Bump.....

This fantastic functional equation problem took me about 4 hours when I tested it Let $P(x, y)$ be the assertion. First, $$ P(1, 1) \to f(1) = 1, \, P(1, y) \to f(1 + y^2f(y)) = f(1 + y). $$Consider $P(1 + x^2f(x), y)$: $$ f(1+x^2f(x) + y^2f(y)) = f(1+yf(1+x^2f(x)))f(1+x^2f(x)) = f(1+yf(1+x))f(1+x) = f(1+x+y^2f(y)). $$By changing the variables $x, y$ we get $$ f(1+x+y^2f(y)) = f(1+y+x^2f(x)). $$ Suppose that there exists $x > 0$ such that $f(1 + x) > 1$. Pick $y > 0$ satisfying $1 + yf(1+x) = y + x^2f(x)$. Then $$ f(1+yf(1+x))f(1+x) = f(1+x+y^2f(y)) = f(1+y+x^2f(x)) \implies f(1+x) = 1, $$a contradiction. Therefore, $f(x) \leq 1, \, \forall x > 1$. Now we consider the following two cases: Case 1. $\exists a > 0$ s.t. $f(1+a) = 1$: $P(1+a, y) \to f(1+y^2f(y)) = f(1+y) = f(1+y^2f(y) + a)$, in particular, $f(1+a+a^2f(a)) = 1$. By induction we can show that $f(1+a+ na^2f(a)) = 1$ for all $n \in \mathbb{N}$. Specially, there exists arbitrarily sufficiently large $A$ such that $f(A) = 1$. Hence for all $y > 0$ there exists $x > 0$ such that $$ f(1+x+y^2f(y)) = 1 = f(1+yf(1+x))f(1+x). $$However, $f(1+x), f(1+yf(1+x)) \geq 1$ as $1+x, 1+yf(1+x) > 1$, so $f(1+x) = f(1+yf(1+x)) = 1 \implies f(1+y) = 1$. Hence $f(x) = 1, \, \forall x > 1$. Plugging in to the original equation we get $f(x+y^2f(y)) = f(x)$. Pick $y = 1$ we have $f(x) = f(x + 1) = 1 \implies \boxed{f(x) \equiv 1}$ which is a solution. Case 2. $\forall a > 0, f(1+a) < 1$: If there exists $x > 0$ such that $f(x) < 1/x$, consider $y = x - x^2f(x) > 0$, then $$ f(1+yf(1+x))f(1+x) = f(1+x+y^2f(y)) = f(1+y+x^2f(x)) \implies f(1+yf(1+x)) = 1, $$a contradiction. Hence $f(x) \geq 1/x, \, \forall x > 0$. The original equation then implies $f(1+yf(x)) \leq xf(x+y^2f(y))$. Since $y^2f(y) \geq y$, for all $y >1 - x$ we have $$ f(1+yf(x)) \leq xf(x+y^2f(y)) < x \qquad (\because x + y^2f(y) \geq x + y > 1). $$ This means, for any given $k > 0$, there exists $M > 0$ such that $f(y) < k$ for all $y > M$. Now, suppose that there exists $a \neq b$ such that $f(a) = f(b)$, then $P(a, 1), P(b, 1) \to f(1+a) = f(1+b)$. Furthermore, $$ f(1+x+a^2f(a)) = f(1+xf(1+a))f(1+a) = f(1+xf(1+b))f(1+b) = f(1+x+b^2f(b)), $$which implies $f$ is periodic at $x > N$ with period $T = |(a^2-b^2)f(a)| > 0$ for some $N > 0$. Arbitrarily choose $x > N$, there exists $M > 0$ such that for all $y > M$ we have $f(y) < 1/x \leq f(x)$. However, there exists $n \in \mathbb{N}$ such that $x + nT > M \implies f(x+nT) <1/x \leq f(x)$, which is a contradiction. Hence $f$ is injective and by $f(1+y) = f(1+y^2f(y))$ we get $\boxed{f(y) = 1/y, \, \forall y > 0}$, which is also a solution.

