I saw that it's a function on two variables, so abandoned it. Never looked at it again. Never.

This is wrong for sure, but I can't find any mistake. I'm surely missing something obvious, can someone point it out ? We will construct a function satisfying both the condition but $f(e,e) = e$, where $e$ denote the euler's constant. Note that $\log(e-1)$ and $\log(e) = 1$ are linearly indepent over $\mathbb{Q}$ since the former is irrational but the latter isn't. Write $\mathbb{R}$ as a $\mathbb{Q}$ vector space with one of the basis being $1$ and $\log(e-1)$. Let $E$ be the set of basis vectors, and for $x \in E$, let $g(x) = 0$ if $x = \log(e-1)$, or $g(x) = x$ otherwise. Now for $y = \sum_{e_i \in E, a_i \in \mathbb{Q}} a_ie_i$, let $g(y) = \sum_{x \in E} a_i g(e_i)$. Observe $g$ is additive. Now define $f(x) = \begin{cases} f(x,y) = e^{g(\log |x| )} & \text{ if } y = e \\ 1 & \text{ otherwise } \end{cases}$, which I think works.

Kayak wrote: This is wrong for sure, but I can't find any mistake. I'm surely missing something obvious, can someone point it out ? We will construct a function satisfying both the condition but $f(e,e) = e$, where $e$ denote the euler's constant. Note that $\log(e-1)$ and $\log(e) = 1$ are linearly indepent over $\mathbb{Q}$ since the former is irrational but the latter isn't. Write $\mathbb{R}$ as a $\mathbb{Q}$ vector space with one of the basis being $1$ and $\log(e-1)$. Let $E$ be the set of basis vectors, and for $x \in E$, let $g(x) = 0$ if $x \in \log(e-1)$, or $g(x) = x$ otherwise. Now for $y = \sum_{e_i \in E, a_i \in \mathbb{Q}} a_ie_i$, let $g(y) = \sum_{x \in E} a_i g(e_i)$. Observe $g$ is multiplicative. Now define $f(x) = \begin{cases} f(x,y) = e^{g(\log |x| )} & \text{ if } y = e \\ 1 & \text{ otherwise } \end{cases}$, which I think works. Can someone solve it in an easier way?

I just got $f((1-x)^k,x)=1$ and left (which is most likely false) and $f(1,x)=1$.

TheDarkPrince wrote: I just got $f((1-x)^k,x)=1$ and left (which is most likely false). Come on, you tried! I didn't. So be happy.

TheDarkPrince wrote: I just got $f((1-x)^k,x)=1$ and left (which is most likely false). I just solved till $f(1,x) = 1$. And $f(\frac{1}{2},\frac{1}{2}) =1$

TheDarkPrince wrote: I just got $f((1-x)^k,x)=1$ and left (which is most likely false) and $f(1,x)=1$. I got something similar(something to the power of n, x =1) but couldn't get anything after that

Math-wiz wrote: Kayak wrote: This is wrong for sure, but I can't find any mistake. I'm surely missing something obvious, can someone point it out ? We will construct a function satisfying both the condition but $f(e,e) = e$, where $e$ denote the euler's constant. Note that $\log(e-1)$ and $\log(e) = 1$ are linearly indepent over $\mathbb{Q}$ since the former is irrational but the latter isn't. Write $\mathbb{R}$ as a $\mathbb{Q}$ vector space with one of the basis being $1$ and $\log(e-1)$. Let $E$ be the set of basis vectors, and for $x \in E$, let $g(x) = 0$ if $x \in \log(e-1)$, or $g(x) = x$ otherwise. Now for $y = \sum_{e_i \in E, a_i \in \mathbb{Q}} a_ie_i$, let $g(y) = \sum_{x \in E} a_i g(e_i)$. Observe $g$ is multiplicative. Now define $f(x) = \begin{cases} f(x,y) = e^{g(\log |x| )} & \text{ if } y = e \\ 1 & \text{ otherwise } \end{cases}$, which I think works. Can someone solve it in an easier way? There are several issues with this. (1) $\mathbb{R}$ over $\mathbb{Q}$ is an infinite dimension vector space. You cannot "construct" a basis for it, merely show existence of one. Showing that there is a basis with $1$ and $\ln(e-1)$ might be non-trivial(or impossible). (2) $g(q\ln(e-1))\ne q g(\ln(e-1))$ for $q\in\mathbb{Q}$ and $q\ne 1$. So $g$ is not linear. Hence you cannot write $g(y)=\sum_{x \in E} a_i g(e_i)$.

