I claim all $f$ are of the following form: take any real $c$, then $f(x)=cx$ for all $x$. Clearly if $f$ is constant, then taking $x+y\ne 3$, we find $f\equiv 0$. Hence assume $f$ is non-constant; let $P(x,y)$ denotes the given assertion. Note that $P(0,y)$ yields (by taking $y\ne 3$), $f(0)=0$. This, together with $P(x,0)$ then yields $f(x^2) = xf(x)$. From here, taking $-x$ in place of $x$, we find $f(-x)=-f(x)$ for all $x$. Next, comparing $P(x,y)$ and $P(x,-y)$, and using $f(-xy)=-f(xy)$ we obtain \[ f(x^2)\pm 2f(xy) = xf(x\pm y) \pm yf(x). \]Summing them and recalling $f(x^2)=xf(x)$, we find \[ f(x) = \frac{f(x+y)+f(x-y)}{2}. \]Now, taking $x=1$ here, we have that $f(y+1)-f(y-1)=2f(1)$ where we used the fact $f(1-y)=-f(y-1)$. We are now ready to conclude. Comparing $P(x,1)$ and $P(x,-1)$ like above, we find \[ 4f(x) = x\bigl(f(x+1)-f(x-1)\bigr) +2f(x) \implies f(x)=xf(1). \]Namely, there is a constant $c$ such that $f(x)=cx$ for all $x$. From here, any $c$ is seen to work as the equation is homogeneous then. Any valid solution is of form $f(x)=cx$, concluding the problem.

grupyorum wrote: I claim either $f\equiv 0$ or $f(x)=x$ for all $x$. Clearly if $f$ is constant, then taking $x+y\ne 3$, we find $f\equiv 0$. Hence assume $f$ is non-constant; let $P(x,y)$ denotes the given assertion. Note that $P(0,y)$ yields (by taking $y\ne 3$), $f(0)=0$. This, together with $P(x,0)$ then yields $f(x^2) = xf(x)$. From here, taking $-x$ in place of $x$, we find $f(-x)=-f(x)$ for all $x$. Next, comparing $P(x,y)$ and $P(x,-y)$, and using $f(-xy)=-f(xy)$ we obtain \[ f(x^2)\pm 2f(xy) = xf(x\pm y) \pm yf(x). \]Summing them and recalling $f(x^2)=xf(x)$, we find \[ f(x) = \frac{f(x+y)+f(x-y)}{2}. \]Now, taking $y=1$ here, we have $f(y+1)-f(y-1)=2f(1)$. We are now ready to conclude. Comparing $P(x,1)$ and $P(x,-1)$ like above, we find \[ 4f(x) = x\bigl(f(x+1)-f(x-1)\bigr) +2f(x) \implies f(x)=xf(1). \]Namely, there is a constant $c$ such that $f(x)=cx$ for all $x$. From here, it is easily seen $c\in\{0,1\}$, yielding the answers claimed at the beginning. You should fix your solution since $f(x)=cx$ hold for any constant $c$. And you still have some mistakes like on step when you get $y=1$ it transform to $f(x+1)+f(x-1)=2f(x)$ and not as you wrote.

dangerousliri wrote: grupyorum wrote: I claim either $f\equiv 0$ or $f(x)=x$ for all $x$. Clearly if $f$ is constant, then taking $x+y\ne 3$, we find $f\equiv 0$. Hence assume $f$ is non-constant; let $P(x,y)$ denotes the given assertion. Note that $P(0,y)$ yields (by taking $y\ne 3$), $f(0)=0$. This, together with $P(x,0)$ then yields $f(x^2) = xf(x)$. From here, taking $-x$ in place of $x$, we find $f(-x)=-f(x)$ for all $x$. Next, comparing $P(x,y)$ and $P(x,-y)$, and using $f(-xy)=-f(xy)$ we obtain \[ f(x^2)\pm 2f(xy) = xf(x\pm y) \pm yf(x). \]Summing them and recalling $f(x^2)=xf(x)$, we find \[ f(x) = \frac{f(x+y)+f(x-y)}{2}. \]Now, taking $y=1$ here, we have $f(y+1)-f(y-1)=2f(1)$. We are now ready to conclude. Comparing $P(x,1)$ and $P(x,-1)$ like above, we find \[ 4f(x) = x\bigl(f(x+1)-f(x-1)\bigr) +2f(x) \implies f(x)=xf(1). \]Namely, there is a constant $c$ such that $f(x)=cx$ for all $x$. From here, it is easily seen $c\in\{0,1\}$, yielding the answers claimed at the beginning. You should fix your solution since $f(x)=cx$ hold for any constant $c$. And you still have some mistakes like on step when you get $y=1$ it transform to $f(x+1)+f(x-1)=2f(x)$ and not as you wrote. Fixed all $c$ part. However, for \[ f(x) = \frac{f(x+y)+f(x-y)}{2}. \]part; I actually let $x=1$ (not $y=1$) which yields $2f(1)= f(y+1)+f(1-y) = f(y+1)-f(y-1)$, where the last step uses $f(1-y)=-f(y-1)$.

