$P(1,x)$ implies $f(x)\equiv 0$ or $f(1+f(y))=f(1)$

Murad.Aghazade wrote: $P(1,x)$ implies $f(x)\equiv 0$ or $f(1+f(y))=f(1)$ I guess your solution is not complete!

K.N wrote: Murad.Aghazade wrote: $P(1,x)$ implies $f(x)\equiv 0$ or $f(1+f(y))=f(1)$ I guess your solution is not complete! Yes.İ am now trying to solve

Sorry for my mistake.

doxuanlong15052000 wrote: For the large $n$, $\frac {y}{(a+1)^n}=0$, so $f(x)=f(0)=c$ for all real $x.$ First of all, $f(0)$ does not exist since $f$ is only defined for positive reals. Also, your argument uses continuity at $0$ which is given neither.

Do I have a mistake?

doxuanlong15052000 wrote: Do I have a mistake? Tintarn just said it

First blood? I'll explain the motivation asap.

Can you explain this part? FEcreater wrote: $f\left(2x+2f\left(1\right)\right)=\frac{f\left(2x\right)^4}{f\left(1\right)^3}$

My pleasure! We have $$ f\left(2x+2f\left(1\right)\right)=\frac{f\left(2x+f\left(1\right)\right)^2}{f\left(1\right)}=\frac{f\left(2x\right)^4}{f\left(1\right)^3} $$

Well done. Nice short solution

$P(x,y):f(y)f(x+f(y))=f(x)f(xy)$ $P(1,x):f(x)f(1+f(x))=f(1)f(x)\implies f(x)=0\lor f(1+f(x))=f(1)$ but since $R_f$ is $\mathbb{R}^{+}$ we have $f(1+f(x))=f(1)\forall x$ $P(x,1):f(x)^2=f(1)f(x+f(1))=f(1)^2\implies f(x)=f(1)\forall x$ hence we get a solution $f(x)=c\forall x\mathbb{R}^{+} ,c\in\mathbb{R}^{+}$

Reynan wrote: $P(x,y):f(y)f(x+f(y))=f(x)f(xy)$ $P(1,x):f(x)f(1+f(x))=f(1)f(x)\implies f(x)=0\lor f(1+f(x))=f(1)$ but since $R_f$ is $\mathbb{R}^{+}$ we have $f(1+f(x))=f(1)\forall x$ $P(x,1):f(x)^2=f(1)f(x+f(1))=f(1)^2\implies f(x)=f(1)\forall x$ hence we get a solution $f(x)=c\forall x\mathbb{R}^{+} ,c\in\mathbb{R}^{+}$ You've proved that $f(1+f(x))=f(1)\forall x$ not $f(x+f(1))=f(1)$

bgn wrote: Reynan wrote: $P(x,y):f(y)f(x+f(y))=f(x)f(xy)$ $P(1,x):f(x)f(1+f(x))=f(1)f(x)\implies f(x)=0\lor f(1+f(x))=f(1)$ but since $R_f$ is $\mathbb{R}^{+}$ we have $f(1+f(x))=f(1)\forall x$ $P(x,1):f(x)^2=f(1)f(x+f(1))=f(1)^2\implies f(x)=f(1)\forall x$ hence we get a solution $f(x)=c\forall x\mathbb{R}^{+} ,c\in\mathbb{R}^{+}$ You've proved that $f(1+f(x))=f(1)\forall x$ not $f(x+f(1))=f(1)$ oh sorry, misread my solution

FEcreater wrote: First blood? I'll explain the motivation asap.

I think P(x , a + f(y)) must be P(a, a+f(y) ) and you get f(x)=f(2x) it is not for all x >0 .

Claim:$\forall a$ so that $f(a)=c$ we have that $f(ax)=f(x)$, where $c:=f(1)$ Proof: $P(x,1):cf(x+c)=f(x)^2$(1) $P(x,a):cf(x+c)=f(x)f(ax)$ using the previous relation we obtain the desired. $P(1,1):cf(1+c)=c^2$ $\implies$ $f(c+1)=c$,$P(c+1,y):f(y)f(c+1+f(y))=f(c+1)f((c+1)y)$ $\implies$ $f(y)f(c+1+f(y))=cf(y)$.Now using (1) $cf(c+1+f(y))=f(f(y)+1)^2$ $\implies$ $f(f(y)+1)=c$ as well as $f(c+1+f(y))=c$ now plug $y=1$ and so $f(2c+1)=c$.Using the claim $f((2c+1)x)=f(x)$. $P(2c+1,y):f(y)f(2c+1+f(y))=cf((2c+1)y)=cf(y)$ and so $f(2c+1+f(y))=f(f(y)+1+c)^2=f(f(y)+1)^4=c=c^4$ and so $c=1$.Now the first relation becomes $f(x+1)=f(x)^2$ and now notice that $f(2)=f(1)^2=1=f(1)$ and so by the claim $f(2x)=f(x)$.Ending is as above : $$f(2x+2)=f(x+1)=f(x)^2=f(2x+1)^2=f(2x)^4=f(x)^4 \implies f(x)=1 \forall x$$

