq y$ and $2^{x-y}$ and $2^x$ are integer powers of $2$ and one apart.

So $x-y=0$ and $x=1$, therefore $y=1$.

Answer 3

Hint If either of $x$ or $x-y$ is negative, multiplying by $2^z$ where $-z$ is the whichever of $x$ and $x-y$ is closer to $- \infty$, we get and equation of the type

$$2^a \pm 1 =2^b \,.$$

Then, as $a \geq 0, b\geq 0$ one of $2^{a}$ or $2^{b}$ has to be odd.

Answer 4

Hint: What powers of two are odd? There is only one.

This should tell you what $x - y$ is and from that you can work out what $x$ and $y$ are.

We know that both $x$ and $y$ are integers and $2^x \gt 1$ so $2^x$ must be an integer because the only integer powers of 2 that are not integers are less than 1 so $2^{x-y}$ must also be an integer.

Answer 5

As an alternative (using factoring), note that $2^{x-y}=\frac{2^x}{2^y},$ so multiplying both sides by the non-zero number $2^y$ yields the equivalent equation $$2^x+2^y=2^x2^y,$$ which in turn yields $$0=2^x2^y-2^x-2^y\\0=2^x(2^y-1)-2^y\\0=2^x(2^y-1)-2^y+1-1\\0=2^x(2^y-1)-(2^y-1)-1\\0=(2^x-1)(2^y-1)-1\\1=(2^x-1)(2^y-1)$$

Note that we cannot have $x=0$ or $y=0$. (Why?) Also $x$ and $y$ must have the same sign, for if not, then $(2^x-1)(2^y-1)$ is negative. If $x,y$ are negative, then $(2^x-1)(2^y-1)$ is positive, but less than $1,$ so we must have $x,y$ positive.

Now, for any positive integers $x,y,$ we have $x,y\ge 1$ (since $x,y$ integers), so $2^x,2^y\ge 2,$ so $2^x-1,2^y-1\ge 1,$ and so $(2^x-1)(2^y-1)\ge 1.$ Equality holds in the last case precisely when it holds in the first case--that is, precisely when $x,y=1.$

Answer 6

Multiply with 2^y on both the sides of the equation.

You will get the equation as :-

2^x + 2^y = 2^x * 2^y

Hence on factoring you will get the equation as

(2^x - 1) (2^y - 1) = 1

I think that should help !