grupyorum wrote: part; I actually let $x=1$ (not $y=1$) which yields $2f(1)= f(y+1)+f(1-y) = f(y+1)-f(y-1)$, where the last step uses $f(1-y)=-f(y-1)$. grupyorum wrote: Now, taking $y=1$ here, we have $f(y+1)-f(y-1)=2f(1)$.

dangerousliri wrote: Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all real numbers $x$ and $y$, $$f(x^2)+2f(xy)=xf(x+y)+yf(x).$$ $P(0,y):$ gives $f(0)=0$ $P(x,0)$ gives $f(x^2)=xf(x)$ (1) $P(1,y)$ gives $f(y+1)=2f(y)-yf(1)+f(1)$ (2) $P(x,1)$ gives$f(x^2)+f(x)=xf(x+1)=x[2f(x)-xf(1)+f(1)]$ which gives $f(x)(x-1)=xf(1)(x-1)\Rightarrow f(x)=xc$ for $\forall x\varepsilon \mathbb{R}-{1}$ now take $x+y=1$ we have $f(1)=c$ So $f(x)=cx$ for every real number $x$

dangerousliri wrote: Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all real numbers $x$ and $y$, $$f(x^2)+2f(xy)=xf(x+y)+yf(x).$$ Nice one , is it yours? We claim that $f(x)=cx$ for any real $c$ works. Let $P(x,y)$ the assertion of the F.E. Claim 1: $f(0)=0$ Proof: By $P(0,x)$ $$3f(0)=xf(0) \; \forall x \in \mathbb R \implies f(0)=0$$Claim 2: $f(x^2)=xf(x)$ and $f$ odd. Proof: By $P(x,0)$ $$f(x^2)=xf(x) \; \forall x \in \mathbb R$$And on this result if we switch $x$ with $-x$ we get $$-xf(-x)=f(x^2)=xf(x) \implies -f(x)=f(-x) \; \forall x \in \mathbb R \implies f \; \text{odd}$$Claim 3: $2f(x)=f(x+y)+f(x-y)$ Proof: By $P(x,y)+P(x,-y)$ $$2xf(x)=xf(x+y)+xf(x-y) \implies 2f(x)=f(x+y)+f(x-y) \; \forall x,y \in \mathbb R$$Claim 4: $f$ is additive on $\mathbb R$ First note that by swiching $x,y$ on Claim 3 u get $2f(y)=f(x+y)-f(x-y)$ and adding this with Claim 3 we get $f(x+y)=f(x)+f(y)$ for every reals $x,y$ so $f$ is additive on $\mathbb R$ Finishing: By $P(x,1)$ we get $$xf(x)+f(x)=xf(x+1)=xf(x)+xf(1) \implies f(x)=xf(1) \; \forall x \in \mathbb R$$Hence $\boxed{f(x)=cx}$ is the only one works and we are done

Let $P(x,y)$ be the assertion. $P(0,4)\Rightarrow f(0)=0$ $P(x,0)\Rightarrow f(x^2)=xf(x)\Rightarrow f$ is odd $P(x,1)\Rightarrow xf(x+1)=xf(x)+f(x)$ $P(1,x)\Rightarrow xf(x+1)=2xf(x)+(x-x^2)f(1)$ Then $2xf(x)+(x-x^2)f(1)=xf(x)+f(x)$, so $f(x)=cx$ for all $x\ne1$ and $c$ is constant. Then by $P(1,1)$, we have $2c=f(2)=2f(1)$ so $\boxed{f(x)=cx}$ which works.

