I only have a soln for a (I guess b was the tough part): f(g(f(x)))=f(x^3)=(f(x))^2, so f(0), f(1), f(-1) are all roots of the eqn x^2-x=0. Then at least 2 of them must be equal. Let's assume f(0)=f(1). This means that g(f(0))=g(f(1)) so 0=1, false. The same happens if f(0)=f(-1) or f(1)=f(-1) because g(f(x))=x^3.

For a) very easy f(x) = x ², g(x) =x f(g(x)) = f(x) =x ² g(f(x))=g(x ² )=x ²

But we should have g(f(x)) = x, moubi !

I think you got that wrong: in b) it's f(g(x))=x^2 and g(f(x))=x^4. Maybe the fact that it was written g(f(x))=(x^2)^2 confused you?

Sorry for a) i didn't see the exponent

Just an idea (not the proof and the complete solutions) I found in my drawers If x > 1the following functions f(x) =2^(2^sqrt(lnx/ln2)) g(x) = 2^((lnx/ln2)^2) verify the conditions on the interval ]1;+ \infty [, i hope.

For b) Partial answer without proof Sorry the exponent was not at the right place for x>1 f(x) = 2^(2.sqtr(lnx/ln2)) g(x) = 2^((lnx/ln2)^2) I hope now ot works

They satisfy the given relations indeed, but how did you come across them ? Adn what about ]-oo, 1[ ?

I found this question in my note, but i don't understand the proof some line missing It was 12 years ago (when i was student) We have to find a clear nice proof. Let's continous to think about it.

I believe that for x>=1 f(x) = 2^(2.sqtr(lnx/ln2)) g(x) = 2^((lnx/ln2)^2) for 0<x<1 f(x) = 2^(-2.sqtr(-lnx/ln2)) g(x) = 2^(-(lnx/ln2)^2) for x=0 f(0)=g(0)=0 for x<0 f(x)=f(-x) g(x)=g(-x) Check me!

And how did you find them ?

i have the solution of this again.... not by me of coz, again, troublesome to type it myth 's correct

So, a solution, anyone ? Post it please ...

There is no solution, there is an answer

I don't know if it will work but from (1) f(g(x)) = x^2 (2) g(f(x)) = x^4, the functions x^2 & x^4 are C^(oo) , (derivable as much as we want). I wonder if from (1), (2) we can get a differentials equations involving only f and an other one with g, and solve them to get f,g

yes there is an answer but I remember thinking about that problem one year ago for half of hour and I just couldn't figure out the function ... so I guess that the question people are asking is just how in the world did you come up with those (not so easy to come up with) functions ?

From (1) and (2) f(x^4)=f^2(x). Let h(x)=f(e^x). Then h(4x)=h^2(x). Let H(x)=ln h(x). Then H(4x)=2H(x) - and what the simplest function satisfies with equation? Of course \sqrt x, and so we try H(x)=c*\sqrt x ==> h(x)=e^(c\sqrt x) ==> f(x)=e^(c\sqrt ln x). By the same reasons g(x)=e^(c'ln^2(x)). If we find c and c' we find f and g.

For the equation f(g(x)) =xx, let y=g(x), then f(y)=xx. Consider x=2^2^u, y=2^2^v. Then 2^2^v =g(2^2^u) => v=log_2 log_2 g(2^2^u) and f(2^2^v) = 2^2^(u+1) => u+1 = log_2 log_2 f(2^2^v). Let G(u) = log_2 log_2 g(2^2^u) and F(v) = log_2 log_2 f(2^2^v), then v=G(u), u+1=F(v). Hence u+1=F(G(u)). SImilarly, g(f(x))=x^4 leads to v=F(u), u+2=G(v), hence u+2=G(F(u)). To get such F and G, we try F(u)=au+b, G(u)=cu+d. Then u+1=a(cu+d)+b=acu+ad+b and u+z=c(au+b)+d=cau+bc+d. Hence we want ac=1, ad+b=1, bc+d=2, taking a=1/2, c=2, b=1, d=0, we get F(u)=u/2 +1, G(u)=2u satisfy F(G(u))=u+1, G(F(u))=u+2. Then g(x)=2^2^G(u)=2^2^2u=2^(2^u)^2=2^(log_2 x)^2 and f(y)=2^2^F(v) = 2^2^(v/2 +1) = 2^[2(2^v)^(1/2)]=2^[2 \sqrt log_2 y] Notice the f function is defined only if log_2 y>=0 or y>=1 In that case, we check f(g(x)=xx and g(f(x)=x^4. To make this work for all x E R, we now extend f,g to the real number as follow: define f(x) =2^[2 \sqrt log_2 y] if x>=1 =1/f(1/x) if 0<x<1 =0 if x=0 =f(|x|) if x<0 and g(x)=2^(log_2 x)^2 if x>=1 =1/g(1/x) if 0<x<1 =0 if x=0 =g(|x|) if x<0 And this function satisfies the conditions

