@Kurt Gödel, the problem isn't that hard as you've made it (actually it's worth an $A1$ level problem)!

Easy to see that $|f(y)|=|f(-y)|$ (just look at $P(x,y), P(x, -y)$. Let $u>v$. Look at $P(v, \sqrt{u-v})$, we get $f(u) \ge f(v)$ so $f$ is incresiang. We will prove $f(0)=0$. $P(0,1)$ gives us $f(1)=f(0)+|f(1)|$. If $f(1)=0$, $f(0)=0$. If $f(1)>0$, then $|f(1)|=f(1)$, so $f(0)=0$. If $f(1)<0$, then $f(1)=-x, x>0$, and $f(0)=-2x$. We know that $|f(-1)|=|f(1)|=x$ and $f(-1) \le f(1)$. If $f(-1)=f(1)$, then $f(1)=f(-1)\le f(0)< f(1)$, impossible, so $f(-1)=x >0>f(1)$, impossible. Thus, we know $f(0)=0$. We look at $P(0,y)$ and get $f(y^2)=|yf(y)|$, so $f(x+y^2)=f(x)+f(y^2)$. $(1)$ If we have an $a$ such that $a \neq 0$ and $f(a)=0$, (we can suppose $a>0$ because $|f(a)|=|f(-a)|$ then $P(0,a)$ gives us $f(a^2)=0$, $P(a,a^2)$ gives us $f(a+a^2)=0$, $P(a+a^2, a)$ gives us $f(2a+a^2)=0$ and so on, so we find infinitely large $Ms$ such that $f(M)=0$, but for any $t \in (0,M)$, $0=f(0) \le f(t) \le f(M) =0$, so $f(t)=0$, so $f$ is constant zero. Use again the fact that $|f(a)|=|f(-a)|$ for the negatives. Now, we have that $f(x)=0$ if and only if $x=0$, so $f(x)=-f(-x)$ (if they were equal, then $f(-x)<f(0)<f(x)$ or vice versa depending on the sign of x, impossible). Now, looking at $P(x-y^2,y)$, we have $f(x)=f(x-y^2)-f(-y^2)$, so $f(x+z)=f(x)+f(z)$ for negative $z$. We know that $f(x+z)=f(x)+f(z)$ holds for nonnegative $z$ ($(1)$), so $f$ is increasing and satisfies Cauchy equation, $f(x)=cx, c \ge 0$

A quick note: If a function is increasing or stays the same over $\mathbb{R}$, it is continuous at one point, which is enough for Cauchy...

Wow...I misread the absolute value signs as greatest integer function and solved it that way. It still is a nice problem when read either way.

From $|yf(y)|=f(x+y^2)-f(x)$, we let $x=0$ so that $|yf(y)|=f(y^2)-f(0)$. Plugging this in, we have $f(x+y^2)=f(x)+f(y^2)-f(0)$. Let $g(x)=f(x)-f(0)$. We have $g(x+y^2)=g(x)+g(y^2)$, so $g(x+y)=g(x)+g(y)$ for all reals $x$ and all nonnegative reals $y$. $g(x+y^2)=g(x)+|yf(y)|$, so we have $g(x)$ is monotonically increasing. Therefore, we have $g(x)=cx$ for all nonnegative $x$. From $g(y-x)=g(y)+g(-x)$, where $y -x\ge 0$, we have $c(y-x)=cy+g(-x)$, so $g(-x)=-cx$. Therefore, we have $g(x)=cx$ for all reals $x$. Since $g$ is monotonically increasing, $c \ge 0$. Now $f(x)=cx+d$, where $c \ge 0$. Plugging this into the original FE, we easily find $d=0$. Therefore, the answer is $f(x)=cx$, for $c \ge 0$.

I claim that $f(0)=0$. $y=1 \implies f(1) = f(0) + |f(1)|$. Suppose FTSOC $f(1) < 0$, which implies that $f(0) < 0$. This means that $f(2) = f(1) + |f(1)| = 0$. Then, $f(3) = f(2) + f(1) = f(1)$. Finally, $f(4) = f(3) + |f(1)| = 0$. However, we also have that $f(4) = f(0) + |2f(2)| = f(0) < 0$, contradiction! Thus $f(1) \ge 0$ and $f(0) = 0$ as desired. Also, plugging in $x=0$ yields, in light of $f(0)=0$, that $f(y^2) = |yf(y)|$. Hence our given functional equation becomes $f(x+y^2)=f(x)+f(y^2)$. It is obvious that the function is increasing or stays the same (in fact, we can prove that if it is not strictly increasing, then it is identically 0, but this isn't exactly useful...), so by Cauchy's functional equation and the condition $f(0)=0$, we know $f(x) = cx$. Plugging back in, we see that solutions are $\boxed{f(x)=cx, c \ge 0}$.

