$P(a,b)$ and $P(-a-b,b)$ imply that $f((a+b)^2)-f(a^2)=(2a+b)f(b).$ Let $Q(a,b)$ denotes the above equation. $Q(a,-2a-b)$ implies that $f((-a-b)^2)-f(a^2)=-bf(-2a-b).$ Hence, $(-2a-b)f(b)=bf(-2a-b)$ for all $a,b\in\mathbb{R}.$ This means that $af(b)=bf(a)$ for all $a,b\in\mathbb{R}.$ In particular, $f(x)\equiv cx$ for some constant $c.$ Plugging back in the equation, we get $c\in\{-1,1\}.$ Thus, $f(x)\equiv x$ or $f(x)\equiv -x,$ both of which are solutions.

a nice one let $P(x,y)$ be the assertion $$f(a^2 +ab+ f(b^2))=af(b)+b^2+ f(a^2) $$ $P(-\frac{f(b^2)}{b},b) \implies f(b^2)f(b)=b^3$ for $b \neq 0$ so $f(1)=1$ (if $f(1)=0$ then $P(0,1) \implies f(0)=f(0)+1$ contradiction) $P(0,b) \implies f(f(b^2))=b^2+f(0)$ plugging $b=1 \implies f(0)=0$ so $f(f(b))=b \forall b \ge 0$ $P(-a,a)\implies f(a^2)=af(a) $ so $f(a)^2=a^2$ so $$f(a)=a \forall a \in \mathbb{R}$$or $$f(a)=-a \forall a \in \mathbb{R}$$(if we have $f(a)=a ,f(b)=-b$ for some $a,b$ it's easy to get contradiction just by $P(a,b)$)

This FE is very suitable for a introductory exercise . It captures all the basic tricks .

TLP.39 wrote: $P(a,b)$ and $P(-a-b,b)$ imply that $f((a+b)^2)-f(a^2)=(2a+b)f(b).$ Let $Q(a,b)$ denotes the above equation. $Q(a,-2a-b)$ implies that $f((-a-b)^2)-f(a^2)=-bf(-2a-b).$ Hence, $(-2a-b)f(b)=bf(-2a-b)$ for all $a,b\in\mathbb{R}.$ This means that $af(b)=bf(a)$ for all $a,b\in\mathbb{R}.$ In particular, $f(x)\equiv cx$ for some constant $c.$ Plugging back in the equation, we get $c\in\{-1,1\}.$ Thus, $f(x)\equiv x$ or $f(x)\equiv -x,$ both of which are solutions. General question, why do you ( and many others) write $f(x)\equiv x$ instead of $f(x)=x$?

The answers are $\boxed{f(x)\equiv x}$ and $\boxed{f(x)\equiv -x}$. It is easy to see that they satisfy the given functional equation, we now show that these are the only solutions. Let $P(a,b)$ denote the given assertion, we prove that $f$ is odd, let $a\ne 0$, from $P(-a,b)$ and $P(a,-b)$, we have \[P(-a,b) : f(a^2-ab+f(b^2))=-af(b)+b^2+f(a^2)\]\[P(a,-b) : f(a^2-ab+f(b^2))=af(-b)+b^2+f(a^2)\]and by comparing these two equations, we get $f(b)=-f(-b)$ for all $b\in \mathbb{R}$. So, we can deduce that $f(0)=0$ by letting $b=0$. We now show that $f$ is an involution, \[P(0,a) : f(f(a^2))=a^2 \ \ \textrm{and} \ \ -f(f(a^2))=f(f(-a^2))=-a^2 \ \ \forall a\in \mathbb{R}\]as desired. Since $f$ is an involution, it is also bijective. Then, we show that $f(a^2)=af(a)$, we have \[P(a,-a) : a^2=f(f(a^2))=-af(a)+a^2+f(a^2) \implies f(a^2)=af(a) \ \ \forall a\in \mathbb{R}\]as desired. Thus, $f(f(a)^2)=f(a)f(f(a))=af(a)=f(a^2) \implies f(a)^2=a^2 \implies f(a)=\pm a$. Now, assume that there exist $a,b$ such that $f(a)=a$ and $f(b)=-b$ where $a,b\ne 0$, we have $f(a^2)=af(a)=a^2$ and $f(b^2)=bf(b)=-b^2$, therefore, \[f(a^2+ab-b^2)=f(a^2+ab+f(b^2))=af(b)+b^2+f(a^2)=-ab+b^2+a^2.\]If $f(a^2+ab-b^2)=a^2+ab-b^2$, then $a=b$, so $-a=-b=f(b)=f(a)=a \implies a=0$ which is impossible. If $f(a^2+ab-b^2)=-a^2-ab+b^2$, then $a=0$ which is also impossible. Hence, $\boxed{f(x)\equiv x}$ and $\boxed{f(x)\equiv -x}$ are indeed the only solutions. $\blacksquare$