@nikolapavlovic. How can you have this : $f(2c+1+f(y))=f(f(y)+1+c)^2=f(f(y)+1)^4=c=c^4$

MAITHANHHUY wrote: @nikolapavlovic. How can you have this : $f(2c+1+f(y))=f(f(y)+1+c)^2=f(f(y)+1)^4=c=c^4$ Honestly,I have no idea.

i think $f(2c+1+f(y))=f(f(y)+1+c) = f(f(y)+1)=c$

bgn wrote: Find all functions $f:\mathbb {R}^{+} \rightarrow \mathbb {R}^{+} $ such that for all positive real numbers $x,y:$ $$f(y)f(x+f(y))=f(x)f(xy)$$ Let $P(x,y)$ denote the assertion of the given FE. Easily check that the constant function works. Otherwise, assume there exists $a>b>0$ such that $f(a)=f(b)$. Comparing $P(x,a)$ and $P(x,b)$, we have $$f(ax)=f(bx)\Rightarrow f(x)=f(dx)\quad\forall x>0,$$where $d=\frac{a}{b}>1$. Hence comparing $P(x,y$) and $P(dx,y)$ gives $$f(dx+f(y))=f(x+f(y))=f(dx+df(y))\Rightarrow f(x+c)=f(x),\quad\forall x>e.$$Therefore comparing $P(x+c,y)$ and $P(x,y)$ gives $$f(xy)=f(xy+cy)\Rightarrow f(x)=f(x+cy),\quad\forall x>e,y>0.$$Plugging $x$ by $cx$ into the above equality and then changing the roles of $x,y$, we have $$f(cx)=f(cy),\quad\forall x,y>\frac{e}{c}\Rightarrow f(x)=f(y)=k,\quad\forall x,y>e.$$For all $y>0$, let $x>\max \{ e,\frac{e}{y}\}$ in $P(x,y)$, then $f(y)=k,\quad\forall y>0$, a contradiction.

mfw $f(a)=f(b) \implies f(ax)=f(bx)$ (never seen this before) The answer is constant functions only which clearly work. Let $P(x,y)$ denote the assertion. Suppose that $f(a)=f(b)$. Then from $P(x,a)$ and $P(x,b)$ we obtain $f(ax)=f(bx)$, which holds for all $x \in \mathbb{R}^+$. Thus let $S$ denote the set of all $r \in \mathbb{R}^+$ such that $f(x)=f(rx)$ for all $x$. Clearly $1 \in S$, and $S$ is closed under multiplication and division. From $P(1,1)$ we obtain $f(1)=f(1+f(1))$, so $1 \neq 1+f(1) \in S$, and therefore we have arbitrarily small (positive) elements of $S$. Now let $r \in S$ be arbitrary and compare $P(x,y)$ with $P(rx,y)$, which yields $f(x+f(y))=f(rx+f(y))$. Thus $$\frac{rx+f(y)}{x+f(y)} \in S.$$For fixed $r<1$ and $f(y)$, by varying in $\mathbb{R}^+$ the above fraction spans the interval $(r,1)$. Since we can pick $r$ to be arbitrarily small it follows that all of $(0,1)$ is in $S$, so then $S=\mathbb{R}^+$ since $r \in S \implies 1/r \in S$, implying $S$ is constant. $\blacksquare$