Can we say that $f$ is odd or even?

I meant from this condition $|f(y)|=|f(-y)|$ ???

Murad.Aghazade wrote: Can we say that $f$ is odd or even? Murad.Aghazade wrote: I meant from this condition $|f(y)|=|f(-y)|$ ??? Certainly not since maybe $f(-x)=f(x)$ for some (but not all) nonzero $x$ (and so $f(x)$ is not odd) and $f(-x)=-f(x)$ for all other $x$ (and so $f(x)$ is not even). Example : $f(x)=x^2$ $\forall x\in\mathbb Q$ and $f(x)=x$ $\forall x\notin\mathbb Q$ (such $f(x)$ is neither odd, neither even, and matches $|f(x)|=|f(-x)|$ $\forall x$)

I decided to try this problem for the time and I will give the "stupidest" solution possible. Substituting $(x,y)=\{(1,0),(-1,0),(-1,1)\}$ it follows that \[f(1)=f(0)+|f(1)|=f(0)+|f(-1)|\]and \[f(0)=f(-1)+|f(1)|.\]By brute force it follows that $f(0)=0$ and thus $f(y^2)=|yf(y)|$. Therefore $f(x+y^2)=f(x)+f(y^2)$ and taking $x=-y^2$ it follows that $f(y^2)=-f(-y^2)$. Thus it follows that $f(x+y)=f(x)+f(y)$ and $f(y)\ge0$ for $y\in\mathbb{R}^+$. Using Cauchy's Functional Equation it follows that $f(x)=cx$ and it follows that $c\ge0$ as $f(y^2)\ge0$. Thus $f(x)=cx$ for $c\ge0$ and these easily check out.

Quote: Using Cauchy's Functional Equation it follows that $f(x)=cx$ and it follows that $c\ge0$ as $f(y^2)\ge0$. Thus $f(x)=cx$ for $c\ge0$ and these easily check out. Would this be considered rigorous enough to get full points? Although It is well known claim do we have to prove it or just saying by Cauchy's is good enough Thanks

Math1331Math wrote: Quote: Using Cauchy's Functional Equation it follows that $f(x)=cx$ and it follows that $c\ge0$ as $f(y^2)\ge0$. Thus $f(x)=cx$ for $c\ge0$ and these easily check out. Would this be considered rigorous enough to get full points? Although It is well known claim do we have to prove it or just saying by Cauchy's is good enough Thanks This should basically sum it up: http://mathcircle.berkeley.edu/archivedocs/2010_2011/lectures/1011lecturespdf/FunctionalEquationsbmc.pdf

Taking $x=0$, we see that $f(y^2)=f(0)+|yf(y)|$. Therefore $$f(x+y^2)=f(x)+|yf(y)|=f(x)+f(y^2)-f(0).$$Define $g(x)=f(x)-f(0)$. Then $g(x+y^2)=g(x)+g(y^2)$. Note also that $g(y^2)=f(y^2)-f(0)=|yf(y)|\ge 0$, so it follows that $g(y)\ge 0$ for $y\ge 0$. Note that $g(a+b)=g(a)+g(b)$ for $a\in \mathbb{R}$ and $b\ge 0$. Clearly $g(0)=0$. Taking $a=-b$ gives $g$ odd. Taking $a\mapsto a-b$, $$g(a)=g(a-b)+g(b)\implies g(a-b)=g(a)-g(b)=g(a)+g(-b).$$Hence, $g$ is cauchy and bounded below on $[0,\infty)$. It follows $g$ is linear, so $g(x)=cx$, and $f(x)=cx+f(0)$ for some constant $c$. We now force $f(0)=0$. Plugging back in and simplifying, $cy^2=|cy^2+yf(0)|$. Clearly $c\ge 0$ by absolute values. If there is some $y\neq 0$ such that $cy^2=cy^2+yf(0)$, we get $f(0)=0$ and the solution $\boxed{f(x)=cx\text{ for any constant }c\ge 0}$. Otherwise, for all $y\neq 0$, we have $cy^2=-cy^2-yf(0)$, impossible since this is a quadratic in $y$ and has at most two solutions.