The answer is $f(x)=x$ and $f(x)=-x$. This can be easily checked to work. Denote $P(a,b)$ the given assertion. Let $f(0)=c$. From $P(a,0)$ we get: $$f(a^2+c)=ac+f(a^2).$$But by $P(-a,0)$ we get: $$f(a^2+c)=-ac+f(a^2).$$Comparing these two equations gives us $c=0$, so $f(0)=0$. Now, comparing $P(1,b)$ and $P(-1,-b)$ we find that $f(b)=-f(-b)$, so $f$ is odd. Then, by $P(0,b)$ we derive: $$f(f(b^2))=b^2.$$Combining this with $f$ odd, we get that $f(f(x))=x$ for all $x$. Therefore $f$ is involutive and also bijective, etc. WIth $P(-b,b)$ we get: $$f(f(b^2))=-bf(b)+b^2+f(b^2).$$Comparing this with $P(0,b)$ we get $f(b^2)=bf(b)$. But if we replace $b$ with $f(b)$ and use $f(f(x))=x$, we get: $$f(b^2)=f(f(b)^2) \implies f(b)^2=b^2 \implies f(x) \in \{x,-x\},$$where we use the injectivity of $f$ here. Now suppose there exist some $r,s \neq 0$ such that $f(r)=r$ and $f(s)=-s$. Since $f$ is odd, WLOG $r,s>0$. From $P(\sqrt{s},\sqrt{r})$ we get: $$f(s+\sqrt{rs}+r)=\sqrt{s}f(\sqrt{r})+r-s.$$Now taking a total of four cases on the value of $f(s+\sqrt{rs}+r)$ and $f(\sqrt{r})$ will give $r=0$, $s=0$, or $r=s$ in each of them, none of which are possible. This leaves $f(x)=x$ and $f(x)=-x$ as the only solutions. $\blacksquare$

Let $P(x,y)$ denotes assertion of given functional equation. Note that $P(1,0)$ and $P(-1,0)$ gives us: $$ f(1+f(0)) =f(0)+f(1)=-f(0)+f(1) \implies f(0) = 0 $$Consequently from $P(0,b)$ it follows $f(f(b^2)) =b^2$ or in other words $f(f(x)) = x$ for all positive reals $x$. Now we seek to gain some more useful identities, thus we take look at $P(a,-a)$: $$ f(f(a^2)) =af(-a) + a^2 +f(a^2) \implies f(a^2) = -af(-a) $$Replacing $a$ by $-a$ gives in last equality yields to: $$ f(a^2) = af(a) = -af(-a) \implies f(-a) = -f(a) $$This helps us since: $$ -f(f(x)) = f(-f(x)) = f(f(-x)) =-x $$Last identity simply implies that we $f(f(x)) =x$ for all negative reals $x$, therefore we conclude that $f(f(x)) =x$ for all real numbers $x$. Now we take $f$ from both sides of $f(a^2) = af(a)$: $$ f(f(a^2)) = f(af(a)) = a^2 $$Finally we replace $a$ by $f(a)$ in the last identity: $$ f(af(a)) = f(a)^2 =a^2 $$This implies that $f(a) =a$ or $f(a) =-a$. Now we are left to escape pointwise trap. Assume that for some non zero real numbers $x,y$ we have $f(x) =x$ and $f(y)=-y$. Consider $P(x,y)$ : $$ f(x^2 + xy + yf(y)) = -xy +y^2 + xf(x) $$This transforms into: $$ f(x^2+ xy -y^2) =-xy+y^2 +x^2 $$If $f(x^2+ xy -y^2) = x^2+xy-y^2$, then $2y^2 = 2xy \implies x=y$ - contradiction, otherwise: $$ -x^2- xy+y^2 = -xy+y^2 +x^2 \implies x=0$$Therefore functions $f(x) =x$ and $f(x) =-x$ are the only ones that work.

