$P(0,y) \implies $ either $f(x) \equiv 1-x$ or $f(0)=0$ if $f(0)=0 \implies P(x,-x) \implies xf(x)=f(x)f(-x)$ and $xf(-x)=-f(x)f(-x)$ which implies that $f(x)=0,-x$ now if $f(\theta)=0 \implies P(\theta,y-\theta) \implies f(\theta \cdot y)=\theta \cdot f(y)$ and if $f(\alpha)=-\alpha \implies P(x-\alpha,\alpha) \implies f(\alpha \cdot x)=\alpha \cdot f(x)$ so $$f(xy)=xf(y)=yf(x)$$done

Why is it done when f(xy)=xf(y)?

Plug in $x=0$. We get that either $f(0) = 0$ or $f(x) = 1-x$ for all real $x$, which is a solution. Now assume $f(0)=0$. Plug in $y=-x$. This gives $f(x)(f(-x)-x)=0$, so for each real $x$, $f(x)=0$ or $f(-x)=x$. So for every real, $f(x)$ can take a value of either $0$ or $-x$. Plugging in $y=0$, we have $f(x^2)=xf(x) = -xf(-x)$, so $f$ is odd. (Also true for $x=0$) Plug in $x=1$. The first and last terms cancel to give that $f(1)=0$ or $f(y)=-y$ for all real $y$, which is a solution. So assume $f(0)=f(1)=0$. Now plug in $y=1-x$. This gives $xf(x) = f(x)(1-x)$. So if we don't have $f(x)=0$, then $f(1-x) = x$. But since we had that $f(1-x) = 0$ or $f(1-x)=x-1$, we definitely have $f(1-x)=0$ and thus $x=0$. However this leads to $f(1)=0$, which is a contradiction. So for all $x$, we have $f(x)=0$ in this case. Thus the solutions are $$f(x)=-x \quad \forall x \in \mathbb{R}$$$$f(x)=1-x \quad \forall x \in \mathbb{R}$$and $$f(x)=0 \quad \forall x \in \mathbb{R}$$

Paradoxes wrote: Why is it done when f(xy)=xf(y)? well in that case we figure out that $\frac{f(x)}{x}=Cons.$ and we knew that $f(x)$ is either $-x$ or 0 for a number $x$ so it tells us that there is only one of them true for all $x$

Here is my solution for this problem Solution $f(x^2 + xy) = f(x)f(y) + yf(x) + xf(x + y)$ Let $x = y = 0$, we have: $f(0) = f^2(0)$ So: $f(0) = 0$ or $f(0) = 1$ If $f(0) = 1$: Let $x = 0$, we have: $f(0) = f(0)f(y) + yf(0)$ or $f(x) = 1 - x, \forall x \in \mathbb{R}$ If $f(0) = 0$: Let $y$ be $- x$, we have: $f(x)f(- x) - xf(x) = 0$ Then: $f(x) = 0$ or $f(x) = - x$ Suppose there exist $a, b \in \mathbb{R}$, $a, b \ne 0$, which satisfy $f(a) \ne 0$, $f(b) \ne - b$ So: $f(a) = - a$, $f(b) = 0$ Let $x = b$, $y = - a$, we have: $f(b^2 - ab) = bf(b - a)$ Let $x = b - a$, $y = a$, we have: $f(b^2 - ab) = - af(b - a) + af(b - a) + (b - a)f(0) = 0$ So: $f(b - a) = 0$ Let $x = a$, $y = b - a$, we have: $f(ab) = - af(b - a) + a(a - b) = a(a - b)$ Let $x = b$, $y = a - b$, we have: $f(ab) = - ab$ Then: $a(a - b) = - ab$ or $a = 0$, conflict with $a \ne 0$ Retry, we see that $f(x) = 1 - x$, $f(x) = 0$, $f(x) = - x$ satisfy the problem In conclusion, we have: $f(x) = 1 - x, f(x) = 0, f(x) = - x, \forall x \in \mathbb{R}$