Nice solution

Extremely hard one :/ $P(1,1): f(1)=1$. $P(x+y^2f(y),z)$ and $P(x,z)$: $f(x+y^2f(y)+z^2f(z))=f(1+zf(x)f(1+yf(x)))f(x)f(1+yf(x))$. After switching $y,z$, this is equal to $f(1+yf(x)f(1+zf(x)))f(x)f(1+zf(x))$. Take $x=1, y=t, z=s$ and we get $f(1+tf(1+s))f(1+s)=f(1+sf(1+t))f(1+t)$. Call this $Q(t,s)$. $P(1+s,t), P(1+t,s)$ and $Q(t,s)$ give $f(1+s+t^2f(t))=f(1+t+s^2f(s))$. Call this $R(t,s)$. Lastly, $P(x+1,y)$ and $R(x,y)$ give $f(1+yf(x+1))f(x+1)=f(1+y+x^2f(x))$. Call this $S(x,y)$. If $f(x)>1$ for some $x>1$, then we can consider $S(x-1, (x-1)^2f(x-1)/(f(x)-1))$, which gives $f(x)=1$. This is a contradiction, and so $f(x)\leq 1$ for any $x>1$. Lemma 1. If $f(a)=f(b)$ for some $a,b\in\mathbb{R}^+$, then

Lemma 2. If $f(a)=f(b)$ for some $a,b\in\mathbb{R}^+$, then

Lemma 3. If $f(s)=1$ for some $s\geq 1$, then $s=1$ unless it holds for all $s$.

By Lemma 3., we can split into two cases: Case 1. $f(x)=1\quad\forall x\geq 1$. Then for all $x\in\mathbb{R}^+$, take $y$ sufficiently larger than $1$. Then $P(x,y)$ gives $f(x+y^2)=f(x)$ and thus $f(x)=1$. Case 2. If $s\geq 1$ and $f(s)+1$, then $s=1$. If $f(x)<1/x$ forsome $x\in\mathbb{R}^+$, then $S(x,x(1-xf(x))):f(1+x(1-xf(x))f(x+1))=1$, which is a contradiction. Therefore $f(x)\geq 1/x$ for all $x\in\mathbb{R}^+$. $P(x,1)$ now gives $1\geq f(x+1)=f(1+f(x))f(x)$, andso $f(1+f(x))\leq 1/f(x)$ for all $x\in\mathbb{R}^+$. Now for all $x\in\mathbb{R}^+$, take $t$ such that $1/t>x^2f(x)$. Then $$y=f(t)-x^2f(x)\geq \frac{1}{t}-x^2f(x)>0,$$and $S(x,y)$ gives $$\frac{f(x+1)}{1+yf(x+1)}\leq f(1+yf(x+1))f(x+1)=f(1+f(t))\leq \frac{1}{f(t)}=\frac{1}{x^2f(x)+y}\leq \frac{1}{x+y}.$$This shows that $f(x+1)\leq 1/x$ for all $x\in\mathbb{R}^+$. We can now show that $f$ is injective. If $f(a)=f(b)$ but $a<b$, then $a^2f(a)=a^2f(b)<b^2f(b)$. By Lemma 2., for all $x>1+a^2f(a)$ and $n\in\mathbb{N}$, we have $f(x)=f(x+n(b^2f(b)-a^2f(a)))$. Take $n$ to infinity and we get a contradiction with the inequalities we have obtained. Once we have the injectivity, we are done as $P(1,y)$ gives $f(1+y)=f(1+y^2f(y))$ and thus $yf(y)=1$, showing that $f(x)=1/x$ for all positive reals $x$.