TheDarkPrince wrote: I just got $f((1-x)^k,x)=1$ and left (which is most likely false) and $f(1,x)=1$. I too got that, didn't write it though in the answer sheet, ... Should I cry?

TheDarkPrince wrote: I just got $f((1-x)^k,x)=1$ and left (which is most likely false) and $f(1,x)=1$. Its true I wrote a huge algebraic equation where it's difficult to spot out an error (I know where the error is) and proved the claim (last 1 min)

TheDarkPrince wrote: I just got $f((1-x)^k,x)=1$ and left (which is most likely false) and $f(1,x)=1$. How does it help to solve the problem?

TheDarkPrince wrote: I just got $f((1-x)^k,x)=1$ and left (which is most likely false) and $f(1,x)=1$. Isn't that quite obvious

I claimed that f is a constant if I am wrong somewhere plz disprove my claim.

Hait wrote: I claimed that f is a constant if I am wrong somewhere plz disprove my claim. Proof ?

I only showed that f(x(1-z)to the power n,z) is constant for all naturals n and fixed z.

Here's some heuristic (and also my thought process in the exam) elling either this is wrong or very hard: A. Condition (ii) tells the ONLY information we get about f(_, z), that is f(1-z, z) = 1. And that's it, the values of f(_, a) and f(_, b) don't talk to each other. B. When you get that info from (ii), the ONLY info you know from (i) is that f(1-z,z) = 1, and the function g(x) := f(x, z) is multiplicative. And nothing else. And you can pin no other value of f, now the only way you can f(x, z) is that you can constrict x as some multiplicative combination of (1-x) and +-1, which is not guaranteed, from where you suspect either the problem is wrong... or you messed up heavily.

Now take cases where z-1 has absolute value less than 1 and greater than 1. And the power n holds for negatives integers. So x(1-z)power n can be made to tend to 0 by taking very large n.

Thus f(x,z )is constant for fixed z.

And f(1,z)=1.

shaxm wrote: Important announcement from the guy who decides whos fit for mathematics and whos not. Also that’s false. HBCSE only decides who is eligible for IMOTC and IMO. If you want to pursue Mathematics it’s your own choice.

p_square wrote: Maybe $f(x, yz) = f(x, y) f(x, z) $ That's exactly the conditon HBCSE had missed. Sorcery

Solution:

Claim 1

Claim 2

Both claims are trivial.

Claim 3

Proof

Now, from claim 2 we have $$f(x,\frac{1}{1-x})=1$$for $x \neq 1$. Call this $P(x)$. Now, $P(\frac{1}{1-x})$ gives $$f(\frac{1}{1-x},\frac{1-x}{-x})=1$$But we also have $f(\frac{1}{1-x},x)=1$ due to claim 2. Hence $f(\frac{1}{1-x},x-1)=1$ and from claim 2, $$f(1-x,x-1)=1=f(y,-y)$$for $y \neq 1$, but it is trivial for $y=1$. Similiarly, $P(\frac{1}{1+x})$ gives $$f(\frac{1}{1+x},\frac{1+x}{x})=1$$But we also have $f(\frac{1}{1+x},x)=1$ due to claims 2 and 3. Hence $f(\frac{1}{1+x},x+1)=1$ and from claim 2, $$f(1+x,x+1)=1=f(z,z)$$for $z \neq 1$, but it is trivial for $z=1$. Hence part (a) is proved. For part (b): $$f(x,y)f(y,x)=f(x,y)f(x,x)f(y,x)f(y,y)$$$$=f(x,xy)f(y,xy)=f(xy,xy)=1$$

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If HBCSE was a youtube channel, then it would have been ripped down the middle due to copy strike.

p_square wrote: Maybe $f(x, yz) = f(x, y) f(x, z) $. Does that give anything? It might have been $f(x, y) f(1-y,x) = 1$. All we can do is guess.. Legend.

C'mon let this thread die. Let that bad memory die. p_square wrote: Maybe $f(x, yz) = f(x, y) f(x, z) $

BTW this thread has some nice problems (including this ) https://mathoverflow.net/a/84018

Actually, in the paper I didn't even notice that the symmetric condition wasn't given and assuming it solved a great deal FE noob. God knows why but for some reason I was always under the impression that 2 variable functions are always symmetric. FE gods noticed the asymmetry and first thing that came to their mind after some jiggery pokery was continuity.