Solved with SatisfiedMagma. Solution. The only possible working functions are the constant ones which trivially work. We now proceed to show that these are the only solutions. Denote $P(x,y)$ as the assertion to the functional equation. $P(1,x)$ give us that $f(f(x) + 1) = f(1)$. If $f$ is necessarily injective, then we have $f \equiv 0$ which is impossible over $\mathbb{R}^+$. Therefore there exists distinct $a,b$ for which $f(a) = f(b)$. $P(x,a)$ and $P(x,b)$ reveal there exists a $r$ for which $f(x) = f(rx)$ for $r = a/b$ with $a>b$. Comparing $P(rx, 1)$ and $P(x,f(1))$ we get that \[f(rx + f(1)) = f(x + f(1)) = f(rx + rf(1)) \implies f(x + f(1)) = f(x + rf(1))\]which means $f$ is eventually periodic with period $T \coloneqq f(1)(r - 1)$. Now compare $P(x,y)$ and $P(x,y+T)$ with large enough $y$ for periodicity to get \[f(xy + yT) = f(xy).\]Setting $x = 1/y$ here would yield $f(yT+1) = f(1)$. This means $f$ is eventually constant for large enough domain values. Going back to the original equation set $x$ to be insanely huge. Then the RHS will be forced to be constant and the $f(x + f(y))$ term will be constant as well. Simplifying the equation, we would get $f$ constant on $\mathbb{R}^+$ as well. $\blacksquare$ @below fixed, thanks for pointing it out

cookie130 wrote: i dont understand any of that Sorry to degrade HSO, but please stop postfarming. If you don’t understand, don’t be afraid to ask for a walkthrough on the solution.

Denote the assertion with $P(x, y)$. Claim: If $f(a) = f(b)$, then $f(ax) = f(bx)$ for all $x$. Proof. Follows from $P(x, a)$ and $P(x, b)$. $\blacksquare$ Let $S$ be the set of reals $c$ such $f(x) = f(cx)$ for all $c$. Note that $S$ is closed under multiplication. By $P(1, x)$ it follows that \[ f(1 + f(x)) = f(1) \]so $1 + f(x) \in S$. As such, $S$ contains arbitarily large elements. Claim: $f$ is constant on $x \ge 1$ Proof. Fix $s \in S$. By $P(sx, y)$ it follows that \[ f(sx + f(y)) = f(x + f(y)) \]so \[ \frac{sx + f(y)}{x + f(y)} \in S \]which has the image of interval $(1, s)$ as $x$ ranges. Taking arbitrarily large $s$ gives the result. $\blacksquare$ Then, by $P(N, x)$ for $N > \min\left(\frac{1}{x}, 1\right)$ gives that $f$ is constant for all reals. Furthermore, all constant $f$ work.

proxima1681 wrote: Comparing $P(rx, 1)$ and $P(x,f(1))$ we get that \[f(rx + f(1)) = f(x + f(1)) = f(rx + rf(1)) \implies f(x + f(1)) = f(x + rf(1))\]which means $f$ periodic with period $T \coloneqq f(1)(r - 1)$ ... This means $f$ is constant in $(1, \infty)$. But since $f$ is periodic as well, this implies $f$ must be constant on $\mathbb{R}^+$ as well. $\blacksquare$ I love your proof, but I don't think this particular step works. The first statement above says that $f$ is periodic, but only on $(f(1),\infty)$. But you can very easily remedy the last step of the proof by looking at $P\left(x,\dfrac{1}{x}\right)$ to prove that it works for $x<1$ too.

Let $f(1)=c$ and $P(x,y)$ be the assertion below $f(y)f(x+f(y))=f(xy)f(x)\quad\forall x,y\in\mathbb{R}^+$. $$P(x+f(z),y)\implies f(x+f(y)+f(z))=\frac{f(yx+yf(z))f(x+f(z))}{f(y)}\overset{P(x,z)}{=}\frac{f(yx+yf(z))f(x)f(xz)}{f(z)f(y)}\quad\forall x,y,z>0.\ (1)$$Switch $y,z$ of the equation $(1)$ we have $$f(xy)f(yx+yf(z))=f(xz)f(zx+zf(y))\quad\forall x,y,z>0.$$Let $y=1$ then $\frac{f(x)^2f(xz)}{f(z)}\overset{P(x,z)}{=}f(x)f(x+f(z))=f(xz)f(zx+zc)\quad\forall x,z>0$ And so $$f(x)^2=f(zx+zc)f(z)\quad\forall x,z>0.$$$P(x,1)\implies f(x)^2=cf(x+c)\quad \forall x>0$ then $f(zx+zc)f(z)=cf(x+c)\quad\forall x,z>0$. Let $z=\frac{1}{x+c}$ we have $f(x)=f(1/x)\quad\forall x>c$. Now let $x>c$, $$P(x,x),P(x,1/x)\implies f(x^2)=c\quad\forall x>c.$$And so $f(x)=c\quad\forall x>c^2$. In $P(x,y)$, let some postive $x_0>\max\{1,c^2\}$ then $P(x_0,y)\implies f(y)=f(x_0y)=\cdots=f(x_0^ny)\quad\forall y>0$. Fix any $0<y\le c^2$ then there exist $n\in\mathbb{N}$ such that $x_0^ny>c^2$ and so $f(y)=c$. Thus the only solution is $$\boxed{f\equiv c\quad\forall x>0} \ \blacksquare.$$

