Please don't post the problems now. Post them after 48 hours. @above Ik that. But let people try the problems first.

bump bump

from $ P(x, \frac{x-1}{x})$ we have $ f(f(x)+\frac{x-1}{x})=1$ . define $A=[\frac{x}{y} | x>y , f(x)=f(y)]$ now if $f(a)=f(b)$ we have from $P(a,y),P(b,y)$ that $f(ay+1)=f(by+1)$ so if $z$ is a element of $A$ we have $ f(y+1)=f(zy+1)$ clerly $z>1$ and $f(y+1)=f(z^n y+1)$ so we have $\frac{z^ny+1}{y+1}$ is also in$A$ by letting $y=0$ and after that $lim_n$ to infinity , we get any $x>1$ is in $A$ . so $f=1$, now this is done . if $f$ is injective then if $f(c)=1$ we have $f(x)=c-1 + \frac{1}{x}$ now by puting it in the original problem we have $c=1$ and $f(x)=\frac{1}{x}$

here is a simillar problem . https://artofproblemsolving.com/community/q1h1299711p16925269

Mr.C wrote: by letting $y=0$ and after that $lim_n$ to infinity , we get any $x>1$ is in $A$ . so $f=1$, now this is done $A=(1,\infty)$ doesn’t immediately imply that $f$ is constant. A few more steps are needed. EDIT: apparently only a few substitutions are enough. Sorry about that. DschinghisKhan wrote: Find all $f:R^+ \rightarrow R^+$ such that $f(f(x) + y)f(x) = f(xy + 1)\ \ \forall x, y \in R^+$ $P(2,\frac12)$ implies that there exists a real number $a$ such that $f(a)=1$. If there exists such $a$ that is not one, then there exists $b\in\{a,\frac1a\}$ such that $b>1$. $P(a,x)$ would implies that $f(x+1)=f(bx+1)$ for all $x>0$. Comparing $P(bx+1,y)$ with $P(x,y)$ then gives $f(bxy+y+1)=f(xy+y+1)$. By calculation, we can see that for any two positive real numbers $m,n$ such that $m+1>n>m>\frac{b}{b-1}$, there exist positive $x,y$ such that $bxy+y+1=n$ and $xy+y+1=m$. Thus, $f$ must be eventually constant. Plugging in large $y$ for each $x$ then gives $f(x)=1$ for all $x$. Else, $f^{-1}(\{1\})=\{1\}$. $P(x,\frac{x-1}{x})$ gives $f(x)=\frac1x$ for all $x>1$. $P(x,1)$ then implies that $f(x)=\frac1x$ for all $x$.

DschinghisKhan wrote: Find all $f:R^+ \rightarrow R^+$ such that $f(f(x) + y)f(x) = f(xy + 1)\ \ \forall x, y \in R^+$ @below: https://artofproblemsolving.com/community/c6h2254883_2020_imoc_problems Quote: Feel free to start individual threads for the problems as usual

This problem was given at the Bulgarian 2021 NMO finals, it was problem 3. I commented on it in my blog.

Wrong, please ignore

duongbgbg32 wrote: Lemma 1. If $x>1,f(x)<1$ and If $x<1,f(x)>1$ If $P(x)=const$, it is so easy. Let $P(x) \ne const$ $P(x, \frac{f(x)-1}{x-1})$, We have: $f(x)=1$ $(><)$ So we done! This lemma is false. The point is, you cannot put $ y=\frac{f(x)-1}{x-1}$ unless $f(x)\neq 1$ because the initial condition is not defined when $y=0,$ it is defined only for $x>0,y>0.$ So, what you have proven is actually: Lemma 1. $f(x)\le1,\forall x> 1$ and $f(x)\ge1, \forall x<1.$ which is Observation 2 in the link I provided. So, the whole point is that it's possible for some $x\ne 1$ to have $f(x)=1.$ What should be proved is that if for some $x_0\ne 1$ we have $f(x_0)=1$ then $f$ is the constant $1.$ But this requires new fresh ideas! @Nofancyname: The same issue with your solution.