Let $P(x,y)$ denote assertion of given functional equation. 1) $P(0,0)$: $f(0)=f(0)^2$, which implies that $f(0)=0$ or $f(0)=1$ Case 1 $f(0)=1$ 2) $P(0,y)$: $f(y)=1-y$ for all real numbers $y$, which clearly works Case 2: $f(0)=0$ 3)$P(x,0)$: $f(x^2)=xf(x)=-xf(-x)$, which implies that $-f(x)=f(-x)$ 4)$P(x,-x)$: $0=f(x)f(-x)-xf(x)=-f(x)^2-xf(x)$, which gives us $f(x)=0$ or $f(x)=-x$ 5) It remains to escape pointwise trap. We assume that $f(a)=-a$ and $f(b)=0$, where $a,b$ are not $0$. $P(a,b)$ gives us: $f(a^2+ab)=-ab+af(a+b)$ Case A $f(a^2+ab)=0=f(a+b)$ Then $ab=0$ - contradiction Case B $f(a^2+ab)=-a^2-ab$ and $f(a+b)=0$ Then $-a^2=0$ - contradiction Case C $f(a^2+ab)=-a^2-ab$ and $f(a+b)=-a-b$ Then $-ab=0$ - contradiction Case D $f(a^2+ab)=0$ and $f(a+b)=-a-b $ Then $-2ab = a^2$, which gives us that $a=-2b$. As a result $f(a^2+ab)=f(2b^2)=0$. $P(b,b)$ gives $f(2b^2)=f(b)^2+bf(b)+bf(2b)$, which is equivalent to $f(2b)=0$, but $f(-a)=f(2b)=-f(a)=a=0$ - contradiction Answers: $f(x)=1-x, f(x)=-x, f(x)=0$

Kimchiks926 wrote: Case A $f(a^2+ab)=0=f(a+b)$ Then $ab=0$ - contradiction Case B $f(a^2+ab)=-a^2-ab$ and $f(a+b)=0$ Then $-a^2=0$ - contradiction Case C $f(a^2+ab)=-a^2-ab$ and $f(a+b)=-a-b$ Then $-ab=0$ - contradiction Case D $f(a^2+ab)=0$ and $f(a+b)=-a-b $ Then $-2ab = a^2$, which gives us that $a=-2b$. As a result $f(a^2+ab)=f(2b^2)=0$. $P(b,b)$ gives $f(2b^2)=f(b)^2+bf(b)+bf(2b)$, which is equivalent to $f(2b)=0$, but $f(-a)=f(2b)=-f(a)=a=0$ - contradiction Answers: $f(x)=1-x, f(x)=-x, f(x)=0$ Instead of splitting into so many cases at the end, splitting into 2 cases $f(1)=1$ or $f(1)=0$ works very well. In fact for these you just need $P(1,y)$

For me it was hard Solution. We claim that the only solutions are $f(x)=-x$ and $f\equiv 0$ and $f(x)=1-x$ for all reals. It's easy to check that they indeed works. Let $P(x,y)$ denote the assertion. We have, by $P(0,0)$ that $f(0)=0$ or $f(0)=1$. We deal with this via cases. Case (a): If $f(0)=1$ then by $P(0.y)$ we have $f(x)=1-x~\forall x\in\mathbb{R}$ as our only solution. $\square$ Case (b): If $f(0)=0$ then by $P(x,0)$ we have $f(x^2)=xf(x)$. Note that this gives $-f(x)=f(-x)$ by replacing $x \to -x$ whence we have $f$ is odd. Now combining $P(x,-x)$ with $f$ odd property, gives $-f(x)^2=xf(x)~(\clubsuit)$.Substitute $x=1$ in $(\clubsuit)$ gives $f(1)=0$ or $f(1)=-1$. We further divide our discussion into following two cases: If $f(1)=-1$. Now assume $f(u)=0$ for some $u$. We claim that this $u$ is nothing but $0$. Notice that by $P(u,1-u)$ we have $$0=f(u)=f(u)f(1-u)+(1-u)f(u)+uf(1)=u(-1)=-u$$whence we get $u=0$ as desired. With this, coming back to $(\clubsuit)$ we have $f(x)=-x$ for $x\neq 0$ but already we have $f(0)=0$. Therefore, here we get $f(x)=-x~\forall x\in \mathbb{R}$ as desired. $\square$ If $f(1)=0$ then via $P(1-x,x)$ we have, again, $$f(1-x)[1-f(x)-x]=xf(1)=0~~~~~~(1)$$We claim that if $\exists v\in \mathbb{R}$ such that $f(v)=1-v$. Then $v=1$. For this via $(\clubsuit)$ we have $$-(1-v)^2-f(v)^2=vf(v)=v(1-v) \implies v=1$$as desired which is already (assumed) given. Coming back to $(1)$ and combining it with previous result gives $f(1-x)=0$ for $x \neq 1$ but then again since $f(0)=0$ is already (assumed) given we have $f(x)=0~\forall x\in \mathbb{R}$ as desired. $\square$ which finishes the problem. $\blacksquare$

