http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1168787770&t=556

Peter wrote: Show that there exists a bijective function $ f: \mathbb{N}_{0}\to \mathbb{N}_{0}$ such that for all $ m,n\in \mathbb{N}_{0}$: \[ f(3mn+m+n)=4f(m)f(n)+f(m)+f(n). \] So $4f(3mn+m+n)+1=(4f(m)+1)(4f(n)+1)$ Let $\mathbb{N}_{a,b}=\{n: n\equiv a\mod b\text{ and }n\in\mathbb{N}\}$ Let $\mathbb{P}_{a,b}=\{n: n\text{ is a prime number congruent to } a\mod b\}$ Let the function $g: \mathbb{N}_{1, 3}\rightarrow \mathbb{N}_{1, 4}$ such that $g(3n+1)=4f(n)+1, \forall n\in \mathbb{N}_{0}$ Then $g((3m+1)(3n+1))=g(3m+1)g(3n+1)$ Define the function $h:\mathbb{N}\rightarrow\mathbb{N}$ Let $h$ be a bijection from $\mathbb{P}_{1, 3}$ to $\mathbb{P}_{1, 4}$ Let $h$ be a bijection from $\mathbb{P}_{-1, 3}$ to $\mathbb{P}_{-1, 4}$ Let $h(xy)=h(x)h(y), \forall x, y\in\mathbb N$ Then $h$ is a bijection from $\mathbb{N}_{1, 3}$ to $\mathbb{N}_{1, 4}$ such that $h(xy)=h(x)h(y), \forall x, y\in\mathbb N_{1, 3}$ By letting $g=h$, $f$ is a bijective function. Hence there exists infinitely many bijective functions $f$.

Interesting Problem !

Let a function $g:3\mathbb{N}_0+1\to 4\mathbb{N}_0+1$ be such that $$g(x)=4f(\frac{x-1}{3})+1 ~\forall x\in 3\mathbb{N}_0+1.$$Surely $g$ is multiplicative. Now for the example of such a $g.$ Let $S_1,S_2,S_3,S_4$ be the sets of primes $3n+1,3n+2,4n+1$ and $4n+3.$ Note that these sets are infinite. Also let $\phi$ be the bijection from $S_1\cup S_2$ onto $S_3\cup S_4$ that maps $S_1$ onto $S_3$ and $S_2$ onto $S_4$ bijectively. Construction: $$g(1)=1 \text{ and } g(x)=\prod \phi(p_i), \text{where } x=\prod p_i.$$Clearly $g$ is bijective and multiplicative. Moreover, it is not hard to see that is well defined as well. So such a $g$ exists implying the claim.