let $p(x,y):f(f(x)+y)f(x)=f(xy+1)$. first we prove that $f(x)$ is decrasing.suppose there exist $a$ ,$b$ such that $a>b$ and $f(a)>f(b)$. if $\frac{af(a)-bf(b)}{a-b}=c$ and $\frac{f(a)-f(b)}{a-b}=t$ we have $c,t>0$ and know $$bt+f(a)=at+f(b)=c$$now $$p(a,bt):f(f(a)+bt)f(a)=f(abt+1)$$and $$p(b,at):f(f(b)+at)f(b)=f(abt+1)$$so we have $$f(c)f(a)=f(c)f(b)=f(abt+1)$$so $f(a)=f(b)$ and its contradiction and $f$ is decrasing. suppose there exist $a>1$ such that $f(a)=1$, $p(a,x):f(x+1)=f(ax+1)$ and we have $f(x+1)=f(a^nx+1)$. if letting $x$ fix and $lim_n$ to infinity we have $f(k)=f(\infty)$ and for decrasing have $\forall x> 1: f(x)=k$ and for $x,y>1:p(x,y):k^2=k$ and $k=1$ now if $y>1$ we have $f(x)=1$ for all $x$ and we are done. suppose there exist $a>1$ such that $f(a)=1$, $p(a,x):f(x+1)=f(ax+1)$ and we have $f(x+1)=f(a^nx+1)$. if letting $x$ big and fix and $lim_n$ to infinity we have $\forall x> 1: f(x)=k$ and again we are done. so if $f(a)=1$ have $a=1$. $p(x,\frac{x-1}{x}):f(f(x)+1-\frac{1}{x})=1$ so $f(x)+1-\frac{1}{x}=1$ and $f(x)=\frac{1}{x}$.

It is obvious that solutions $f(x)=\frac{1}{x}$ and $f(x)=1$ satisfy the conditions. The only constant solution is $f \equiv 1$, so assume that $f$ is not constant from now on. I will prove $f(x)=1 \Leftrightarrow x=1$. Denote the given expression by $P(x, y)$. For $x>1$, $P( x, \frac{x-1}{x})$ gives there is $x$ such that $f(x)=1$. Assume for some $\alpha \ne1$, $f(\alpha)=1$. $P(\alpha, x) \Rightarrow f(x+1)=f(\alpha x+1) \Rightarrow f(x+1)=f(\alpha^kx+1)$ for all nonnegative integer $k$s. (1) Let $f(x_0)>1$ for some $x_0>1$. Take $y$ such that $f(x_0)+y=x_0y+1$. $P(x_0, y)$ gives a clear contradiction $\Rightarrow$, if $x_0>1$, then $f(x_0) \leq 1$. (2) Similarly let $f(x_0)<1$ for some $x_0<1$. Take $y$ such that $f(x_0)+y=x_0y+1$. $P(x_0, y)$ gives the contradiction $\Rightarrow$, if $x_0<1$, then $f(x_0) \geq 1$.(3) Assume that $\alpha >1$. $P(x, \frac{\alpha-1}{x}) \Rightarrow f(x)f(f(x)+ \frac{\alpha-1}{x})=1$ If $a-1=1+ \epsilon>1$, then take $x \in (1, 1+\epsilon)$. We can conclude that $x$ and $f(x)+ \frac{\alpha-1}{x}$ are both grater than $1$, so $f(x)=1$ for $x \in (1, 1+\epsilon)$ and $f(x)+\frac{\alpha-1}{x}$ will go to values in the interval $(2, 2+\epsilon)$ and their $f$s are equal to $1$ as well because of (2). We know that $f((1, 1+ \epsilon))=1 \Rightarrow f(x+1)=f(t^kx+1)$ for $t\in (1, 1+\epsilon)$ and $x \in (1, 1+\epsilon) \subset \mathbb{R^+}$ (1) Also since $f((2, 2+\epsilon))=1 \Rightarrow f((2, 1+(1+\epsilon)^k) \Rightarrow f((2, \infty))=1$. Now taking $y$ large enough in $P(x, y)$, we have $f$ gives $1$ for all $x$ which is the solution we found earlier. If $\alpha >1$, take $x=\alpha-1$ and $k$ large enough to have an $x_0 >2$ such that $f(x_0)=1$. Contradiction. $ f(x)=1 \Rightarrow x \leq1$. $P(x, \frac{x-1}{x})$ for $x>1$ gives $f(x)+\frac{x-1}{x} \leq 1 \Rightarrow f(x) \leq \frac{1}{x}$ for $x>1$. Assume $\alpha <1$ now. 1 gives $f(x+1)=f(\alpha^kx+1)$. Take $x=\alpha^{-k}$ and $k$ large enough to have $f(2)=f(\alpha^{-k}+1) \leq \frac{1}{\alpha^{-k}+1}$ which can be made small enough. Absurd. $f(x)=1 \Leftrightarrow x=1$. Because of $P(x, \frac{x-1}{x}$, we conclude that $f(x)=\frac{1}{x}$ for $x>1$ and the rest can be done by taking $y$ large enough in the original equation.