Cute problem The solutions are $f(x)=-x$, $f(x)=1-x$ and $f(x)=0$ for all real numbers. Plug in $x=y=0$, to get that $f(0)=f(0)^2$ Our first case scenario is that of $f(0)=0$ Plug in $y=0$ to get that $f(x^2)=xf(x)$, this implies that $f$ is an odd function. Now plug in $y=-x$ to get that $f(x)f(-x)-xf(x)=0$, this gives us that $f(x)(-f(x)-x)=0 \implies f(x)(f(x)+x)=0$. Now let's claim that for the vast majority of the real numbers we have that $f(x)=0$, but there exists some $k$ such that $f(k)=-k$. We denote a set $S_1=\{x \mid f(x)=0\}$, and we denote a set $S_2=\{x \mid f(x)=-x\}$. Plug in so that $x$ is in $S_1$ and $y$ so that it is in $S_2$, now by quick checking we see that $f(x)=0$, for all real numbers $x$. Now let's claim that for the vast majority of the real numbers we have that $f(x)=-x$, but there exists some $k$ such that $f(k)=0$. We denote a set $S_1=\{x \mid f(x)=0\}$, and we denote a set $S_2=\{x \mid f(x)=-x\}$. Plug in so that $x$ is in $S_1$ and $y$ so that it is in $S_2$, now by quick checking we see that $f(x)=-x$, for all real numbers $x$. Our second case scenario is that of $f(0)=1$. Set $y=0$, to get that $f(x^2)=(x+1)f(x)$, now we set $x=-1$ to get that $f(1)=0$. From the above relation we have that: $$(x+1)f(x)=(1-x)f(-x) \implies f(-x)=\frac{x+1}{1-x}f(x)$$Now we set $y=-x$ to get the following: $$1=\frac{x+1}{1-x}f(x)^2-xf(x)+x$$now by solving this quadratic we have that $f_1=1-x$, and that $f_2=\frac{x-1}{x+1}$. Now assume that we have two sets $S_1$ and $S_2$ such that $S_1=\{x \mid f(x)=1-x\}$ and $S_2=\{x \mid f(x)=\frac{x-1}{x+1}\}$, now by plugging in all sorts of combinations we get that the only function (in this case) is $f(x)=1-x$.