Peachy problem, whose pit is the following claim: Claim: Either $f\equiv 0$ or $f$ is strictly increasing. Proof: First, $f(a)\geq f(b)$ if $a>b$ by taking $(b,\sqrt{a-b})$. Now, suppose that it is not strictly increasing, so $f(a)=f(b)$ for some $a>b$. Taking $y=\sqrt{a-b}$ gives $$f(x+(a-b))=f(x) \Rightarrow f(n(a-b))=f(0)$$for all integers $n$. Since $n(a-b)$ can take on arbitrarily large and small values, and the function is nondecreasing, it must be constant, and it can be checked that $0$ is the only constant solution, as desired. (claim ends here) Now, take the not always $0$ case. Taking $(x,y)$ and $(x,-y)$ gives $|f(y)|=|f(-y)|$, and if $y>0$, this gives $f(y)=|f(y)|=-f(-y)$ from increasing. Now, we can drop the absolute value sign. Now, this is just problem 8.3 from https://web.evanchen.cc/handouts/FuncEq-Intro/FuncEq-Intro.pdf which is pretty straightforward

v_Enhance wrote: Determine all functions $f:\mathbb{R}\to\mathbb{R}$ such that for every pair of real numbers $x$ and $y$, \[f(x+y^2)=f(x)+|yf(y)|.\] $f(x+y^2)=f(x)+|yf(y)|$ Let $P(x,y)$ be the assertion Let $f(0)=c$ $P(0,y)=>$ $f(y^2)=c+|yf(y)|$ $=>$ $|yf(y)|=f(y^2)-c$ Substitute it in the original equation $=>$ $f(x+y^2)=f(x)-c+f(y^2)$ Subtract $c$ from the two sides $=>$ $f(x+y^2)-c=f(x)-c+f(y^2)-c$ Now define a new function $g$ such that: $g(x)=f(x)-c$ $=>$ $g(x+y^2)=g(x)+g(y^2)$ Let $y^2=z$ $=>$ $g(x+z)=g(x)+g(z)$ It is Cauchy Therefore, $g(x)=dx$ $=>$ $f(x)=dx+c$ Substitute it in the original equation and we will get that $c=0$ $=>$ $f(x)=dx$ Now, we have to show that $f$ is continuous $P(0,y)=>$ $f(y^2)=|yf(y)|$ $=>$ $f$ is continuous Therefore, $f(x)=dx$ ∀ $x$ ∈ ℝ And we won a pizza