moubi wrote: I found this question in my note, but i don't understand the proof some line missing It was 12 years ago (when i was student) That's strange ! This question was proposed for IMO 1997, and now you're saying you saw it twelve years ago

not all questions proposed at the IMO are original !! for example the following question was used at a selection test in 2000 (Romania) and was submitted on the jury in '99 Romania. prove that there are an infinity of positive integers a,b,c,d such that a^3+b^3 | c^3 + d^3. fortunately, the olympiad committe being formed by Romanians (that IMO was in Romania) they noticed the same problem could be found in an old (1984) Romanian book. so it is indeed possible that Moubi solved that problem 12 years ago !

The question you're talking about was a bit different Valentin : Prove that for each positive n, there exist positive integers a, b, c and d such that n = (a + b)/(c + d). Am I mistaken or is this very tricky and difficult ? I don't feel comfortable with the official solution. "Your version" of the above problem is ... well ... hmm, original at least, but a bit trivial

yeah you are right Arne - although I don't agree with the problem being very difficult - the official answer was found by many Romanian contestants that year (almost all 6 who made it to the team) so i guess it's not that hard to come. However, it is possible, although not probable, that some of the people who solved back then had knowledge of the shortlist 1999. About the problem, hmm indeed you are right - i've just created a new problem .... let's solve it ... maybe put some more conditions like {0,1} \cap {a,b,c,d} = \emptyset

You know, I'm just an unexperienced Belgian guy ... so for me this is far from easy. Indeed a was exaggerating a bit when I said this is "very" difficult. But it's tricky and I don't like it (Did you solve it Valentin ?) About your problem : Putting c = ka, d = kb we get an infinite family of solutions. But I guess this problem can't be solved properly. There are too much possibilities, you've got four variables !

hmmm ... nope i didn't solve it back then - I was too busy preparing to go to Korea for the StarCraft BroodWar BattleNet Finals - so I didn't trained properly for the IMO TST's

Huh ? What's the StarCraft ... ?

StarCraft BroodWar the greatest strategy game ever - created by Blizzard. It's a RTS game ...

b) Let g(x)= x^lnx if x is bigger than or equal to 1 g(x)= x^(-lnx) if x is bigger than 0 and smaller than 1 g(x)= (-x)^ln(-x) if x is smaller than or equal to -1 g(x)= (-x)^(-ln(-x)) if x is bigger than -1 and smaller than 0 g(x)=0 if x=0 and f(x)=y^2 ,where y is such that g(y)=x or g(-y)=x. these functions f and g are satisfying the given condition(it can proved by let y=f(x)^(1/2)...) I like this problem it is not very hard but it is nice

$f,g :R \rightarrow R$ үүд нь $f(g(x)) =x^2$ ба $g(f(x))=x^k$ байх функцууд байг. а) Хэрэв $k=3$ бол ийм функцууд олдох уу? б) Хэрэв $k=4$ бол яах вэ?

Old-Classics Revisited: a) $f(g(x)) = x^2$ and $g(f(x)) = x^3$ Note that: $f$ is injective function: $f(a)=f(b) \implies g(f(a)) = g(f(b)) \implies a^3 = b^3 \implies a=b$. Now notice that: expanding $f(g(f(x)))$ in 2 different ways gives us: $f(x^3)=f(x)^2$. Plugging $x=0, 1, -1$, we must have each of the numbers $f(0), f(1), f(-1)$ to equal its square. Due to PHP, two amongst them must be equal, which contradicts $f$ is injective. No such functions exists. b) $f(g(x)) = x^2$ and $g(f(x)) = x^4$ First consider the problem over $[1, \infty)$. Notice that, we have: $f(x)^2=f(x^4)$. Plugging $f(x) = a^{h(log_b(x))}$ gives us: $h(4t)=2h(t)$. Thus: $h(x) = c \sqrt{x}$. Since we have two independent variables $a, c$ for function $f$, we let $c=1$ (the value of $a$ would adjust itself for $f(x)$). Thus: $f(x) = a^{\sqrt{\log_b(x)}}$. Similarly, $g(x) = c^{(\log_d x)^2}$. Plugging back gives: $a=2, b=2, c=16, d=2$. Now, we will extend over real line. Basically, for $x \in (0, 1)$, we let $f(x) = \frac{1}{f(1/x)}$ and $g(x) = \frac{1}{g(1/x)}$. Further, for $x \in (-\infty, 0]$, let $f(-x)=f(x)$ and $g(-x)=g(x)$. Therefore, such functions exists.