Let $P(x,y)$ be the assertion $f(x^2+xy+f(y^2))=xf(y)+y^2+f(x^2)$. $\textbf{1. }f(x^2)=xf(x),f(f(x))=x$ $P(-x,x)\Rightarrow f(f(x^2))=-xf(x)+x^2+f(x^2)$ $P(0,x)\Rightarrow f(f(x^2))=x^2+f(0)$ So $f(x^2)=xf(x)+f(0)$. Plugging in $x=1$ yields $f(0)=0$, hence $f(x^2)=xf(x)$ and $f$ is odd. We immediately get $f(f(x^2))=x^2$ and then $f(f(x))=x$ by oddness. $\textbf{2. }f(x)^2=x^2$ Since $f(f(x^2))=x^2$, we have $f(xf(x))=x^2$, and taking $x\mapsto f(x)$ and comparing yields $f(x)^2=x^2$. $\textbf{3. }f(x)=x\forall x\text{ or }f(x)=-x\forall x$ Assume FTSOC that $\exists a,b$ such that $f(a)=a$, $f(b)=-b$, and $ab\ne0$. Note that $f(a^2)=af(a)=a^2$, and we similarly have $f(b^2)=-b^2$. $P(a,b)\Rightarrow f(a^2+ab-b^2)=-ab+b^2+a^2\Rightarrow (a^2+ab-b^2)^2=(-ab+b^2+a^2)^2\Rightarrow 2(a-b)^2=0$, so $a=b$. This means that $a=f(a)=f(b)=-b=-a$, so $a=0$, contradiction. Then the only solutions are $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$, which both work.

Nice Choose $b \ne 0$. $P(\frac{-f(b^2)}{b},b) \implies f(b^2)f(b)=b^3$. Now put it $b=1 \implies f(1)^2=1 \implies f(1)= \pm 1$. Now put in $b=-1$ to get $f(1)f(-1)=-1$. So $f(-1) = \mp 1$. Now $P(1,0) \implies f(f(1))=1+f(0)$. In any case, $f(f(1))=1 \implies f(0)=0$. Now $P(0,b) \implies f(f(b^2))=b^2$. Also by putting in $a=-b$, we get $f(x^2)=xf(x) \forall x$. Recall that we had obtained $f(b^2)f(b)=bf(b)^2=b^3 \implies f(b) = \pm b$. If $f(a)=a, f(b)=-b$ for some $a,b \ne 0$, we get a contradiction by $P(a,b)$. So, $f(x)=+x$ and $f(x)=-x$ are the solutions, which clearly work $\Box$

Let $P(a,b)$ denote the assertion. $P(a,0)$ and $P(-a,0)$ gives $f(0) = 0$. Comparing $P(a,b)$ and $P(-a,-b)$ gives $f$ is odd. $P(0,b) \implies f(f(b^2)) = b^2$ and so $f(f(x)) = x$ for all $x\geq 0$. But since $f$ is odd, we have $f(f(-x)) = f(-f(x)) = -f(f(x)) = -x$ and so $f(f(x)) = x$ for all $x\in \mathbb{R}$, so $f$ is bijective. $P(-b,b) \implies b^2 = f(f(b^2)) = -bf(b) + b^2 + f(b^2) \implies f(b^2) = bf(b)$ for all $b \in \mathbb{R}$. So, $f(a^2) = af(a) \implies f(f(a)^2) = af(a) = f(a^2)$ and since injective, we have $f(a)^2 = a^2 \implies f(a) \in \{a, -a\}$ for all $a\in \mathbb{R}$. Clearly, $f(a) = a \ \forall a \in \mathbb{R}$ and $f(a) = -a \ \forall a \in \mathbb{R}$ are solutions. Lets assume for $a , b \neq 0$, $f(a) = a$ and $f(b) = -b$. Then $P(a,b)$ gives us $a=b$ or $a=0$, contradiction. So we have our 2 solutions $\boxed{f(a) = a \ \forall a \in \mathbb{R}}$ and $\boxed{f(a) = -a \ \forall a \in \mathbb{R}}$.

Physicsknight wrote: Determine all the functions $f:\mathbb R\mapsto\mathbb R$ satisfies the equation $f(a^2 +ab+ f(b^2))=af(b)+b^2+ f(a^2)\,\forall a,b\in\mathbb R $ Cute Let \(P(a,b)\) be the assertion of the functional equation. Then. \(P(0,b)\) gives us that \(f(f(b^2))=b^2+f(0)\). \(P(-b,b)\) also gives us that \[f(f(b^2))=-bf(b)+b^2+f(b^2)\]so \(f(b^2)=bf(b)+f(0)\). Put \(b=1\) in this to get \(f(0)=0\). Thus, \(f(b^2)=bf(b)\). Now, \(P(\frac{-f(b^2)}{b},b)\) gives us that \[f(b)f(b^2)=b^3\]or \(f(b)^2=b^2\), implying that \(f(b)=\pm b\). If \(f(n)=n\) for some \(n\) and \(f(m)=-m\) for some \(m\), \(P(m,n)\) gives us the desired contradiction.