Throughout the solution we denote $c=f(0), d=f(1).$ Let the assertion of the problem be $P(x, y).$ We start with $P(0, 1),$ which gives us $d=c+\lvert d\rvert \iff c=d-\lvert d\rvert.$ Case 1: $d\geq 0$ We have $c=0.$ Now $P(0, y)$ gives $f(y^2)=\lvert yf(y)\rvert\geq 0$ for all $y^2\geq 0.$ Next observe that $$f(x+y^2)=f(x)+\lvert yf(y)\rvert=f(x)+f(y^2)\hspace{1cm} (*)$$for all $x, y\in \mathbb{R}.$ Denote the assertion $(*)$ as $Q(x, y).$ Note that $Q(-y^2, y)$ yields $f(-y^2)=-f(y^2),$ so $f(-t)=-f(t)$ for all $t\in\mathbb{R}.$ We claim that $f$ satisfies Cauchy's functional equation$.$ Let $a, b\in\mathbb{R}$ be two arbitrary real numbers$.$ If one of them is non-negative (say $b$)$,$ we can use $Q(a, \sqrt{b})$ to obtain $f(a+b)=f(a)+f(b).$ Suppose that $a, b<0.$ We then have $$f(a)+f(b)=-f(-a)-f(-b)=-f(-a-b)=f(a+b)$$by the above discussion$.$ Now look at $f(y^2)=\lvert yf(y)\rvert.$ Using additivity and changing $y\mapsto y+1,$ we obtain $$\lvert yf(y)\rvert +2f(y)+d=f(y^2+2y+1)=\lvert (y+1)f(y+1)\rvert=\lvert yf(y)+f(y)+dy+d\rvert.$$ Assume that $y\geq 0.$ Recall that $f(y)=\lvert \sqrt{y}f(\sqrt{y})\rvert\geq 0,$ thus $yf(y)\geq 0$ and $yf(y)+f(y)+dy+d\geq 0.$ From here we obtain $$yf(y)+2f(y)+d=yf(y)+f(y)+dy+d$$and so $f(y)=dy$ for all $y\geq 0.$ Now for $y>0,$ we have $f(-y)=-f(y)=-dy.$ We conclude that $f(x)=dx$ for all $x\in\mathbb{R},$ whatever is the value of $d=f(1)\geq 0.$ It is easy to verify that this is a solution$.$ Case 2: $f(1)<0$ This time we have $c=d-\lvert d\rvert =2d$ and $P(0, y)$ gives $$f(y^2)=c+\lvert yf(y)\rvert=2d+\lvert yf(y)\rvert\hspace{0.5 cm} \text{for all $y\in\mathbb{R}$}$$Plugging back in the original assertion$,$ we have the new assertion $Q(x, y)$ $$f(x+y^2)=f(x)+f(y^2)-2d$$for all $x, y\in\mathbb{R}.$ Let $g(t):=f(t)-2d$ for all $t\in\mathbb{R}.$ We can rewrite the assertion $Q(x, y)$ as $$g(x+y^2)=g(x)+g(y^2) \hspace{0.5 cm}\text{for all $x, y\in\mathbb{R}$}$$We also have $g(0)=f(0)-2d=0.$ Now $Q(-y^2, y)$ yields $g(-y^2)=-g(y^2),$ so $g(-t)=-g(t)$ for all $t\in\mathbb{R}.$ We will prove that $g$ satisfies Cauchy's functional equation$.$ Let $a, b$ be two arbitrary real numbers$.$ If one of them is non-negative (say $b\geq 0$)$,$ we can use $Q(a, \sqrt{b})$ to deduce $g(a+b)=g(a)+g(b).$ Assume that $a, b<0.$ Now $$g(a)+g(b)=-g(-a)-g(-b)=-g(-a-b)=g(a+b).$$ Now we proceed with $g(y^2)=f(y^2)-2d=\lvert yf(y)\rvert=\lvert y(g(y)+2d)\rvert$ (note that $g(y^2)\geq 0$ for all $y^2\geq 0$)$.$ Again change $y\mapsto y+1$ and use additivity to obtain (taking into account $g(1)=f(1)-2d=-d$) $$\lvert yg(y)+2dy\rvert+2g(y)-d=g(y^2+2y+1)=\lvert yg(y)+g(y)+dy+d\rvert\hspace{1 cm} (*)$$We will first do a slight detour$,$ namely using $P(1, 1)$ to get $f(2)=f(1)+\lvert f(1)\rvert=0.$After that consider $P(2, \sqrt{u})$ for an arbitrary $u\geq 0$ to obtain $f(2+u)=\lvert uf(u)\rvert\geq 0.$ Now let $y>2.$ We have $g(y)+2d=f(y)\geq 0,$ so $g(y)+d\geq -d>0.$ Now $$y(g(y)+2d)\geq 0$$and $$yg(y)+g(y)+dy+d\geq yg(y)+dy+d=y(g(y)+d)+d\geq y(-d)+d=d(1-y)>0.$$Therefore in $(*)$ we have $$2g(y)-d+yg(y)+2dy=yg(y)+g(y)+dy+d$$$$\iff g(y)=-dy+2d \hspace{1 cm}\text{for all $y>2$}$$Plugging back$,$ we get $0\leq f(y)=2d+g(y)=4d-dy=d(4-y)$ for all $y>2.$ But setting $y=3,$ we arrive at a contradiction$.$ $\blacksquare$

Let $P(x,y)$ be the assertion $f(x+y^2)=f(x)+|yf(y)|$. $P(0,x)\Rightarrow|xf(x)|=f(x^2)-f(0)$ Let $g(x)=f(x)-f(0)$, then the assertion becomes $Q(x,y):g(x+y^2)=g(x)+g(y^2)$, since $P(0,x)$ becomes $g(x^2)=|xf(x)|$, or $g(x)\ge0\forall x\ge0$. Now $Q(-x^2,x)$ easily gives that $g$ is odd, and so $g$ is additive. Since $g$ is bounded, $g$ is linear, then so is $f$. Testing gives $\boxed{f(x)=cx},c\in\mathbb R_{\ge0}$.