The only solution is $c\in \mathbb R^+$. Let's assume it is not constant. Let $f(a) = f(b)$, $b>a>0$. Then comparing $P(x,a)$ and $P(x,b)$, \[ f(ax) = f(bx) \implies f(x) = f(rx), r=\frac{b}{a} > 1. \] Now, $P(x,y)$ and $P(rx,y) \implies f(x + f(y)) = f(rx+f(y))$. Now $k = \dfrac{rx + f(y)}{x + f(y)} \iff x = \dfrac{f(y)(k-1)}{r-k}$. $f(x) = f(kx)$ for all $k \in (1,r)$. Now we can increase $r \rightarrow r^2$ and we can increase indefinitely to cover the whole $\mathbb R^+$. Also, $f(x) = f(kx) = f(k^2x) = \cdots \implies f$ is a constant, contradiction. Then $f$ is injective. $P(x,x) \implies f(x+f(x)) = f(x^2) \implies f(x) = x^2 - x$. But putting $x = 1$, we get $f(1) = 1^2 - 1 = 0$, contradiction again, and we are done.

P(1,y) gives:f(y)=f(y+f(1)) P(x,y) and P(x,y+f(1)) gives:f(xy)= f(xy +xf(1)) Now, take x=(b-a)/f(1) and y=af(1)/(b-a), for b>a. So,f(a)=f(b), for all b>a. Hence, f(x)=c. One can check that this works.

Call the assertion $P(x,y)$. We claim that the only solution is $f\equiv c$, where $c$ is a positive constant, which clearly works. Note that $P(1,x)$ gives $f(1+f(x))=f(1)$. Hence, the function is clearly not injective. Suppose there are distinct $x_1, x_2$ such that $f(x_1)=f(x_2)$. The above equations warrants the existence of at least few such pairs. Comparing $P(x,x_1)$ and $P(x,x_2)$ gives us that $f(xx_1)=f(xx_2)$. It follows that if $r=\frac{x_1}{x_2}\ne1$, we have $f(x)=f(rx)$ for every $x>0$. Let $S$ be the set of all positive real numbers $r$ for which $f(x)=f(rx)$. WLOG assume $r>1$. Claim: If $r\in S$, then $[1,r]\in S$. Proof. Compare $P(x,y)$ and $P(rx,y)$ to get that $f(rx+f(y))=f(x+f(y))$. So, for any $x,y$ we have $\frac{rx+f(y)}{x+f(y)}\in S$. Let $1<s<r$ be a positive real. Then, we can set $x=\frac{(s-1)f(y)}{r-s}$ so that $\frac{rx+f(y)}{x+f(y)}=s$. The claim follows. $\blacksquare$ Note that since $f(x)=f(rx)=f(r^2x)=\cdots=f(r^nx)$ for any positive integer $n$, we have that $r^n\in S$ and hence $[1,r^n]\in S$ for any $n$. Since $n$ can be arbitrarily large, we have that $S=[1,\infty)$. So, for any $x, y$ with $0<x\leq y$, we have $\frac yx\in S$ which means $f(x)=f(y)$. It follows that the function is constant function. $\blacksquare$

The answer is $f(x) = \boxed{c}$, which works. Denote the given assertion as $P(x,y)$. It is clear that all constant $f$ work, so we will suppose otherwise henceforth. Claim: $f$ is injective Proof: Suppose otherwise; that is, suppose there exist distinct $a$ and $b$ such that $f(a)=f(b)$. Comparing $P(x,a)$ and $P(x,b)$ yields $f(ax)=f(bx)$, so we have $f(x) = f(rx)$, where $r = \tfrac{a}{b}$. Then, comparing $P(x,y)$ and $P(rx,y)$ yields \[f(x+f(y)) = f(rx+f(y)).\] Therefore, $f$ is periodic with period $p = |(r-1)x|$. Comparing $P(x,y)$ and $P(x+p,y)$ yields \[f(xy) = f(xy+yp),\] which means we can vary $y$ to make the period arbitrary. In other words, $f$ is constant, a contradiction. $\square$ If $f$ is injective, plugging in $P(1,1)$ yields \[f(f(1)+1) = f(1) \implies f(1) = 0,\] a contradiction. Hence, no nonconstant $f$ exists, leaving only our solution set.

