Nmd_22 wrote: 4. Let $\mathbb{N}$ denote the strictly positive integers. A function $f$ : $\mathbb{N}$ $\to$ $\mathbb{N}$ has the following properties which hold for all $n \in$ $\mathbb{N}$: a) $f(n)$ < $f(n+1)$; b) $f(f(f(n)))$ = 4$n$ Find $f(2022)$. a) implies $f(n)\ge n$ b) implies $f(n)\ne n$ and so $f(n)\ge n+1$ b) also implies $f(4n)=4f(n)$ 1) $f(n)=n+1$ $\forall n\in[1,3]$ $f(f(f(1)))=4$ and $f(n)\ge n+1$ imply $f(1)=2$, $f(2)=3$ and $f(3)=4$ Q.E.D. 2) $f(n)=n+4^n$ $\forall n\in[4^n,3\times 4^n]$ If $f(n)=n+c_n$ $\forall n\in[a_n,b_n]$, then $f(4a_n)=4f(a_n)=4a_n+4c_n$ $f(4b_n)=4f(b_n)=4b_n+4c_n$ And since $f(4b_n)-f(4a_n)=4b_n-4a_n$, we get $f(n)=n+4c_n$ $\forall n\in[4a_n,4b_n]$ Hence the conclusion (using starting situation in 1) above) Q.E.D. 3) $f(n)=4n-2\times 4^{n+1}$ $\forall n\in[3\times 4^n,4^{n+1}]$ Let $n\in[4^n,2\times 4^n]$ $f(n)=n+4^n$ (using 2) above And since $n+4^n\in[4^n,3\times 4^n]$, $f(f(n))=f(n+4^n)=n+2\times 4^n$ And so $f(n+2\times 4^n)=f(f(f(n)))=4n$ $\forall n\in[4^n,2\times 4^n]$ Which is $f(n)=4(n-2\times 4^n)$ $\forall n\in[3\times 4^n,4^{n+1}]$ Q.E.D. 4) $\boxed{f(2022)=3046}$ $2022\in[4^5,3\times 4^5]$ and so, using 2), $f(2022)=2022+4^5=3046$ Q.E.D.