Let $P(x,y)$ denote the assertion that \[f(x+y^2)=f(x)+|yf(y)|.\] $P(0,x): f(x^2)=f(0)+|xf(x)|$. So $f(x^2)-f(0)=|xf(x)|$. Now set $g(x)=f(x)-f(0)$. Since $f(x^2)-f(0)\ge0$, $g(x)$ is bounded on the interval $[0,\infty)$. So the assertion $Q(x,y)$ is: $g(x+y^2)=g(x)+g(y^2)$. So for any $k\ge0$, $g(x+k)=g(x)+g(k)$. $Q(-x^2,x): 0=g(-x^2)+g(x^2)$, which implies that $g$ is odd. Claim: $g(x+l)=g(x)+g(l)$ for any $l\in\mathbb{R}$. If $k\ge0$, then $g(x+k)=g(x)+g(k)=g(x)-g(-k)$. Now we take the negative of both sides, $g(-k-x)=g(-k)-g(x)$, so $g(-k+(-x))=g(-k)+g(-x)$, as desired. So $g$ satisfies Cauchy. Since $g$ is bounded over the interval $[0,\infty)$, $g(x)=cx$, for some constant $c$. So $f(x)=cx+f(0)$. Claim: $f(0)=0$. $P(0,x): cx^2+f(0)=f(0)+|cx^2+xf(0)|$ so $cx^2=|cx^2+xf(0)|$, so we have two cases. Case 1: $cx^2=-xf(0)-cx^2$ Here, $2cx^2=-xf(0)$, so if $x\ne0$, then $2cx=-f(0)$. If $c\ne0$, then $f(0)$ has many different values, which is not possible. If $c=0$, then $f(0)=0$. Case 2: $cx^2=cx^2+xf(0)$. So $xf(0)=0$, which implies $f(0)=0$. Thus, the answer is $\boxed{f(x)=cx}$, for $c\ge 0$. All such solutions work.

I think this works?? $P(0,y) \implies f(y^2)=f(0)+|yf(y)| (\spadesuit)$. Also, $P(-y^2,y) \implies f(0)=f(-y^2)+|yf(y)|$. Now we have, $$f(y^2)=f(0)+f(0)-f(-y^2) \implies f(y^2)+f(-y^2)=2f(0)$$. Letting $y=-1$ in $\spadesuit$, we get $f(1)=f(0)+|-f(-1)|$. Now if $f(-1)$ is non negative, then we get $f(1)=f(0)+f(-1)$. But we also have $f(1)+f(-1)=2f(0)$, and solving the equation we get $f(1)=\frac{3f(0)}{2}, f(-1)=\frac{f(0)}{2}$. This means that $f(0)$ must have been non negative since $f(-1)$ is non negative. Consequently, $f(1)$ is non negative and letting $y=1$ in $\spadesuit$, we obtain $f(1)=f(0)+|f(1)|=f(0)+f(1)$ giving $f(0)=0$. In the case when $f(-1)$ is negative, we get $f(1)=f(0)-f(-1)$ or that $f(1)+f(-1)=f(0)=2f(0)$ so $f(0)=0$. In either case, we get $f(0)=0$. This implies that $f(x)+f(-x)=0 \hspace{.1 cm}\forall x$. We also get $f(y^2)=|yf(y)|$ which gives us that $f(y^2)=yf(y)$ if $y$ is positive, and $f(y^2)=-yf(y)$ if $y$ is negative. Now choose $y>0$ and we get that $f(x+y^2)=f(x)+f(y^2)$ which holds true for all x. Multiplying $-1$ to both sides, we obtain $f(-x-y^2)=f(-x)+f(-y^2)$ so that the $f$ is additive for all real numbers. But we also have $f(y^2)=|yf(y)|$ which is positive so the function is bounded on a non trivial interval. In conclusion, $f(x)=ax$ for real number $a$, and putting this into the equation, it is easy to obtain that $a$ must be non negative, and consequently our answer if $f(x)=ax \forall a \in \mathbb{R}_{\geq 0}$