We claim $f$ constant, which works. Suppose $f(a)=f(b)$. Then $P(x,a),P(x,b)$ give $f(xa)=f(xb)$. Let $1+f(1)=c>1$. Then $P(1,1)$ gives $f(1)=f(c)$. Thus $f(x)=f(cx)$. Now $P(x,y),P(cx,y)$ for fixed $y$ give $f(x+f(y))=f(cx+f(y))=f(x+\tfrac{f(y)}c)$. Thus $f$ is periodic with period $p=f(y)(1-\tfrac1c)$ for $x>\tfrac{f(y)}c=r$. Now $P(x,y),P(x+p,y)$ for $x>r$ give $f(xy)=f(xy+py)$ so for any $1\le k<1+\tfrac pr$ we have $f(kx)=f(x)$. Now if $a<b$ then $f(a)=f(a^{\frac{t-1}t}b^{\frac 1t})=f(a^{\frac{t-2}t}b^{\frac 2t})=\dots=f(b)$ for $t$ large enough so that $(\tfrac ba)^{\frac 1t}<1+\tfrac pr$. Thus $f$ is constant.

The only solution is $f\equiv k$ for some constant $k$. These work. Now we show they are the only solutions. Suppose $f$ wasn't constant. Claim: $f$ is injective. Proof: Suppose $f(a) = f(b)$ for some $a < b$. $P(x,a)$ compared with $P(x,b)$ gives $f(ax) = f(bx)$, so $f(x) = f(cx) \forall x > 0$, where $c = \frac ba >1$. $P(x,y)$ compared with $P(cx, y)$ gives\[ f(x + f(y)) = f(cx + f(y))\]Thus, $f(c(x + f(y))) = f(cx + f(y))$, so $f(cx + f(y)) = f(cx + f(y) + f(y)(c-1))$. Thus, for any $x > f(y)$, we have $f(x) = f(x + f(y) (c-1))$. Now, fix constants $r,d$ so that for all $x \ge r$, we have $f(x) = f(x + d)$ (possible by above). For $y \ge r$, $P(x,y)$ compared with $P(x, y + d)$ gives that $f(xy) = f(xy + xd)$. For any $x \le 1$, varying $y \ge r$ gives that $f(y) = f(y + xd) \forall y \ge r$. Iterating this gives that $f(y) = f(y + nxd) \forall y \ge r$ and integers $n >0$, which clearly implies $f$ is eventually constant. Suppose it is eventually constant at $k$ and $f(x) = k\forall x \ge R$. If we set $y \ge R$, we have $k f(x + k) = k f(x) \implies f(x) = f(x+k)$. Iterating gives that $f(x) = f(x + nk)$ for any positive integer $n$. Thus, for any $x > 0$, setting $nk > N$ gives that $f(x) = f(x + nk) = k$, so $f$ is constant, a contradiction. $\square$ Now setting $x = 1$ gives $f(f(y) + 1) = f(y)$, so $f(y) + 1 = y$, contradiction for $y \le 1$.

Let $P(x, y)$ denote the assertion: \[P(1, y)\]\[f(1+f(y))=f(1)\]\[P(x, 1+f(y))\]\[f(1)f(x+f(1))=f(1)f(x(1+f(y)))\]\[P(x, 1)\]\[f(1)f(x+f(1))=f(1)f(x)\]Thus $f(x)=f(kx)$ when $k=1+f(y)$ \[P(kx, y)\]\[f(y)f(kx+f(y))=f(y)f((xy))\]Thus $f(x)=f(kx+f(y))$ thus let $f(y)=c$, $f(x)=f(kx)$ and $f(x)=f(kx+c)$ so $f(kx)=f(kx+c)$. Thus $f(x)=f(x+c)$ for any $x$. \[P(x+c, y)\]\[f(y)f(x+f(y))=f(y)f(xy+yc)\]Thus $f(xy+yc)=f(xy)$ for any $x, y$ so we get that $x$ must be constant.