VMO 2021 right?

lixiancvp wrote: VMO 2021 right? that right.

OK here is a proper solution. It is easy to see that the only linear solution is $f\equiv 0$. Assume $f$ is non-linear. $P(x,y):f(x)f(y)=f(xy-1)+xf(y)+yf(x)$ $P(x,0):f(x)f(0)=f(-1)+xf(0)$. Since $f$ is non-linear, $f(0)=f(-1)=0$. $P(x,-1): f(-x-1)=f(x)$ (1) $P(1,1): f(1)=0$ or $2$. If $f(1)=0$ then $P(x,1): f(x-1)+f(x)=0$. So $f(x)=-f(x-1)=-f(-x)$ from (1) Put back into (1), $f(x-1)=f(x)$. $P(x-1,y): f(x)f(y)=f(xy-y-1)+xf(y)-f(y)+yf(x)$ Compare with the original equation, $f(xy-y-1)=f(y)+f(xy-1)$ So $f(x-y)=f(y)+f(x)$. Put $y=x$ and $f(x)=0$, contradiction. Thus $f(1)=2$. $P(x,1): f(x)=f(x-1)+2x$ so $f(-x)=f(x)-2x$ $P(x,-y): f(x)(f(y)-2y)=f(xy)-yf(x)+x(f(y)-2y)$ $f(x)f(y)-2yf(x)=f(xy)-yf(x)+xf(y)-2xy$ $f(x)f(y)-f(xy)=yf(x)+xf(y)-2xy$ Let $g(x)=f(x)-x$ we have $g(x)g(y)=g(xy)$. In the original equation $g(x)g(y)=g(xy-1)+2xy+1$ So $g(x-1)+2x-1=g(x)$. So $g(x^2)+2g(x)+1=(g(x)-2x+1)(g(x)+2x+1)=g(x-1)g(x+1)=g(x^2-1)=g(x^2)+2x^2+1$. Since $g(x^2)=g(x)^2$ we have $g(x)=x^2$ and $f(x)=x^2+x$.

let $$p(x,y)=f(x)f(y)=f(xy-1)+yf(x)+xf(y)$$let $f(x)=x+g(x)$ re write as $$p(x,y)=g(x)g(y)+1=g(xy-1)+2xy$$now we clearly have $$g(x)g(y)=g(xy)g(1)$$for non_zero $x,y$ which results in either $g(0)=0$ or $g(x)=c$ the second one won't happen , so $g(0)=0$ now from $p(1,1)$ we have $g(1)^2=1$ we take cases : $g(1)=1$ then clearly $$g(xy)=g(x)g(y)$$and also $$g(x+1)=g(x)+2x+1$$as $g(1)=1$ we clearly get $g(n)=n^2$ for integers. which results in $g(x)=t(x^2)$ from here it is easy to drive $g(x)=t(x^2)=x^2$ so $$f(x)=x+x^2$$now if $g(1)=-1$ we have $$g(x)g(y)=-g(xy)$$and $$g(xy-1)+2xy=1+g(x)g(y)$$or $$g(x-1)+g(x)+2x=1$$let $g(x)=-t(x)$ hence $$t(xy)=t(x)t(y)$$and $$t(x)+t(x-1)=2x-1$$as $t(1)=1$ hence $t(x+2)=t(x)+2$ easy to drive again that $$t(x)=x$$and $g(x)=-x$ hence $f(x)=0$ and done.

Let $P(x,y)$ be the assertion $f(x)f(y)=f(xy-1)+yf(x)+xf(y)$ $P(x,0)\implies f(x)f(0)=f(-1)+xf(0)\quad\forall x\in\mathbb{R}$ So if $f(0)\neq 0$ we have $f$ must be linear which is not a solution so $f(0)=0$. $P(1,1)\implies f(1)=2\quad\text{or}\quad f(1)=0$ If $f(1)=0$ let $g(x)=f(x)-x$ so we have $$g(x)g(y)=g(xy-1)+2xy-1\quad\forall x,y\in\mathbb{R}$$$g(1)=-1$ so we have $g(x-1)=-g(x)-2x+1$ so $g(x)g(y)=-g(xy)$ Caculate $g(x^2-1)$ $$g(x^2-1)=-g(x-1)g(x+1)=-(-g(x)-2x+1)(-g(x)-2x-1)=-(g(x)+2x)^2+1$$$$g(x^2-1)=-g(x^2)-2x^2+1=g(x)^2-2x^2+1$$So $2(g(x)+x)^2=0\quad\forall x\in\mathbb{R}$ So $g(x)=-x\quad\forall x\in\mathbb{R}$ Thus $$\boxed{\text{S1}:\ f(x)=0\quad\forall x\in\mathbb{R}}$$If $f(1)=2$ Now let $g(x)=f(x)-x$ so $P(x,y)$ becomes $$g(x)g(y)=g(xy-1)+2xy-1\quad\forall x,y\in\mathbb{R}(2)$$We have $f(0)=0,f(1)=2$ so $g(0)=0$ and $g(1)=1$ Let $R(x,y)$ be the assertion of equation $(2)$ so $R(x,1)\implies g(x-1)=g(x)-2x+1\quad\forall x\in\mathbb{R}$ So $g(x)g(y)=g(xy)\quad\forall x,y\in\mathbb{R}$ Caculate $g(x^2-1)$ we have $$g(x^2-1)=g(x+1)g(x-1)=(g(x)+2x+1)(g(x)-2x+1)=(g(x)+1)^2-4x^2$$$$g(x^2-1)=g(x^2)-2x^2+1=g(x)^2-2x^2+1$$By $2$ equation we get that $g(x)=x^2\quad\forall x\in\mathbb{R}$. Thus $$\boxed{\text{S2}:\ f(x)=x^2+x\quad\forall x\in\mathbb{R}}$$

Does VMO stand for Vietnam Math Olympiad?

lahmacun wrote: Does VMO stand for Vietnam Math Olympiad? yes that's true

What is difficulty level of this contest like 2.round difficulty or 1.round.