I will give the outline of my solution. Firstly $y=-y$ gives $|f(y)|=|f(-y)|$.Also $x=0$ and $x=-y^2$ and adding ,gives $f(-y^2)+f(y^2)=2f(0)$.Since $y^2$ is surjective over the non-negative reals,if there exists a non negative real $a$,such that $f(a)=-f(a)$,gives $f(0)=0$.If there doesnt exist such,then $f(x)=f(0)$,for every non-negative $x$.Only $0$ works here.We work with the latter first. Now in our equation let $x$ be an arbitrary negative.Then choose $y>0$ st $x^2+y$ is positive giving $f(x)=0$ for every real $x$. Now ,consider the former,$f(0)=0$,which gives $f(y^2)=|yf(y)|$.Now,choose $x=x,y=\sqrt{y^2-x},y>x>0$.This gives $f(y^2)=f(y^2-x)+f(x)$,Again as $y^2$ is surjective over non-negatives,$f$ is non-negative over this interval.Also if $a>b$,then $f(a)=f(a-b)+f(b)$,which gives $f(a) \ge f(b)$.Now,since $f$ is additive, increasing and non-negative over the non-negative reals,we have $f(x)=ax,a\ge0$ for $x\ge 0$.Now,in our equation let $x$ be negative and choose $y>-x>0$ which gives $f(x)=ax,a\ge0$ for all real $x$,which incorporates the first solution as well.

Setting $x=0$ gives $f(y^2)-f(0)=|xf(x)|.$ Now set $f(x)=g(x)+f(0)$ to obtain $$P(x,y): g(x+y^2)=g(x)+g(y^2).$$Note that this is bounded in $[0,+\infty)$ also $g(x)\geq 0$ if $x\geq 0.$ We get $g$ is odd \ i.e. $g(-x)=-g(x)$ from $P(-x^2,x).$ Combining all of these we get that $g$ satisfies Cauchy's FE and thus $g(x)=cx.$ It gives $f(x)=cx+f(0).$ Now we can directly check and get $f(0)=0$ and $f(x)=cx,$ which fits. Note that (as @below points out) $c$ is non-negative.

ZETA_in_olympiad wrote: Setting $x=0$ gives $f(y^2)-f(0)=|xf(x)|.$ Now set $f(x)=g(x)+f(0)$ to obtain $$P(x,y): g(x+y^2)=g(x)+g(y^2).$$Note that this is bounded in $[0,+\infty)$ also $g(x)\geq 0$ if $x\geq 0.$ We get $g$ is odd \ i.e. $g(-x)=-g(x)$ from $P(-x^2,x).$ Combining all of these we get that $g$ satisfies Cauchy's FE and thus $g(x)=cx.$ It gives $f(x)=cx+f(0).$ Now we can directly check and get $f(0)=0$ and $f(x)=cx,$ which fits. Wrong. $f(x) = -x$ doesn't work.

RevolveWithMe101 wrote: ZETA_in_olympiad wrote: Setting $x=0$ gives $f(y^2)-f(0)=|xf(x)|.$ Now set $f(x)=g(x)+f(0)$ to obtain $$P(x,y): g(x+y^2)=g(x)+g(y^2).$$Note that this is bounded in $[0,+\infty)$ also $g(x)\geq 0$ if $x\geq 0.$ We get $g$ is odd \ i.e. $g(-x)=-g(x)$ from $P(-x^2,x).$ Combining all of these we get that $g$ satisfies Cauchy's FE and thus $g(x)=cx.$ It gives $f(x)=cx+f(0).$ Now we can directly check and get $f(0)=0$ and $f(x)=cx,$ which fits. Wrong. $f(x) = -x$ doesn't work. Okay but show where I wrote that as I don't see it. Thanks.

ZETA_in_olympiad wrote: RevolveWithMe101 wrote: Wrong. $f(x) = -x$ doesn't work. Okay but show where I wrote that as I don't see it. Thanks. Gladly. ZETA_in_olympiad wrote: Now we can directly check and get $f(0)=0$ and $f(x)=cx,$ which fits.

ZETA_in_olympiad wrote: Setting $x=0$ gives $f(y^2)-f(0)=|xf(x)|.$ Now set $f(x)=g(x)+f(0)$ to obtain $$P(x,y): g(x+y^2)=g(x)+g(y^2).$$Note that this is bounded in $[0,+\infty)$ also $g(x)\geq 0$ if $x\geq 0.$ We get $g$ is odd \ i.e. $g(-x)=-g(x)$ from $P(-x^2,x).$ Combining all of these we get that $g$ satisfies Cauchy's FE and thus $g(x)=cx.$ It gives $f(x)=cx+f(0).$ Now we can directly check and get $f(0)=0$ and $f(x)=cx,$ which fits. You forgot to include that $c\geq 0$.

