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IMO 2019 Problem 1

July 18, 2019 Leave a comment Go to comments

Problem 1. Let {\Bbb{Z}} be the set of integers. Determine all functions {f:\Bbb Z \rightarrow \Bbb Z} such that, for all integers {a} and {b},

\displaystyle f(2a)+2f(b) = f(f(a+b)).

Solution: As usual with this kind of functional equations the first thing that comes into mind is to pick simple cases of {a} and {b}.

For {a = 0} we have {f(f(b)) = 2f(b)+f(0)} and for {b=0} we have {f(f(a))=f(2a)+2f(0)}. In the end, this gives {f(2a) = 2f(a)-f(0)}. Using the formulas for {f(f(a))} and {f(2a)} found above, the initial equation becomes

\displaystyle 2f(a)-f(0)+2f(b) = 2f(a+b)+f(0)

which amounts to

\displaystyle f(a)+f(b) = f(a+b)+f(0).

Picking {g(x) = f(x)-f(0)} we have

\displaystyle g(a)+g(b) = g(a+b)

and since {g: \Bbb{Z} \rightarrow \Bbb{Z}} it is classical that {g(x) = mx} for some {m \in \Bbb{Z}}. In the end we find that {f(x) = mx+n} with {m,n \in \Bbb{Z}}.

Plugging this into the initial equation we find that {m^2=2m} and {3n = mn+n}. This quickly gives the solutions {m=0,n=0} and {m=2,n\in \Bbb{Z}}.

Categories: Algebra, IMO, Olympiad Tags: , ,
  1. Soham Roy
    August 13, 2019 at 11:43 am
    Reply

    What are these kind of function series finding problems called?

    • beni22sof
      August 14, 2019 at 3:30 pm
      Reply

      If I remember correctly such problems are called functional equations.

  2. CH
    November 18, 2020 at 4:14 am
    Reply

    Hi Beni, I am new to IMO and I have a query for this solution. When I try to solve this question, the first thing I can think of is the “continuity” of this function f. As the question asks us to find ALL the functions f which can satisfy for all integers a and b, will there be different sets of function f when a and b are of particular values? Say for (a,b) = (0,0), there may be a function f1 that works, but this function f1 may not work for (a,b) = (1,1)? With this hestiation in mind, I am not comfortable to mix equations getting from putting different particular values of a and b, because there may be a chance that the function f is indeed function f1 (for a1,b1 set) and function f2 (for a2, b2 set).

    I have browsed around the web and found that my query was not being asked. Is it too obvious to ignore it or may be I am wrong to think so in the first place? Really hope that you can give me an insight!

    Thanks
    CH

    • beni22sof
      November 18, 2020 at 7:51 am
      Reply

      Since the functional equation is valid for all integers a, b any function should satisfy all equations coming from all the possible values you may give to a and b.

      I hole this answers your question.

      • CH
        November 18, 2020 at 9:02 am

        Thanks so much for your reply!!

        I know it is difficult for me to explain myself but please allow me. From my understanding of the condition, this does not necessarily mean that you can put any figures of a and b into it and do some operations, mainly because we do not know if f(x) is discrete or not. What we know about function f is that its takes integers and produce integers.

        Now a and b are like parameters to the determine which function f we are talking about. That means we have lots of different f, depending the values of a and b. And more importantly, these different f may also fulfill the given functional equation and our task is to find them all out.

        For example if the function is something like this:
        for (a=0 and b=0), f(x) = 0
        for (a=0 and 0<b=1), f(x) =100
        for (a=/=0, and any b), f (x)=-100

        the above set of f(x) satisfy the requirement of function f, which is ” its takes integers and produce integers”. but you cannot use the results for a=0 to put into operations with cases that a does not equal to 0. You can do so only because you have assumed that the function is not “discrete” (something like I made up above), right?

        My query is that why it looks like that we can ignore the cases that I presented above, the discrete function f?

        Thank you very much and any feedback is very much welcomed!

      • beni22sof
        November 18, 2020 at 12:18 pm

        It is not clear to me why the example you give is a function of one variable. You seem to assign different values for different variables a and b, which means that you are in fact considering two variables.

  1. November 17, 2022 at 2:50 pm
    ≫ Pregunta 1 de la OMI de 2019: tenga en cuenta sus decisiones

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