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If $f(f(x)) = x^2 + 2$, then find $f(11)$? Given that if $a>b$ then $f(a)>f(b)$

I got this question from a study group of which I am part of. There the question was described as Let $x,f(x),a,b$ be positive integers and if $a>b$ then $f(a)>f(b)$ and $f(f(x)) = x^2 + 2$ then what is $f(11)$?

I tried by substituting $x= 1$ and $3$ and got $f(f(1)) = 3$ and $f(f(3))=11$ but don't know how to proceed further.

David G. Stork
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Siddarth Singh
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  • What have you Tried ? Show us your thoughts . (Hint :- try to put specific values of $x$ , how about putting $x = 11$) ? – Anonymous Sep 22 '20 at 05:24
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    I must say... granted the OP hasn't shown work, but does this problem really deserve to be downvoted? I think it is a very interesting problem. And I tried inserting $x = 11$, but it didn't help much. $x=3$ seemed to be more promising, but then too... – David G. Stork Sep 22 '20 at 05:59
  • I upvoted to partially reverse downvote, because I agree with David Stork's comment. I favor (instead) verbally encouraging the OP to edit his query. – user2661923 Sep 22 '20 at 06:07
  • First of all, what is the background of the question. That is, is this question from a book, class, contest, or where? If from a book or class, what pertinent theorems or recent solved problems do you think might be pertinent here? Please edit your query to provide background. ...see next comment. – user2661923 Sep 22 '20 at 06:10
  • This site isn't about "interesting problems" (there are lots of those around the www), it's about useful questions provoking helpful answers. –  Sep 22 '20 at 06:12
  • I also favor Anonymous' comment. Please try to make up any function that satisfies the constraints, and (for example) evaluate $f(x)$ for $x \in {0,1,2,3,4}.$ Does any pattern present itself? As you are working, please periodically edit your query to (also) show your work. Personally, I regard 30 minutes to an hour as a reasonable investment in exploration time. – user2661923 Sep 22 '20 at 06:12
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    @user2661923: That's the first thing I tried, and found $n = \frac{\log (2)}{W(\log (2))}$ where $W$ is the ProductLog. OK... so now what? My "hint" to you: Don't give "hints" unless you're sure your hint leads to a solution. Are you sure in this way? If so I suggest you post your full answer and save us all a lot of time. – David G. Stork Sep 22 '20 at 06:36
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    @DavidG.Stork You are right - I have deleted my comment. In fact, I wasn't sure that it would lead anywhere, but I was simply desparate to help. To repeat, I agree, I should not have given that hint. – user2661923 Sep 22 '20 at 06:38
  • The question can't be answered because domain and range of $f$ are not clear. If $f$ has to be defined for all integers (or even all reals), there is no solution at all. If it's a function from natural numbers to natural numbers (including 0 or not) existence isn't trivial, but not hard to show. There's no unique solution, but $f(11)$ is uniquely defined. Not defining the domain alone makes the question useless. –  Sep 22 '20 at 06:42
  • @ProfessorVector: So if we assume $f: \mathbb{Z} \to \mathbb{Z}$, then $f(11)$ is uniquely defined? Wow... very cool. I think the OP (or you) should edit the question to clarify, so we can go on to find $f(11)$. I, for one, find this a useful question... in that a particular value is uniquely defined, even if the function is not. Your answer was... in your terms... "helpful," even though you haven't given the full logical justification for your answer. – David G. Stork Sep 22 '20 at 06:48
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    @ProfessorVector very interesting comment. What do you mean "...makes the question useless"? That is, assuming that you are right, can't the question be meaningful as long as it can be shown that $f(11)$ is uniquely defined. Why does the question become useless just because $f(x)$ itself is not uniquely defined? – user2661923 Sep 22 '20 at 06:48
  • Reading is a very useful capability: there is no such function defined on all integers. So the answer depends on that missing information, and that makes the question useless. As long as the OP doesn't add some context, I won't give any "justification". –  Sep 22 '20 at 06:54
  • If I interpret the OP's context correctly, it looks like $f(1)$ has to be 2, and $f(2)$ has to be 3. However, even if this is true, I have no idea how to use this insight. – user2661923 Sep 22 '20 at 07:07
  • If you don't mind dropping the implied constraint that $f:\mathbb{Z}\rightarrow\mathbb{Z}$, then a monotone increasing half-iterate $h:\mathbb{R}^+\rightarrow\mathbb{R}^+$ (functional-square-root) exists in the form of infinitely nested radicals such that $h(h(3))=11$ and $h(h(x))=x^2+2$ for $x>0$, and I've written about it here https://mathematica.stackexchange.com/a/230577/72682 . – flinty Sep 22 '20 at 22:19

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So the function has to be defined for positive integers. Then, we must have $f(x)>x$, because $f(x)<x$ would imply $f(f(x)<f(x)<x$, but $x^2+2>x$, and $f(x)=x$ is equally impossible. Replacing $x$ by $f(x)$ shows that $f(f(f(x))=f(x)^2+2=f(x^2+2)$. Since $x<f(x)<f(f(x))=x^2+2$, we must have $1<f(1)<1^2+2=3$, i.e. $f(1)=2$. Then, $f(2)=f(f(1))=1^2+2=3$, $f(3)=f(f(2))=2^2+2=6$ and $f(11)=f(3^2+2)=f(3)^2+2=38$.

  • I upvoted, very nice. I couldn't quite make the leap that f(3) = f(f(2)) = 6. Also, I must separately compliment your comments, which I found to be stylish. In fact, I am (also) forced to admit that there have been times in my life when it either was or would have been useful to be able to read. – user2661923 Sep 22 '20 at 07:17
  • f(1)=2 is not convincing unless it's given that f is integer valued!! – Lawrence Mano Sep 22 '20 at 23:33
  • @Lawrence Mano As is often the case, reading the question helps: "Let $x,f(x),a,b$ be positive integers..." –  Sep 23 '20 at 05:57
  • Oh yes. Thank you. Your solution is 100% correct and elegant. Thank you once again for the answer. – Lawrence Mano Sep 23 '20 at 11:51