Subs $x=y=0$, $f(0)=0$ Subs $y=0$, $f(x^2)=xf(x)$ The problem equivalent to $xf(x)+2f(xy)=xf(x+y)+yf(x)\dots (1)$ Subs $y=1$ to $(1) \rightarrow (x+1)f(x)=xf(x+1)$ Subs $x=1$ to $(1) \rightarrow f(1)+2f(y)=f(y+1)+yf(1) \rightarrow f(x)=xf(1) \rightarrow f(x)=cx, c\in\mathbb R$

dangerousliri wrote: Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all real numbers $x$ and $y$, $$f(x^2)+2f(xy)=xf(x+y)+yf(x).$$ Proposed by Dorlir Ahmeti, Kosovo Isn't Dorlir Ahmeti from Albania?

$p(x,y)$:$f(x^2)+2f(xy)=xf(x+y)+yf(x)$ $p(0,0):f(0)=0$ $p(x,0):f(x^2)=xf(x)$ $p(x,1):\frac{x}{x+1}=\frac{f(x)}{f(x+1)}$ $p(1,x):2f(y)-f(y+1)=(y-1)f(1)$ $2-\frac{f(y+1)}{f(y)}=2-\frac{y+1}{y}=\frac{y-1}{y}=\frac{(y-1)f(1)}{f(y)}$$\rightarrow$ $f(x)=cx$

Let $P(x,y)$ denote the given assertion. $P(0,0)$ implies $f(0)=0$ and moreover $P(x,0)$ implies $f(x^2)=xf(x).$ Multiplying $P(1,x)$ with $x$ and then subtracting from $P(x,1)$ yields $f$ is linear and any linear function works.

The answer is $\boxed{f(x) = cx}$ for any real constant $c$. We can check that they work. We now prove they are the only solutions. Let $P(x,y)$ denote the given assertion. $P(0,0): f(0) = 0$. $P(x,0): f(x^2) = xf(x)$. $P(-x,0): f(x^2) = -xf(-x)$. So $xf(x) = -xf(-x)\implies f(x) = -f(-x)\forall x\ne 0$. Since $f(0) =0$, $f$ is odd. $P(-x,y): f(x^2) - 2f(xy) = -xf(y-x) - yf(x)$. Adding to the original FE gives \[2f(x^2) = 2xf(x)= x(f(x+y) - f(y-x))\implies f(x+y) - f(y-x) =2f(x), \]or \[f(x+y) + f(x-y) = 2f(x)\] Setting $x=y$ here gives $f(2x) = 2f(x)$. For any reals $a,b$ with $a\ne b$ and $a\ne -b$, setting $x=\frac{a+b}{2}$ and $y=\frac{a-b}{2}$ gives \[f(a) + f(b) = 2f\left(\frac{a+b}{2}\right) = f(a+b)\] $f(a)+f(b) = f(a+b) $ also holds if $a=b$ or $a=-b$. So for all $a,b\in \mathbb{R}$, $f(a+b) = f(a)+f(b)$, which implies $f$ is additive. $P(x+1,0): f(x^2 + 2x + 1) = xf(x) + f(x) + xf(1) + f(1)$. So \[xf(x) + 2f(x) + f(1) = xf(x) + f(x) + xf(1) + f(1)\implies f(x) = xf(1) ,\]so indeed $f(x) = cx$ for some constant $c$.

