Show for any integer $k \geq 1$ can be uniquely expressed as $k = 2^x + i2^{x+1}$ for $i,x \geq 0$ and $i,x \in \mathbb{N}$
My attempt was to prove it inductively: $k = 1$, true assume true for $k = n$ i.e. $n = 2^x + i2^{x+1}$ then we know $k+1$ is either odd or even, say $k+1 = 2n = 2(2^x + i2^{x+1}) = 2^{x+1} + i2^{x+2}$ hence true for even $k+1$, now for odd $k+1 = 2n+1 = 2^{x+1} + i2^{x+2} + 2^0 $
I'm not really sure what to do about the "$2^0$" term and/or how to prove the uniqueness.
This is for a basic course so please avoid complex solutions if possible
thank you
Begin by writing $k=2^sm$, where $m$ is odd and $s\ge 0$. If $s=0$ then $k$ is odd, and we can uniquely write $k=1+2i$ for some $i\ge 0$, i.e. $k=2^0+i2^1$. If $s>0$ then we use our proof on odd numbers to uniquely write $m=1+2i$. Now $k=2^s(1+2i)=2^s+i2^{s+1}$. This must be unique since $s$ is the highest power of $2$ that divides $k$ (taking both sides mod $2^s$ and again mod $2^{s+1}$ shows this).