Internat. Math Olympiad exercise. Weird function.

AI Thread Summary
The discussion revolves around a problem from the International Math Olympiad involving a function defined by the equation f(x + xy + f(y)) = (f(x) + 1/2)(f(x) + 1/2). Participants express confusion regarding the function's domain and codomain, initially thought to be both R, which leads to contradictions in the problem's statements. There is a consensus that critical information may be missing, such as continuity of the function. The quadratic formula is applied incorrectly, leading to further misunderstandings about the function's behavior. Clarification on the function's properties and correct interpretation of the problem is sought to proceed with solving for f.
GBarboza
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Homework Statement



A friend of mine tried to classify for the IMO a few days ago (he didn't do so well). A problem he had to solve was:

f(x + xy + f(y)) = (f(x) + 1/2)(f(x) + 1/2)

I didn't really understand what he said later. First he told me to find the values of X and Y for which this function makes sense, but later I asked him if it said anything more and he told me the domain and codomain were both R, which is a bit weird. As far as I know, the domain is all possible values of X.

Homework Equations



I probably misunderstood. I would appreciate it if anyone could tell me what it probably was, and, even better, tell me how to proceed.

The Attempt at a Solution



I got as far as f(x) = -1/2 +- f(x + xy + f(y)) using the quadratic formula... which checks out. My dad says I'm missing information. I don't really understand all this much.
 
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In the problem, you are given an identity satisfied by f, and your goal is to solve for f.


I do believe you have misstated the problem, and suspect you have left out information. (e.g. it is not uncommon to include that f is continuous)

I think you have misstated, because there is a contradiction: if you substitute
x = -f(1)
y = 1​
then you get
f(-f(1)) = (f(-f(1)) + 1/2)(f(-f(1)) + 1/2)​
However, this contradicts your statement that f is a function from R to R, because the equation t = (t + 1/2)(t + 1/2) has no real solutions.

Also, the problem you stated contradicts what you said you got via the quadratic formula.
 
Thanks for replying.

Right, I'm pretty sure that the function being from R to R is the part that I got wrong. He told me that later in the day(or, at least, that's what I managed to make of the three or so words he said); earlier he had told me to find the possible values of x, which is what I can't do. :/

Oh, and, oops, I meant f(x) = -1/2 +- sq( f( x + xy + f(y) ) ).
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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