Can $2^k + 2^j$ be expressed as $2^n$?

Question

If we are given $j,k \geq 0, j> k$ and $j,k$ are integers, can $2^j + 2^k$ be ever expressed as $2^n$ where $n \geq 0$ and is an integer?

What I said:

Suppose it can. Then for some $0 \leq n \in \mathbb{N}: 2^j + 2^k = 2^n$. We know that $j>k$ so $j=k+s$ for some $0 < s \in \mathbb{N}$. So: $2^j + 2^k = 2^{k+s} + 2^k = 2^k \cdot 2^s + 2^k = 2^k \cdot (2^s + 1)$. We see that $2^s + 1$ can be an even number only if $s=0$ but we know that $s > 0$, thus we have that $2^j + 2^k$ which is an even number is equal to an odd number (even * odd), thus it can't be.

Is that correct?

Answer 1

An even number times an odd number is still even... but you can finish this proof by dividing both sides of your equation $2^k(2^s+1)=2^n$ by $2^k$.

Answer 2

No, $2^k + 2^j$ has two bits set while $2^n$ has only one.