Solution: We claim that $f(x)=1-x$, $f(x)=-x$ and $f(x)=0$ for all real $x$ are the only solutions. It is easy to see that they all satisfy the given functional equation. We now prove that they are the only solutions. Let $P(x,y)$ be the given assertion, we have \[P(0,0) \Rightarrow f(0)=0 , 1 \]Case 1: $f(0)=1.$ Using $P(0,x)$, we get \[f(0)=f(0)f(x)+xf(0) \Rightarrow f(x)=1-x \ \forall x\in \mathbb{R}.\]Case 2: $f(0)=0.$ Using $P(x,-x)$, we get \[f(0)=f(x)f(-x)-xf(x)+xf(0) \Rightarrow f(x)=0, -x \ \forall x\in \mathbb{R}\] Assume that there exist $a,b \ne 0$ such that $f(a)=0, f(b)=-b$, $P(a,b)$ gives \[f(a^2+ab)=af(a+b)\]Case a: $f(a^2+ab)=0.$ This implies that $a=0$ or $f(a+b)=0$, the former is impossible since $a \ne 0$, so $f(a+b)=0$. However, evaluating $P(b,a)$ gives \[f(b^2+ab)=bf(a+b)-ab=-ab\]and so $-b^2-ab=-ab$ or $0=-ab$, in either case, we will end up $a=0$ or $b=0$ which is impossible. Case b: $f(a^2+ab)=-a^2-ab.$ Thus implies that $-a-b=f(a+b)$, and $P(b,a)$ gives \[f(b^2+ab)=bf(a+b)-ab=-2ab-b^2\]and so either $-b^2-ab=-2ab-b^2$ or $0=-2ab-b^2$, the former implies $a=0$ or $b=0$ which is impossible. So, $b=-2a$, $f(2a^2)=0$, $f(b)=f(-2a)=2a$, evaluating $P(a,a)$ gives \[0=f(2a^2)=f(a)^2+af(a)+af(2a) \Rightarrow f(2a)=0\]but this means $2a=f(-2a)=-f(2a)=0 \Rightarrow a=0$, a contradiction. Hence, $f(x)=0$, $f(x)=-x$ and $f(x)=1-x$ are indeed the only solutions.$\blacksquare$

Amir Hossein wrote: Determine all functions $f : \mathbb R \to \mathbb R$ such that for all real numbers $x$ and $y$, $$f(x^2 + xy) = f(x)f(y) + yf(x) + xf(x+y).$$Proposed by Walther Janous, Austria Let $P(x,y)$ the assertion: $P(1,x)$ $$f(1+x)=f(1)f(x)+xf(1)+f(x+1) \implies f(1)(f(x)+x)=0 \implies f(1)=0 \; \text{or} \; f(x)=-x$$. $P(0,x)$ $$f(0)=f(0)f(x)+xf(0) \implies f(x)=1-x \; \text{or} \; f(0)=0$$. Here we can have $3$ thinks: $f(0)=0$ , $f(1)=0$ or $f(0)=0$ and $f(1)=0$ at the same time. Note that $f(x)=1-x$ works for $f(1)=0$ then done in this case. Note that $f(x)=-x$ works for $f(0)=0$ (Note this also works for another sol) then done for this case. Now we will work for $f(1)=f(0)=0$. $P(x, -x)$ $$f(0)=f(x)f(-x)-xf(x)+xf(0) \implies f(x)f(-x)=xf(x)$$Note that $f(-x)=x$ its part of $f(x)=-x$ then $f(x) \equiv 0$. Here is the list of all solutions . $f(x) \equiv 0$. $f(x)=-x$. $f(x)=1-x$. For all $x \in \mathbb R$ thus we are done

A very good instructive problem about pointwise trap(FE pointwise trap can get excessively hard, such as 2016 J3) The answers $f\equiv -x, 1-x, 0$. These work. Denote the assertion by $P(x,y)$. $P(0,0)$ gives that $f(0)=0$ or $f(0)=1$. Case 1: $f(0)=1$ Setting $P(0,y)$ gives that $f(x)=1-x.$ Case 2: $f(0)=0$ $P(x,0)$ gives that $f(x^2)=xf(x),$ so $f$ is odd. Then $P(x,-x)$ gives that $f(x)(f(x)+x)=0,$ so $f(x)=0, -x$. To avoid pointwise trap, suppose that $f(a)=a$ and $f(b)=-b$ for $a,b \neq 0$. From $P(a,b),$ we get that $f(a^2+ab)=af(a+b),$ so either both $f(a^2+ab)=-a^2-ab$ and $f(a+b)=-a-b$ or $f(a^2+ab)=f(a+b)=0.$ $P(b,a)$ gives that $f(b^2+ba)=f(b)a+bf(a+b)=-ab+bf(a+b).$ If $f(a+b)=0,$ we get that $f(b^2+ab)=-ab,$ so this means that $b^2+ab=ab,$ which means that $b=0,$ absurd. Therefore, $f(a^2+ab)=-a^2-ab$ and $f(a+b)=-a-b.$ Substituting, we get that $f(b^2+ab)=-b(2a+b)$. If $f(b^2+ab)=-b^2-ab,$ we get $ab=0,$ absurd, so we have that $f(b^2+ab)=-b^2-ab=0.$ This means that $b=-2a.$ To finish it off, we now have that $f(a)=0$ and $f(-2a)=2a.$ $P(a,a)$ gives that $f(2a^2)=af(2a)=-2a^2$($f$ is odd). But from $P(2a, -a),$ we get that $f(2a^2)=f(2a)f(-a)-af(2a)+2af(a)\implies f(2a^2)=-af(2a)=2a^2.$ Therefore, $2a^2=-2a^2,$ absurd. We are done.

