2013 Balkan MO SL A6 : Three variables functional equation!

Here is an interesting three variable functional equation:

Let SS be the set of positive real numbers. Find all functions f:S3→Sf: S^3 \rightarrow S such that, for all positive real numbers x,y,zx,y,z and k,k, the following three conditions are satisfied:
(a) xf(x,y,z)=zf(z,y,x),xf(x,y,z)=zf(z,y,x), \\ (b) f(x,ky,k2z)=kf(x,y,z),f(x,ky,k^2z)=kf(x,y,z), \\ (c) f(1,k,k+1)=k+1.f(1,k,k+1)=k+1.

#Algebra

Note by ChengYiin Ong
5 months ago

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To avoid spoiler, I'll hide my solution in the reply section:

ChengYiin Ong - 5 months ago

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The answer is f(x,y,z)=y+y2+4xz2xf(x,y,z)=\frac{y+\sqrt{y^2+4xz}}{2x}, it is easy to see that it satisfies the three given condition. Now, we'll show that it is the only solution. First, by setting (x,y,z)→(1,t,t+1)(x,y,z) \rightarrow (1,t,t+1) in (b) where t∈St \in S, we get f(1,kt,k2(t+1))=kf(1,t,t+1)=k(t+1)f(1,kt,k^2(t+1))=kf(1,t,t+1)=k(t+1) where k∈Sk\in S. Now, if we let a=kta=kt and b=k2(t+1)b=k^2(t+1), we can express k(t+1)k(t+1) in terms of aa and bb, so k=bt+1  ⟹  a=bt+1t  ⟹  bt2−a2t−a2=0  ⟹  t=a2+a4+4a2b2b  ⟹  k=a2+4b−a2k=\sqrt{\frac{b}{t+1}} \implies a=\sqrt{\frac{b}{t+1}}t \implies bt^2-a^2t-a^2=0 \implies t=\frac{a^2+\sqrt{a^4+4a^2b}}{2b} \implies k=\frac{\sqrt{a^2+4b}-a}{2} therefore, f(1,a,b)=a+a2+4b2, ∀a,b∈S.f(1,a,b)=\frac{a+\sqrt{a^2+4b}}{2}, \ \forall a,b \in S. Then, from (a), we have f(b,a,1)=f(1,a,b)b=a+a2+4b2b.f(b,a,1)=\frac{f(1,a,b)}{b}=\frac{a+\sqrt{a^2+4b}}{2b}. Also, from (b), we have f(b,aj,j2)=jf(b,a,1)=j(a+a2+4b2b).f(b,aj,j^2)=jf(b,a,1)=j\left(\frac{a+\sqrt{a^2+4b}}{2b}\right). For the final step, let (x,y,z)=(b,aj,j2)(x,y,z)=(b,aj,j^2) where j∈Sj\in S and we get that a=yz,j=za=\frac{y}{\displaystyle\sqrt z}, j=\sqrt z, so f(x,y,z)=z(yz+y2z+4b2b)=y+y2+4xz2x, ∀x,y,z∈S.f(x,y,z)=\sqrt z\left(\frac{\frac{y}{\displaystyle\sqrt z}+\sqrt{\frac{y^2}{z}+4b}}{2b}\right)=\boxed{\frac{y+\sqrt{y^2+4xz}}{2x}, \ \forall x,y,z \in S}.

ChengYiin Ong - 5 months ago
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