For $x>1$, taking $y=\frac{x-1}{x}$ gives $f\left(f(x)+\frac{x-1}{x}\right)=1$. Now this implies $\exists \alpha$ such that $f(\alpha)=1$. If the only real satisying that it's image is $1$ is $1$, then it is easy to conclude $f(x)=\frac{1}{x} $ $\forall x\in\mathbb{R}^+$, which is a solution. Now assume the contrary, that there is some $\alpha \neq 1$ such that $f(\alpha)=1$.Taking $x=\alpha \implies f(y+1)=f(\alpha y+1) \implies f(y+1)=f(\alpha^ny+1)$ for $n$ natural. As $\alpha\neq 1$, taking the limit $n\to+\infty$, this implies that for all $N\in\mathbb{R}$ there exist $a,b\in\mathbb{R}^+$ such that $f(a)=f(b)$ and $\frac{b}{a}>N$. Knowing this, we'll show that for any $1<c_1<c_2$, then $f(c_1)=f(c_2)$. Substitute in the original equation $x=a$, $y=y_1$, and then $x=b$, $y=y_1$, giving $f(ay_1+1)=f(by_1+1)$. Then substitute $x=ay_1+1$, $y=y_2$, and $x=by_1+1$, $y=y_2$, to get that $f(ay_1y_2+y_2+1)=f(by_1y_2+y_2+1)$. We want to show we have a solution for positive reals to $ay_1y_2+y_2+1=c_1$ and $by_1y_2+y_2+1=c_2$. Notice that we need that $y_2=\frac{bc_1-ac_2}{b-a}-1>0$, which is equivalent to $\frac{b}{a}>\frac{c_2-1}{c_1-1}$, so taking $N=\frac{c_2-1}{c_1-1}$ such $a,b$ exist. And now $y_1=\frac{c_1-1-y_2}{ay_2}$, so we only need $c_1>1+y_2 \iff c_1<c_2$ after expanding, which is true by assumption. Thus $f(x)$ is constant for all $x>1$, and from here it is easy to conclude that $f(x)=1$ is true for all $x\in\mathbb{R}^+$.

Now, because I don' t like f:R+->R+, I will call this equation P(x, y) for f:(0, ∞)->(0, ∞) Claim 1: f is strecly decreasing Proof: P(x, (f(x)-1)/x-1)) and in this situation f(x) want to be over 1 This give us f(f(x)+(f(x)-1)/x-1). f(x)=f(x.(f(x)-1)/x-1 + 1) f(f(x).x-1/x-1).f(x)=f(f(x).x-1/x-1) Therefore: f(x)=1 for all x, which is one solution Now, x can be selected such that x>1, therefore for f(x)≠1-must to be f(x)<1 Now, f(x.y+1)=f(x).f(f(x)+y)<f(x), because f(f(x)+y)<1 Therefore, if y->(x+y-1)/x, i.e f(x+y)<f(x) and we are ready Claim 2: f(x) is infectivity Proof: for claim 1 this is clearly true, because decreasing Claim 3: f(x)=f(1)/x Proof: P(x, (x+y-1/x)) f(f(x)+(x+y-1)/x).f(x)=f(x+y) Now: f(f(x)+(x+y-1)/x).f(x)=f(f(x)+(x+y-1)/x +x+y-f(x)-(x+y-1)/x)=A.f(f(x)+(x+y-1)/x)=f(f(f(x)+(x+y-1)/x)+(f(x)+(x+y-1)/x +x+y-f(x)-(x+y-1)/x -1)/(f(x)+(x+y-1)/x)).f(f(x)+(x+y-1)/x) After we remove f(f(x)+(x+y-1)/x) Therefore: f(x)=A By infectivity x=f(f(x)+(x+y-1)/x)+(x+y-1)/f(x)+(x+y-1)/x x->1, 1=f(f(1)+y)+y/y+f(1) 1-y/y+f(1) =f(f(1)+y) f(1)/f(1)+y =f(f(1)+y) y->y-f(1), in this case we have accepted that y>f(1) f(1)/y =f(y), the finish of the claim 3 Claim 4: f(x)=1/x for all x Now, just moving f(f(x)+y).f(x)=(f(1)/f(1)/x +y).f(1)/x =f(1)^2./f(1)+x.y =f(x.y+1)=f(1)/x.y+1 <-> f(1)/f(1)+x.y =1/x.y+1 <-> f(1).x.y=x.y, therefore f(1)=1 f(x)=f(1)/x=1/x, proof of the claim Now, all solutions are f(x)=1 and f(x)=1/x