Find $f(3)$ if $f(f(x))=x^{2}+2$

Question

Let $a,b,f(x),x$ be positive integers such that If $a>b$ then $f(a)>f(b)$ and $f(f(x))=x^{2}+2$ . Find $f(3)$

My approach:

Replacing $x$ with $f(x)$ in the equation gives $f(f(f(x))) = f(x)^2 + 2$, but $f(f(x)) = x^2 + 2$ so $$f(x^2+2) = f(x)^2 + 2$$ how do i proceed after this. Please help. Thanks alot!

Answer 1

Note that $f(f(1))=3$ so there is some natural number $n$ such that $f(n)=3$.

If $f(1)>3$ then there could be no solution to $f(n)=3$ so we must have $f(1)\in \{1,2,3\}$.

If $f(1)=3$ then we have $3=f(f(1))=f(3)$, a contradiction.

If $f(1)=1$ then we would have $3=f(f(1))=f(1)$, a contradiction.

Thus $f(1)=2$.

It follows that $$f(2)=f(f(1))=1^2+2=3$$ from which we deduce that $$f(3)=f(f(2))=2^2+2=6$$ and we are done.

Answer 2

Note that $f$ is by definition strictly monotonic. So consider: If $f(3)<3$ then we can only have $f(3)=1$ or $f(3)=2$. Else if $f(3)=3$ we have $f(3)=f(f(3)) = 3^2+2=11$, so this cannot be. Finally if $f(3)>3$ we have $f(3) < f(f(3))=11$.

$f(3)=1,2$ would imply $f(1)=f(f(3)) = 11>f(3)$.

Generally we get that $f(x)$ cannot be $x$, else $x=x^2+2$. Also $f(f(1)) = 3$, $f(f(2)) = 6$, $f(f(3)) = 11$. But as $f(3)>3$ (and $f(x)>3$ for $x>3$) this means $f(2)=3$.

But then $f(3) = f(f(2)) = 6$. Thus we’re done.