Li4 wrote: Determine all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ satisfying \[f\bigl(x + y^2 f(y)\bigr) = f\bigl(1 + yf(x)\bigr)f(x)\]for any positive reals $x$, $y$, where $\mathbb{R}^+$ is the collection of all positive real numbers. Proposed by Ming Hsiao. Let $P(x,y)$ denotes the equation, and $\mathcal{S}$ denotes the set $\{k\in\mathbb{R}^+\mid f(k+y)=f\left(k+y^2f(y)\right)\,\forall\, y\in\mathbb{R}^+\}$. Firstly, $P(1,1)$ implies that $f(1)=1$. Substituting $x$ with $1$ thus gives $f\bigl(1 + y^2 f(y)\bigr) = f\bigl(1 + y\bigr)$ for all $y$. Thus, $1\in\mathcal{S}$. Moreover, if $f$ is injective, then $\boxed{f(x)\equiv\frac1x}$, which works. We'll assume that $f$ is not injective henceforth. For $a,b$ such that $f(a)=f(b)$, $P(a,y)$ and $P(b,y)$ imply that $f(a+y^2f(y))=f(b+y^2f(y))$ for all $y$. Thus, $k\in\mathcal{S}\implies k+y^2f(y)\in\mathcal{S}$ for all $y$. In particular, we have $1+y^2f(y)\in\mathcal{S}$ for all $y\in\mathbb{R}^+$. Claim 1 : If $k\in\mathcal{S}$, then $$f\bigl(1 + yf(k+x)\bigr)f(k+x)=f\bigl(1 + xf(k+y)\bigr)f(k+y)$$for all $x,y\in\mathbb{R}^+$. We'll denote this equation by $Q(k,x,y)$. Proof : We know that $k+x^2f(x)\in\mathcal{S}$ for all $x\in\mathbb{R}^+$. Thus, $f(k+y+x^2f(x))=f(k+y^2f(y)+x^2f(x))$ for all $x,y\in\mathbb{R}^+$. Switching variables gives $f(k+y+x^2f(x))=f(k+x+y^2f(y))$ and thus $P(x+k,y)$ and $P(y+k,x)$ give the desired result. Claim 2 : If $k>l\in\mathcal{S}$, then $$f(1+yf(k+x))f(k+x)=f(1+(y+k-l)f(l+x))f(l+x)$$for all $x,y\in\mathbb{R}^+$. Proof : $Q(k,x,y)$ and $Q(l,x,k-l+y)$. Claim 3 : For any $k>l\in\mathcal{S}$, we have $f(k+x)\le f(l+x)$ for all $x\in\mathbb{R}^+$. Moreover, if the equality holds for some $x,k,l$, then $f$ is eventually periodic and the equality holds for all $x,k,l$. Proof : If $f(k+x)>f(l+x)$ for some $x,k,l$, then there exists a $y\in\mathbb{R}^+$ such that $yf(k+x)=(y+k-l)f(l+x)$. The previous claim then gives $f(k+x)=f(l+x)$, a contradiction. If $f(k+x)=f(l+x)$ for some $x,k,l$, then $f$ is eventually periodic with period $(k-l)f(k+x)$. If for some $x_1,k_1,l_1$, we have $f(k_1+x_1)\neq f(l_1+x_1)$, then there exists arbitrarily large $y$ such that $yf(k_1+x_1)-(y+k_1-l_1)f(l_1+x_1)$ is a multiple of $(k-l)f(k+x)$. The previous claim then, again, gives $f(k_1+x_1)=f(l_1+x_1)$, which is a contradiction. Thus, we are done. Now, back to the main problem. Since $f$ is not injective, there exists different $c_1,c_2$ such that $f(c_1)=f(c_2)$. $P(c_1,1)$ and $P(c_2,1)$ give $f(c_1+1)=f(c_2+1)$. Thus, $$f(1+c_1^2f(c_1))=f(1+c_1)=f(1+c_2)=f(1+c_2^2f(c_2))$$ Again, $P(1+c_1^2f(c_1),1)$ and $P(1+c_2^2f(c_2),1)$ gives $f(2+c_1^2f(c_1))=f(2+c_2^2f(c_2))$. Since $c_1^2f(c_1)\neq c_2^2f(c_2)$ and $1+c_1^2f(c_1),1+c_2^2f(c_2)\in\mathcal{S}$, there exists an equality case for the inequality in Claim 3 and thus $f(k+x)=f(l+x)$ for all $k>l\in\mathcal{S}$ and $x\in\mathbb{R}^+$. In particular, $f(2+y^2f(y))=f(2)$. Finally, $\left(2,\frac{y}{f(2)}\right)$ gives $f(1+y)=1$ for all $y\in\mathbb{R}^+$. $P(x,1)$ then gives $\boxed{f(x)\equiv 1}$. Done.

ltf0501 wrote: Consider $P(1 + x^2f(x), f(y))$: $$ f(1+x^2f(x) + y^2f(y)) = f(1+yf(1+x^2f(x)))f(1+x^2f(x)) = f(1+yf(1+x))f(1+x) = f(1+x+y^2f(y))f(1+x). $$By changing the variables $x, y$ we get $$ f(1+x+y^2f(y))f(1+x) = f(1+y+x^2f(x))f(1+y). $$ Hi ltf0501, I'm not so sure this works. Firstly, it seems like you are using $P(1+x^2f(x),y)$, but that is a small issue. The issue lies in the last equality, because $f(1+yf(1+x))f(x+1)=f(1+x+y^2f(y))$, not $f(1+x+y^2f(y))f(x+1)$. Hence the next line should be $f(1+x+y^2f(y))=f(1+y+x^2f(x))$, which matches with the solution 1/x. I'm not sure if this line was used somewhere else in the proof. @below yep

