2 Distinct Bijective functions

Question

I have that A is a set of $2k^2$ so it equals $\{2,8,18,32,50...\}$ How do you Construct two distinct bijections $f, g : \mathbb{Z}^{+} \to A$.

I was able to get $f(x)=2x^2$ what would $g(x)$ be?

Thank You

Answer 1

As written DKal's comment is an answer to the question. It is a function, as it assigns a unique element of the range to each element of the domain. It is a different function, as it disagrees with $f(x)$ for $x \in \{1,2\}$ Many posters (and it seems you) believe that a function has to have a "simple formula". This is simply not true. In particular, there is resistance to definition by cases. We could show DKal's example as $$g(x)=\begin {cases} 8&x=1\\1&x=2\\2x^2& \text{otherwise} \end {cases}$$ To follow up on Thomas Andrews' suggestion define $$p(x)=\begin {cases} x+1 & x \text { odd}\\ x-1 & x\text { even} \end {cases}$$ Then $x \leftrightarrow f(p(x))$ is a different bijection than $x \leftrightarrow f(x)$