RevolveWithMe101 wrote: ZETA_in_olympiad wrote: RevolveWithMe101 wrote: Wrong. $f(x) = -x$ doesn't work. Okay but show where I wrote that as I don't see it. Thanks. Gladly. ZETA_in_olympiad wrote: Now we can directly check and get $f(0)=0$ and $f(x)=cx,$ which fits. Yes, you're right. However it would be more appreciated if you would point out the error (like Quidditch did) or at least be specific instead of saying things like "wrong"

Solved with Arjun Gupta. The only functions that work are $f(x) = ax$ with $a \geqslant 0$ Note that the function is clearly increasing. Next, taking $x = 0$ gives $f(y^2) =f(0) + |yf(y)|$. So we can rewrite the main equation as $f(x+y) = f(x) + f(y) + f(0)$ when $y > 0$. Let $g(x) = f(x)+f(0)$ so that we have $g(x+y) = g(x) + g(y)$ for $y \ge 0$. But also, $f(0)+|yf(y)| = f(y^2) = f((-y)^2) = f(0) + |yf(-y)|$, so we have either $f(y) = f(-y)$ or $f(y) = -f(-y)$. If it is the first case, then due to increasing, $f$ must be constant on $-y$ to $y$ and as a consequence zero from $0$ to $y$. But by induction and quasi-additivity, we can prove all nonnegative reals are mapped to zero, and since $f(-x) = \pm f(x)$, this means all reals go to zero, so $f(x) \equiv 0$, which is a solution. So assume $f(y) = -f(-y)$, so we have $g$ is completely additive. Since $g(x) \geqslant f(0)$ for $x \geqslant 0$, it is also bounded in a nontrivial interval, and so is linear. So $f(x) = ax + b$, checking gives only $f(x) = ax$ works for $a$ nonnegative (due to the $|yf(y)|$ term). These are the solutions claimed, so we're done.

The answer is $f(x)=cx$ only, which clearly work. Let $P(x,y)$ denote the assertion. By comparing $P(x,y)$ with $P(x,-y)$ we find that $|f(y)|=|f(-y)|$ for all $y \neq 0$. From $P(0,y)$ we find that $f(y^2)=f(0)+|yf(y)|$. From $P(-y^2,y)$ we find that $f(0)=f(-y^2)+|yf(y)| \implies 2f(0)=f(y^2)+f(-y^2)$. Therefore, if there exists some $y \neq 0$ such that $-f(y)=f(-y)$, we conclude that $f(0)=0$. Otherwise, $f(y)=f(-y)$ for all $y$, hence $f$ is constant, and it is clear that only $f \equiv 0$ works, which we already handled. Thus assume that the former case applies, hence $f(0)=0$. From $f(0)=0$ we then obtain that $-f(y)=f(-y)$ for all $y$, and that $f(y^2)=|yf(y)|$. Hence $f(x+y^2)=f(x)+f(y^2)$, or $f(x+y)=f(x)+f(y)$ for $y \geq 0$. By imposing $x \geq 0$ on this as well and using the fact that $f$ sends nonnegative reals to nonnegative reals, it follows that $f(x)=cx$ for some fixed $c$ whenever $x \geq 0$, hence from evenness it follows that $f(x)=cx$ for all $x \in \mathbb{R}$.