gghx wrote: ltf0501 wrote: Consider $P(1 + x^2f(x), f(y))$: $$ f(1+x^2f(x) + y^2f(y)) = f(1+yf(1+x^2f(x)))f(1+x^2f(x)) = f(1+yf(1+x))f(1+x) = f(1+x+y^2f(y))f(1+x). $$By changing the variables $x, y$ we get $$ f(1+x+y^2f(y))f(1+x) = f(1+y+x^2f(x))f(1+y). $$ Hi ltf0501, I'm not so sure this works. Firstly, it seems like you are using $P(1+x^2f(x),y)$, but that is a small issue. The issue lies in the last equality, because $f(1+yf(1+x))f(x+1)=f(1+x+y^2f(y))$, not $f(1+x+y^2f(y))f(x+1)$. Hence the next line should be $f(1+x+y^2f(y))=f(1+y+x^2f(x))$, which matches with the solution 1/x. I'm not sure if this line was used somewhere else in the proof. Oh! Thank you so much. This is just a typo. It should be what you state above. The remaining parts seem still correct, right?

This problem was nice and fun, IMO. By the way, I'm taking a break from these to learn new things, especially non math. So I won't be here for a while.

ZETA_in_olympiad wrote: This problem was nice and fun, IMO. By the way, I'm taking a break from these to learn new things, especially non math. ] nooo! we need your functional equations solutions!

Let $P(x, y)$ be the assertion above. $P(1,1)$ gives us that $f(1) = 1$. Putting $P(x, 1)$ and $P(1, x)$, we get the below two equations, \[f(x+1) = f(1+f(x))f(x) \quad (1)\]\[f(x^2f(x)+1) = f(x+1) \quad (2)\]Hence now taking $P(x^2f(x)+1, y)$, we get \[f(1+x^2f(x) + y^2f(y)) = f(1+yf(x^2f(x)+1))f(x^2f(x)+1)\]and if we take symmetry $x \leftrightarrow y$, we get \[f(1+yf(x^2f(x)+1))f(x^2f(x)+1) = f(1+xf(y^2f(y)+1))f(y^2f(y)+1)\]which by $(2)$, we get \[f(1+yf(x+1))f(x+1) = f(1+xf(y+1))f(y+1)\]And by subbing back into original, we get that \[f(1+x+y^2f(y)) = f(1+y+x^2f(x))\] $\textbf{Claim:}$ $f(x) \leq 1$ for all $x \in (1,\infty)$. $\textit{Proof}$ Now if we assume there exists some $x$ such that $f(x+1) > 1$, we can take $y = \frac{x^2f(x)}{f(x+1)-1}$ to get in \[f(1+yf(x+1))f(x+1) = f(1+x+y^2f(y)) = f(1+y+x^2f(x))\]we get that $f(1+yf(x+1)) = f(1+y+x^2f(x))$, giving $f(x+1) = 1$ which is a contradiction. Hence we have that $f(x) \leq 1$ for all $x > 1$. $\square$ Using the inequality in the original $P(x,y)$, we obtain \[f(x) \geq f(x+y^2f(y))\]\[ \implies f(x+1) \geq f(x+1+y^2f(y)) = f(1+y+x^2f(x)) \quad (3)\] $\textbf{Claim:}$ If $f$ is non-constant, $f$ is injective. $\textit{Proof}$ If we assume $f$ is not injective, i.e $f(a) = f(b), a \neq b$, then notice we have that $f(1+y+a^2f(a)) = f(1+y+b^2f(b))$ as we have that $f(a+1) = f(b+1)$ from $(1)$, which means we have a period of $a^2f(a)-b^2f(b) \neq 0$ for values above $y > 1 + b^2f(b)$. But observing the inequality $(3)$, we can make attain any pair of values $(x+1, x^2f(x)+y+1)$ as we can first adjust $x+1$, and then adjust our $y$ to be any arbitrarily large value using our period value. Hence we have that $f(x) = C$ for $x > 1+b^2f(b)$. Now take $x$ to be arbitrarily large in $P(x,y)$. This would give us that $f(1+yf(x)) = 1$ for all $y \in \mathbb{R^+} \implies f(x) = 1$ for all $x > 1$. Now then with $(1)$ trivially gives us that $f \equiv 1$. $\square$ Hence we have injectivity, which trivially using $(2)$ gives us $f \equiv \frac{1}{x}$.