This question is in ILL 1977

One liner part 1 $f(x,y)=f(x,y)f(x,1)\implies f(x,1)=1\implies 1=f(x,1)=f(x,-1)^2\implies f(x,-1)=1\implies f(x,x)=f(x,-1)f(x,-x)=f(x,-x)$.

Jupiter_is_BIG wrote: One liner part 1 $f(x,y)=f(x,y)f(x,1)\implies f(x,1)=1\implies 1=f(x,1)=f(x,-1)^2\implies f(x,-1)=1\implies f(x,x)=f(x,-1)f(x,-x)=f(x,-x)$. Let me show you what real one liner is- $[f(x,x)]^2=f(x,x^2)=[f(x,-x)]^2\implies f(x,x)=f(x,-x)$

I finally overcame my irritation on this problem and solved it after one year. div5252 wrote: Let $f$ be a function defined from $((x,y) : x,y$ real, $xy\ne 0)$ to the set of all positive real numbers such that $ (i) f(xy,z)= f(x,z)\cdot f(y,z)$ for all $x,y \ne 0$ $ (ii) f(x,yz)= f(x,y)\cdot f(x,z)$ for all $x,y \ne 0$ $ (iii) f(x,1-x) = 1 $ for all $x \ne 0,1$ Prove that $ (a) f(x,x) = f(x,-x) = 1$ for all $x \ne 0$ $(b) f(x,y)\cdot f(y,x) = 1 $ for all $x,y \ne 0$ Solution- First note that $f(x,y)=f(x,-y)=f(-x,y)=f(-x,-y)$ as all are equal to fourth root of $f(x^2,y^2)$ Note that $f(x,1)=f(1,x)=1 \forall x$ .Now let $x\in\mathbb{R}-\{0,1\}$ and let $z=\frac{1}{x}$ . Then we have $f(z,1-z)f(z,\frac 1{1-z})=f(z,1)=1 \implies f(z,\frac 1{1-z})=1$ and $f(z,\frac 1{1-z})f(\frac 1z,\frac 1{1-z})=f(1,\frac 1{1-z})=1 \implies f(\frac 1z,\frac 1{1-z})=1\implies f(\frac 1z,\frac 1{z-1})=1$ So $f(x,\frac{x}{1-x})=1$ But then $f(x,x)=f(x,\frac{x}{1-x})f(x,1-x)=1$ Now part (b) is easy as $f(x,y)f(y,x)=f(x,y)f(x,x)f(y,y)f(y,x)=f(x,xy)f(y,xy)=f(xy,xy)=1$ Hence proved.

Note that $f(a,b)=f(a,b).f(a,1)$. Which implies that $f(a,1)=1$ by similar argument $f(1,a)=1$. $f(a,1)=f(a,-1).f(a,-1)=1$ which forces to $f(a,-1)=1$ by similar argument$f(-1,a)=1$. Now ,$f(x,x)=f(x,-x).f(x,-1)=f(x,-x)$ so we get $f(x,x)=f(x,-x)$. Note that $f(x,x)=f(x,\frac{x(x-1)}{(x-1)} )$. $=f(x,\frac{x}{x-1}) .f(x,x-1)=f(x,\frac{x}{x-1}).f(x,1-x).f(x,-1)$. $=f(x,\frac{x}{x-1}).f(x,x)$ .So since $f(x,x)\in R^+$. Hence $f(x,\frac{x}{x-1})=1$. Equivalently $f(x,\frac{x}{1-x})=1$. Put $\frac{1}{s}$ in the above relation get , $f(\frac{1}{s},\frac{s}{1-s} ) =f(\frac{1}{s},\frac{1}{1-s} ) .f(\frac{1}{s},s)\cdots (*)$ Again $f(s,\frac{1}{s-1}) .f(\frac{1}{s},\frac{1}{1-s} )=1$. So we have $f(\frac{1}{s},\frac{1}{s-1})=1$. From $(*)$ we get $f(s,\frac{1}{s} )=1$ . Finally $f(s,\frac{1}{s} ).f(s,s)=f(s,s)$. Which forces $f(s,s)=f(s,1)=1$. So we get $f(x,x)=f(x,-x)=1$. $\underline{Part(ii)}$. $f(x,y).f(y,x)= f(x,x).f(y,y).f(y,x).f(x,y)$ $=f(y,xy).f(x,xy )=f(xy,xy)=1$.

here, the solution of the very similar 1977 LL.