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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!

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    $\begingroup$ Since $f$ is increasing set $f(1)=a$ then $f(a)=3$ there are only $1,2$ values below. can you infer some upper bound for $a$ ? $\endgroup$
    – zwim
    Nov 8, 2021 at 19:50
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    $\begingroup$ For what it's worth, the problem is poorly written. One of the premises is that if $x \in \Bbb{Z^+}$, then $f(x) \in \Bbb{Z^+}.$ This premise is not obvious from the problem's presentation - i.e. the statement "let ... $f(x), x$, be positive integers". $\endgroup$ Nov 8, 2021 at 19:55
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    $\begingroup$ I'm not seeing how you made the observation $f(x^2+2)=(f(x))^2+2$.... $\endgroup$
    – Mike
    Nov 8, 2021 at 19:56
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    $\begingroup$ The answer of lulu is conclusive. Another way of summarizing the answer of lulu is that $f(1)$ must be a positive integer, $f(1)$ must be $> 1$, and $f(1)$ must be $< 3$. $\endgroup$ Nov 8, 2021 at 19:57
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    $\begingroup$ @Mike, since $f(f(f(x)))=f(x)^{2}+2$ but $f(f(x))=x^{2}+2$ so $f(x^{2}+2)=f(x)^{2}+2$ $\endgroup$ Nov 8, 2021 at 19:58

2 Answers 2

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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.

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  • $\begingroup$ @CalvinLin Good point. Let me see if I can repair that... $\endgroup$
    – lulu
    Nov 8, 2021 at 23:43
  • $\begingroup$ @CalvinLin I've rewritten to avoid that substitution. Thanks for pointing out the gap. $\endgroup$
    – lulu
    Nov 8, 2021 at 23:46
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    $\begingroup$ @CalvinLin Another gap should be observed as well...All we've done with this sort of reasoning is to show what the value must be assuming $f(x)$ exists. I assume that this is what the OP intended, and I didn't think about constructing an actual solution to the functional equation. $\endgroup$
    – lulu
    Nov 8, 2021 at 23:48
  • $\begingroup$ BrianMoehring pointed out that the substitution is actually valid. I misunderstood what OP did (in part because the details weren't listed out). $\endgroup$
    – Calvin Lin
    Nov 9, 2021 at 19:16
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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.

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  • $\begingroup$ Thank you so muchhhh!! $\endgroup$ Nov 8, 2021 at 20:04
  • $\begingroup$ @BrianMoehring Could be? You appear to know better, so it’s probably that way. $\endgroup$
    – Lazy
    Nov 8, 2021 at 20:17
  • $\begingroup$ I've been surprised about regional uses of words before (e.g. "positive"), so I tend not to assume. $\endgroup$ Nov 8, 2021 at 20:19
  • $\begingroup$ @BrianMoehring My problem here is that I’m not a native english speaker, so I’m not used to using each possible term in english. In german we just say monoton. $\endgroup$
    – Lazy
    Nov 8, 2021 at 20:23
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    $\begingroup$ @BrianTung I believe that in French, positive means non-negative, EG 0 is both positive and negative. I stand with Moehring on this "regional uses of words". $\endgroup$
    – Calvin Lin
    Nov 8, 2021 at 23:51

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