$P(0,x)\Rightarrow\boxed{f(x)=1-x}$ or $f(0)=0$. (assume the latter) $P(x,-x)\Rightarrow xf(x)=f(x)f(-x)\Rightarrow xf(x)=f(x)f(-x)=-xf(-x)$, so $f$ is odd and thus $f(x)^2=-xf(x)$. This means $f(x)\in\{0,-x\}$. Assume that some $a,b\ne0$ exist with $f(a)=0,f(b)=-b$. Then: $P(b,a)\Rightarrow f(b^2+ab)=-ab+bf(a+b)$ If $f(a+b)=0$, then $f(b^2+ab)=-ab\notin\{0,-b^2-ab\}$. Then $f(a+b)=-a-b$, so $f(b^2+ab)=-2ab-b^2$. If $-2ab-b^2=-b^2-ab$, we have a contradiction, so for our assumption to hold, we need $-2ab-b^2=0$, or $b=2a$. This holds for all $a,b$ such that $f(a)=0$ and $f(b)=-b$ $P(a,a)\Rightarrow f(2a^2)=af(2a)=-2a^2$, but setting $b=2a^2$ we have $2a^2=2a$, hence $a=1$. Since the function $f(x)=\begin{cases}0&\text{if }x=1\\-x&\text{if }x\ne1\end{cases}$ doesn't work, we need either $\boxed{f(x)=0}$ or $\boxed{f(x)=-x}$. All boxed solutions can be verified to work.

Amir Hossein wrote: Determine all functions $f : \mathbb R \to \mathbb R$ such that for all real numbers $x$ and $y$, $$f(x^2 + xy) = f(x)f(y) + yf(x) + xf(x+y).$$Proposed by Walther Janous, Austria Pretty easy.

Let P(x, y) be an assertion P(0, 0): f(0)=f(0)^2 if f(0)=0, P(x, -x) gives that f(x)=0 or f(x)=-x for all real x if f(0)=1, P(0, y) gives that f(y)=1-y for all real y all three function indeed satisfy

rama1728 wrote: Amir Hossein wrote: Determine all functions $f : \mathbb R \to \mathbb R$ such that for all real numbers $x$ and $y$, $$f(x^2 + xy) = f(x)f(y) + yf(x) + xf(x+y).$$Proposed by Walther Janous, Austria Pretty easy.

Your solution is incorrect. While proving claim 3, you show that for each real $x,$ (individually) $f(x)=0$ or $f(x)=-x.$ You cannot directly conclude that $f\equiv 0$ or $f\equiv -\text{id}.$ You have to see what happens when for some $a,b\neq 0, \ f(a)=-a$ and $f(b)=0.$ JasperE3 very nicely dealt with this case in his solution.

$P(0,y) : f(0) = f(0)f(y) + yf(0) \implies f(0) = f(0)(f(y)+y)$ so either $f(0) = 0$ or $f(y) = 1-y$. By checking $f(y) = 1-y$ we can see it's one of the answers so lets assume $f(0) = 0$. $P(x,-x) : 0 = f(x)f(-x) + -xf(x) \implies f(x)(f(x) - x)=0$ so either $f(x) = 0$ or $f(x) = -x$. Assume there exists $a,b$ which are not $0$ and $f(a) = -a, f(b) = 0$. $P(b-a,a) : f(b^2 - ab) = (b-a)f(0) = 0$. $P(b,-a) : f(b^2 - ab) = bf(b-a) \implies f(b-a) = 0$. $P(a,b-a) : f(ab) = a(a-b) = a^2 - ab$. $P(b,a-b) : f(ab) = bf(a) = -ab$. so we have $a^2 - ab = -ab$ or $a = 0$ which gives contradiction. Answers : $f(x) = 1-x, f(x) = 0, f(x) = -x$.