Physicsknight wrote: Determine all the functions $f:\mathbb R\mapsto\mathbb R$ satisfies the equation $f(x^2 +xy+ f(y^2))=xf(y)+y^2+ f(x^2)\,\forall x,y\in\mathbb R $ 19:01 Let $P(x,y)$ denote the given assertion. $P(0,0): f(f(0))=f(0)$. $P(x,0): f(x^2+f(0))=xf(0)+f(x^2)$. $P(0,x): f(f(x^2))=x^2+f(0)$. Thus, $f(f(x))=x+f(0)$ for all nonnegative reals $x$. Case 1: $f(0)\ge 0$. Then $f^3(0)=2f(0)\implies f(0)=2f(0)\implies f(0)=0$. Thus, $f(f(x))=x$ for all nonnegative reals $x$. $P(x,-x): f(f(x^2))=x^2=xf(-x)+x^2+f(x^2)\implies -xf(-x)=f(x^2)$. $P(-x,x): f(f(x^2))=x^2=-xf(x)+x^2+f(x^2)\implies xf(x)=f(x^2)$. So $xf(x)=-xf(-x)$. Since $f(0)=0$, $f$ is odd and an involution. The FE can be rewritten as $f(x^2+xy+yf(y))=xf(y)+y^2+xf(x)$. Let $Q(x,y)$ denote this new assertion. $Q(-f(x),x): f((-f(x))^2)=-f(x)^2+x^2+(-f(x))f(-f(x))=-f(x)^2+x^2+f((-f(x))^2)\implies x^2=f(x)^2$. So $f(x)=x$ or $f(x)=-x$ for each $x$. If $f(a)=a$ and $f(b)=-b$ with $a,b\ne 0$, then $Q(a,b): f(a^2+ab-b^2)=a^2-ab+b^2$. If $a^2+ab-b^2=a^2-ab+b^2$, then $a-b=b-a\implies a=b$, a contradiction. If $-a^2-ab+b^2=a^2-ab+b^2$, then $a^2=0$, a contradiction. So the only solutions here are $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$, which work. Case 2: $f(0)<0$. $P(-x,x): f(f(x^2))=-xf(x)+x^2+f(x^2)$. Now we know $f(f(x^2))=x^2+f(0)$, so $f(0)=-xf(x)+f(x^2)$. Set $x=1$ here to achieve $f(0)=0$, a contradiction.

$P(a,-a) : f(f(a^2)) = af(-a) + a^2 + f(a^2)$ $P(a,0) : f(f(a^2)) = a^2 + f(0)$ so $f(a^2) + af(-a) = f(0)$. $P(-a,a),P(-a,0) : f(a^2) - af(a) = f(0)$ so now we have $f(a) = -f(-a)$. $a = 1 : f(1) - 1f(1) = f(0) \implies f(0) = 0$ so $P(a,0) : f(f(a^2)) = a^2 + f(0) = a^2$ and $f(a^2) = af(a)$. Note that $f(f(a^2)) = a^2$ covers all nonnegative numbers so for all $n \ge 0$ we have $f(f(n)) = n$. we had $f(a^2) = af(a)$ now put $a = f(a)$ so we have $f(f(a)^2) = f(a)f(f(a)) = af(a) = f(a^2) \implies f(f(f(a)^2)) = f(f(a^2))$. Note that both $f(a)^2 , a^2$ are nonnegative so $f(a)^2 = a^2 \implies f(a) = \pm a$ for all nonnegative $a$ but back we proved $f(a) = -f(-a)$. so we have $f$ for negative values as well. Answers : $f(a) = \pm a$.

Can we change the title of this post? Having the solution set (however obvious) of a functional equation as the title seems like a major and completely unnecessary giveaway.

Let $P(a,b)$ denote the given assertion. $P(1,0)-P(-1,0)\implies f(0)=0.$ $P(0,x)\implies f(f(x^2))=x^2.$ $P(x,-x)\implies f(x^2)=xf(x).$ So $f(f(x))=x~\forall x$ by oddness. Also $f(xf(x))=x^2.$ Setting $x=f(x) \implies f(x)=\pm x.$ Note that existence of $u,v:f(u)=u, f(v)=-v$ is absurd by $P(u,v).$ Thus $f\equiv \text{Id}$ and $f\equiv -\text{Id}$, indeed work.

Here is my solution: https://calimath.org/pdf/RMM-SL2019-A1.pdf And I uploaded the solution with motivation to: https://youtu.be/3RjpZHIIVEo

Eazy P(0,b) + P(a,0)+P(-f(b^2)/b ,b)+P(a,-a)+P(1) kill