Find all functions $f:\mathbb{R}^+\to\mathbb{R}^+$ such that $f(x+yf(x)+y^2) = f(x)+2y$ for every $x,y\in\mathbb{R}^+$.
Problem
Source: 2019 Thailand Mathematical Olympiad P3
Tags: algebra
MarkBcc168
22.05.2019 13:57
The answer is $f(x)=2\sqrt{x+c}$ for any constant $c\geqslant 0$. It's straightforward to verify that they are all work.
Now we present several solutions which show that these are all possible solutions.
Solution 1 (Substitution, MarkBcc168): Plug in $(x,y) = \left(a, \frac{\sqrt{f(a)^2+4b}-f(a)}{2}\right)$ gives
$$f(a+b) = \sqrt{f(a)^2+4b} \implies f(a+b)^2=f(a)^2+4b$$This is enough to conclude that $f(x)^2 = 4x+c$ for some constant $c$. This implies the set of solutions mentioned above.
Solution 2 (Squaring, psi241): Squaring the entire equation gives
$$f(x+yf(x)+y^2)^2 = f(x)^2 + 4(yf(x)+y^2).$$By Intermediate Value Theorem, function $g(y) := yf(x)+y^2$ is surjective on $\mathbb{R}^+$ as $\lim\limits_{y\to 0} g(y)=0$ and $\lim\limits_{y\to\infty} g(y)=\infty$. Thus we get
$$f(x+t)^2 = f(x)^2+4t$$for any $x,t\in\mathbb{R}^+$. This is enough to conclude the solution.
Solution 3 (Injectivity): First, we prove that $f$ is injective. Suppose that $f(x+t)=f(x)$ for some $x,t\in\mathbb{R}^+$. Arguing as in above solutions, we can choose appropriate $y$ such that $yf(x)+y^2=t$. Using this choice of $x,y$ in the equation gives $y=0\implies t=0$ which is contradiction.
Now it's easy to finish. Plug in $y=\tfrac{f(t)}{2}$ gives
$$f\left(x+\frac{f(x)f(t)}{2} + \frac{f(t)^2}{4}\right) = f(x)+f(t) = f\left(t+\frac{f(x)f(t)}{2} + \frac{f(x)^2}{4}\right).$$By injectivity, $4x+f(t)^2 = 4t+f(x)^2$ so we are done.
sbealing
25.05.2019 18:57
Let $x_1,x_2 \in \mathbb{R}^+$ then we can find $y_1,y_2 \in \mathbb{R}^+$ such that: $$x_1+y_1 f(x_1)+y_1^2=z=x_2+y_2 f(x_2)+y_2^2$$so long as: $$f(x_i)^2-4(x-z) \geq 0 \Leftrightarrow z \geq x-\frac{f(x_i)^2}{4}$$for $i=1,2$ so for sufficiently large $z$ such $y_i$ exist From our functional equation we get: $$f(x_1)+2y_1=f(x_2)+2 y_2$$squaring gives: $$f(x_1)^2+4 y_1+4y_1^2=f(x_1)^2-4x_1=f(x_2)^2-4x_2$$as for any $x_1,x_2$ we can find a sufficiently large $z$ so the above holds we get $f(x)^2-4x=c$ for some constant $c$ so: $$\boxed{f(x)=\sqrt{4x+c}}$$and for all $c \geq 0$ this gives a valid $f$
green_leaf
15.10.2020 18:17
Let $P(x, y)$ be the assertion. Claim: Injective Proof: Assume $f(a) = f(b)$, WLOG let $a \leq b$. Let $a = x, b = x + yf(x) + y^2$ for some $y$ (it is easy to see that such a $y$ always exists and is positive by taking $a \leq b$). $P(x, y)$ gives $f(a) = f(b) + 2y \implies y = 0$ since $f(a) = f(b)$. So $b = x = a$. So we have injective. Now for the main problem taking $P\left( x, \frac{-f(x)}{2} \right)$ gives $f \left( x - \frac{f(x)^2}{4} \right) = 0 \implies x - \frac{f(x)^2}{4} = c$ for constant $c$ (because $f$ is injective). So $f(x) = 2 \sqrt{x - c}$ and we can see that this works.
Mr.C
15.10.2020 19:33
you should look at the domain . here is mine. let$p(x,y)=f(x+yf(x)+y^2)=f(x)+2y$ now if $f(a)=f(b)=t$ then we have from $p(a,y)-p(b,y)=f(a+ty+y^2)=f(b+ty+y^2)$ so since $y^2+yt$ is surgective we have $f(y+a)=f(y+b)$ so there exists a $l$ such that $f(y+l)=f(y)$ for big enough $y$ now we have from $p(x,y+l)-p(x,y)=f(x+(y+l)f(x)+y^2+2yl+l^2)=2l+f(x+yf(x)+y^2)$ just take $y$ such that $l+2y+f(x)$ is an positive integer and we get $l=0$ so $a=b$ now we have from $p(x,\frac{f(y)}{2})=f(x+\frac{f(x)f(y)}{2}+\frac{f(y)^2}{4})=f(y)+f(x)=f(y+\frac{f(x)f(y)}{2}+\frac{f(x)^2}{4})$ so $4y+f(x)^2=4x+f(y)^2$ so $f(y)^2=4y+f(1)^2-4$ let $f(1)^2-4=4k$ so $f(y)=2\sqrt{y+k}$ and done! . . the problem it self wasn't that hard but once you know the solution you are done. as most people might go for proving there is no such function and just walk around a circle