Case1. $f(x)-is$ $constant$ $\iff$ $f(x)=c$ , $c-cosntant$ $c=c^2+xc+yc \iff c(c+x+y-1)=0 \implies c=0 \vee c+x+y-1 \implies c=0$ So $f(x)=c=0 \implies f(x)=0 \ \forall x\in \mathbb{R}$ Case2. $f(x)-is$ $not$ $constant$ $\iff$ $f(x) \neq 0$ Let $P(x,y)$ denote the given assertion. $P(0,0) \implies f(0)=f(0)^2 \implies f(0)(f(0)-1)=0 \implies f(0)=0 \vee f(0)-1=0 \implies f(0)=0 \vee f(0)=1$ Case2.1 $f(0)=0$ $P(x,-x) \implies 0=f(x)f(-x)-xf(x) \iff f(x)f(-x)=xf(x)$ $...(1)$ $P(-x,x) \implies 0=f(x)f(-x)+xf(-x) \iff f(x)f(-x)=-xf(-x)$ $...(2)$ Combining $(1)$ with $(2)$ we get: $f(x)f(-x)=f(x)f(-x) \implies xf(x)=-xf(-x) \implies xf(-x)=-xf(x) \implies f(-x)=-f(x) \ \forall x\in \mathbb{R}/{0}$ So we have $f(x)f(-x)=f(x)*(-f(x))=-f(x)^2 \implies f(x)f(-x)=-f(x)^2 \ \forall x\in \mathbb{R}/{0}$ Combining with $(1)$ $xf(x)=-f(x)^2$ . We can divide with $f(x)$ because $f(x) \neq 0$ . So we get $f(x)=-x \ \forall x\in \mathbb{R}/{0}$ Combining with $f(0)=0$ we get: $f(x)=-x \ \forall x\in \mathbb{R}$ Case2.2 $f(0)=1$ $P(0,x) \implies f(x)=1-x\ \forall x\in \mathbb{R}$

Look $P(0,0)$; $f(0)=f(0)^2$ $There$ $are$ $2$ $cases$; $f(0)=0$ or $f(0)=1$ You can search these cases (and also there will be subcases) For $P(0,x)$; $f(0)=f(0)*f(x)+x*f(0)=f(0)*(f(x)+x)$ $f(0)=0$ pays. Then if $f(0)=1$ $f(x)+x=1$ and $f(x)=1-x$ Then for $P(x,-x)$; $f(x^2-x^2)=f(0)=f(x)*f(-x)-x*f(x)+x*f(0)$ $IF$ $f(0)=0$ $f(x)*f(-x)=x*f(x)$ From there we get; $f(x)\neq0$ $-->$ $f(-x)=x$ $-x--->x$ $f(x)=-x$ $Also$ $f(x)=0$ After finding you have to check them (in the main functional equation)

answers are 0, -x and 1-x, which trivially work. P(x,y) is assertion its trivial to see that 0, -x, 1-x are only linear solutions, so assume f isn't linear P(0,y) yields f(0) = f(0) f(y) + yf(0). since f isnt linear, f(0) = 0. P(x,0) yields f(x^2) = xf(x). P(1,y) yields f(1) f(y) + yf(1) = 0, so f(1) = 0 P(x,1-x) yields f(x) = f(x) f(1-x) + (1-x) f(x) so for each x either f(x) = 0 or f(1-x) = x. P(x,-x): f(x)f(-x) - x f(x) = 0, so comparing x and -x here gives f(x) = -f(-x) for x\ne 0, but it's also true for x = 0, so f is odd now going back gives that -f(x)^2 - xf(x) = 0 so f(x) is 0 or -x for each x. now if f(x) neq 0 for some x we then have f(1-x) = x, now since -(1-x) neq x we have x = 0 , absurd thus its constant at 0 and were done