Let $P(x,y)$ be the assertion $f(x)f(y)=f(xy-1)+yf(x)+xf(y)$. 1) $P(0,0):$ $f^{2}(0)=f(-1)$; 2) $P(-1,-1):$ $f^{2}(-1)=f(0)-2f(-1)$; 1) and 2) give us $f^{4}(0)=f(0)-2f^{2}(0)$. Assume that $f(0)\neq 0$. Then $f^{3}(0)=1-2f(0)$ (a). 3) $P(0,1):$ $f(0)f(1)=f(-1)+f(0)=$ $f^{2}(0)+f(0)\Rightarrow f(1)=f(0)+1$ 4) $P(1,1):$ $f^{2}(1)=f(0)+2f(1)\Rightarrow$ ${(f(0)+1)}^{2}=f(0)+2f(0)+2$. Then $f^{2}(0)-f(0)-1=0$ (b). Solving (a) and (b) together we can see that there is no such real value of $f(0)$. Thus $f(0)=0$. Moreover, $f(-1)=0$ and $f^{2}(1)=2f(1)$. Thus $f(1)=0$ or $f(1)=2$. $P(x,-1)$ gives us $f(-x-1)=f(x)$. So original equation can be rewritten as $f(x)f(y)=f(-xy)+yf(x)+xf(y)$ (c). Putting in (c) $-x-1$ instead of $x$ we can get $f(x)f(y)=f(xy+y)+yf(x)-xf(y)-f(y)$ (d). Let's compare (c) and (d) to get $f(-xy)=f(xy+y)-(2x+1)f(y)$. Substituting $y=\frac{1}{x}$ into the last equation we get $f(\frac{x+1}{x})=(2x+1)f(\frac{1}{x})$. Now put there $\frac{1}{-x-1}$ instead of $x$: $f(-x)=(\frac{2}{-x-1}+1)f(x)$ (e). Using (e) in (c) we can get $f(x)f(y)=(\frac{2}{-xy-1}+1)f(xy)+yf(x)+xf(y)$. Now putting there $y=1$ we have $f(x)(f(1)-2+\frac{2}{x+1})=xf(1)$. From there if $f(1)=0$ then $f(x)=0$ for every real $x$. If $f(1)=2$ then $f(x)=x^{2}+x$ which is a solution too.

\[f(x)f(y)=f(xy-1)+yf(x)+xf(y)\]Answers are $f\equiv 0$ and $f(x)=x^2+x$ which clearly work. If $f$ is constant, then $f\equiv 0$, suppose that $f$ is non-constant. $y=0$ yields $f(x)f(0)=f(-1)+xf(0)$ or $f(0)(f(x)-x)=f(-1)$ which implies $f(0)=0$. Pick $x=y=1$ to cnnclude that $f(1)^2=2f(1)$ hence $f(1)\in \{0,2\}$. Case $I$: If $f(1)=0$, plugging $y=1$ gives $0=f(x-1)+f(x)$ thus, by induction $f(n)=0$ for all integers. Choosing $y=n$ yields $f(xn-1)=-nf(x)$. Plug $xn-1,m$ to get \[0=f(xnm-m-1)+mf(xn-1)\iff f(xnm-m-1)=-mf(xn-1)=nmf(x)=f(-xmn-1)\]We can replace $x+1$ with $xmn$ which gives $f(x-m)=f(-x-2)$. By rewriting this, we get $f(x)=f(-x+k)$ for any integer $k$. Pick $k=0$ to verify that $f(x)=f(-x)$ hence $f(x)=f(x+k)$. Compare $x,y$ with $x+n,y$ \[f(xy-1)+xf(y)=f(xy+ny-1)(x+n)f(y)\iff f(xy-1)=f(xy+ny-1)+nf(y)\]By replacing $x+1$ with $xy$, we see $f(x)=f(x+ny)+nf(y)$. However, $f(x)=f(2x)+f(x)$ or $f(2x)=0$ yields $f\equiv 0$.$\square$ Case $II$: If $f(1)=2$, then let $g(x)=f(x)-x$. \[g(\frac{x}{y})g(y)=g(x-1)+2x-1\]We have $g(1)=1$. Comparing $x,y$ with $x,1$ implies $g(x)=g(\frac{x}{y})g(y)$ hence $g$ is multiplicative. Let $g(x)=h(x)+x^2$. For $y=1$, we have $g(x)=g(x-1)+2x-1$ or $h(x)=h(x-1)$. Thus, $h(x)=h(x+n)$ for each integer $n$. \[(h(x)+x^2)(h(y)+y^2)=x^2y^2\iff h(x)h(y)=y^2h(x)+x^2h(y)\]Also comparing $x+n,y$ with $x,y$ gives \[(x+n)^2h(y)=h(x+n)(h(y)-y^2)=h(x)(h(y)-y^2)=x^2h(y)\]Hence $h\equiv 0\iff g(x)=x^2\iff f(x)=x^2+x$ as desired.$\blacksquare$