This is from the good all days when stuff that died to Cauchy were just flying around. The answers are $f(x) = kx$ for all $x \in \mathbb{R}$ and any fixed constant $k\geq 0$. It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. We denote by $P(x,y)$ the assertion that $f(x+y^2)=f(x)+|yf(y)|$. We start off with the following key observation. Claim : The function $f$ is non-decreasing. Proof : Note that any any $\epsilon >0$, $P(x,\sqrt{\epsilon})$ gives, \[f(x+\epsilon) = f(x)+|\epsilon f(\epsilon)|\geq f(x)\]which proves the claim. Now, say $f(0)=c$. We have from $P(0,x)$ that \[f(x^2)=f(0)+|xf(x)|=|xf(x)|+c\]Thus, for all real numbers $x$, $|xf(x)|=f(x^2)-x$. Substituting this into the given equation we have, \[f(x+y^2)=f(x)+f(y^2)-c\]We then shift to the function $g(x)=f(x)-c$. This gives, \[g(x+y^2)=g(x)+g(y^2)\]Thus, it is clear that by considering $x>0$, $g$ is additive over the positive reals. Further, $g$ is bounded over this interval since $f$ and in turn $g$ is non-decreasing. Thus, by Cauchy, it follows that $g$ is linear. Plugging in $g(x)=kx+l$ into the equation it is not hard to see that $l=0$. Thus, $g(x)=kx$ for a fixed constant $k$ for all positive reals $x$. Now, for any $x\in \mathbb{R}$, considering sufficiently large $y$ (i.e such that $x+y^2 > 0$) we have that \[g(x)=g(x+y^2)-g(y^2)=k(x+y^2)-ky^2=kx\]so $g(x)=kx$ for all real $x$. This implies that $f(x)=kx+c$ for all reals $x$. Now, we are almost there. Considering $P(x,y)$ for positive reals $y$ and plugging in the form we have for $f(x)$, \[k(x+y^2)+c = kx+c + |ky^2 + yc|\]we have two cases, if $ky^2+c<0$, \[k(x+y^2)+c = kx +c -ky^2 - c = kx-ky^2\]which implies that $2ky^2 = yc$ for all positive reals $y$ which is a very clear contradiction since this implies $2ky=c$ for all positive reals $y$. Further, if $ky^2+c \geq 0$, \[k(x+y^2)+c = kx+c + |ky^2+yc| =kx+ky^2 + c+ yc \]which implies that $c=0$. Thus, $f(x)=kx$ for some real constant $k$, which again upon substitution, \[kx+ky^2 = kx + |ky^2|=kx+|k|y^2\]requires $k=|k|$ so $k\geq 0$ as desired. Thus, all solutions are indeed of the claimed forms and we are done.

Put $x=0$ to get $f(y^2)=f(0)+|yf(y)|$ and $x=-y^2$ to get $f(0)=f(-y^2)+|yf(y)|$. The first equation implies $|f(y)|=|f(-y)|$ so $|f(0)-|yf(y)||=|f(0)+|yf(y)||$. This implies $f(0)=0$ so $f$ sends non negative reals to non negative reals and negative reals to non positive reals. Now we have that $f(x+y)=f(x)+f(y)$ for non negative reals and $f$ is bounded below by $0$ in this interval so by cauchy, $f$ is linear. Now it can be easily verified that $f(x)=kx$ for some constant $k$.

I claim the answer is only $f(x) = kx$ for nonnegative $k$, it is obvious to see that this works. If $f(1)$ is nonnegative, substituting $x = 0, y = 1$ yields $f(0) = 0$. Then taking $x = 0$ yields $f(y^2) = \mid y f(y) \mid$, so the equation resolves to $f(x + y^2) = f(x) + f(y^2)$, or $f(a + b) = f(a) + f(b)$ for all nonnegative $b$ and real $a$. Then taking $x = -y^2$ gives $-f(-y^2) = \mid yf(y) \mid = f(y^2)$, so $f$ is odd, implying that $f(a + b) = f(a) + f(b)$ holds for all nonpositive $b$ and real $a$, so it just holds for all $a,b$. Finally, it is obvious to see that $a > b$ implies $f(a) \ge f(b)$ by setting $x = b, y = \sqrt{a - b}$, so nondecreasing + additive implies linear, since $f(1)$ is nonnegative the function must have form $f(x) = kx$ for nonnegative $k$. Now, if $f(1)$ is nonpositive we have $f(2) = f(1 + 1^2) = f(1) - f(1) = 0 $. Then taking $x = -2, y = 2$ gives $f(2) = f(-2)$. Taking $x = -2, y = \sqrt{2}$ gives $f(0)= \mid \sqrt{2}f(\sqrt{2}) \mid$, taking $x = 0, y = \sqrt{2}$ gives $-f(0) = \mid \sqrt{2} f(\sqrt{2}) \mid$, so $f(0) = -f(0)$ and $f(0) = 0$. We can follow the same logic from the previous step to get $f$ linear, which clearly forces $f(x) = 0$.