Thanks @geometry6 for pointing out $P(0,0)\implies f(0)=0$ ( There is no case that $f(0)=-2$ )

$$f(x^2)+2f(xy)=xf(x+y)+yf(x) \ldots (\alpha)$$In $(\alpha) x=y=0:$ $$\Rightarrow f(0)=0$$In $(\alpha) y=0:$ $$\Rightarrow f(x^2)=xf(x)$$Replacing in $(\alpha) :$ $$\Rightarrow xf(x)+2f(xy)=xf(x+y)+yf(x)\ldots (\beta)$$In $(\beta) x=1:$ $$\Rightarrow f(1)(1-x)+2f(x)=f(x+1)$$$$\Rightarrow xf(1)(1-x)+2xf(x)=xf(x+1)\ldots(\theta)$$In $(\beta) y=1:$ $$\Rightarrow (x+1)f(x)=xf(x+1)\ldots(\gamma)$$By $(\theta)$ and $(\gamma)$: $$\Rightarrow (1-x)f(x)=xf(1)(1-x)$$$$\Rightarrow f(x)=xf(1), \forall x\in \mathbb{R}-\{1\}$$$$\Rightarrow f(x)=xf(1), \forall x\in \mathbb{R}$$$$\Rightarrow f(x)=cx, \forall x\in \mathbb{R}_\blacksquare$$

$\textbf{Answer:}$ $f(x)=xf(1) \forall x \in \mathbb{R}$ $\textbf{Solution:}$ Let $P(x,y)$-denote the given assertion. $P(0,0) \implies 3f(0)=0 \implies f(0)=0$ $P(x,0) \implies f(x^2)=xf(x)$ $...(*)$ $P(x,y) \implies f(x^2)+2f(x)=xf(x+y)+yf(x) \stackrel{(*)}{\implies} xf(x)+2f(x)=xf(x+y)+yf(x)...Q(x,y)$. $Q(x,x) \implies xf(x)+2xf(x)=xf(2x)+xf(x) \implies 2xf(x)=xf(2x) \implies 2f(x)=f(2x) \forall x \in \mathbb{R} / \{0\}.$ Combining with $f(0)=0$ we get: $2f(x)=f(2x) \forall x \in \mathbb{R}$ $\textbf{Claim:}$ $f$-additive in $\mathbb{R}$ $\textbf{Proof:}$ $Q(x,y)-Q(y,x) \implies xf(x)+2f(xy)-yf(y)-2f(xy)=xf(x+y)+yf(x)-yf(x+y)-xf(y) \implies $ $xf(x)-yf(y)+xf(y)-yf(x)=xf(x+y)-yf(x+y) \implies f(x)(x-y)+f(y)(x-y)=f(x+y)(x-y) \implies$ $ (f(x)+f(y))(x-y)=f(x+y)(x-y) \implies f(x)+f(y)=f(x+y) \forall x,y \in \mathbb{R} / \{x\neq y\}.$ Combining this with $2f(x)=f(2x) \iff f(x)+f(x)=f(x+x)$ we get that $f(x)+f(y)=f(x+y) \forall x,y \in \mathbb{R}$. So $f$-is additive in $\mathbb{R} \square$. $Q(x,1) \implies xf(x)+2f(x)=xf(x+1)+f(x) \implies xf(x)+f(x)=xf(x+1) \stackrel{\text{additive}}{=} x(f(x)+f(1))=xf(x)+xf(1) \implies xf(x)+f(x)=xf(x)+xf(1) \implies f(x)=xf(1) \forall x \in \mathbb{R} \blacksquare$. This can easily be verified that works

Let $P(x,y)$ be the assertion $f(x^2)+2f(xy)=xf(x+y)+yf(x)$ $P(0,0) \implies f(0)=0$ $P(x,0) \implies f(x^2)=xf(x)$ $2f(xy)=xf(x+y)+yf(x)-xf(x)=yf(x+y)+xf(y)-yf(y)$ $\implies x(f(x)+f(y)-f(x+y))=y(f(x)+f(y)-f(x+y))$ If $f(x)+f(y)-f(x+y) \neq0$ then $x=y$ $\implies$ contradiction So $f(x+y)=f(x)+f(y)$ $P(x,1) \implies xf(x)+2f(x)=xf(x+1)+f(x)=xf(x)+xf(1)+f(x)$ $\implies f(x)=xf(1) \implies f(x)=cx$ for constant $c\in\mathbb R$

Another standard Vanilla FE! The answer is $f(x) = cx$ for a fixed constant $c \in \mathbb{R}$. It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. Let $P(x,y)$ denote the assertion $f(x^2)+2f(xy)=xf(x+y)+yf(x)$. Note that, $P(0,1)$ gives, \[3f(0)=f(0)\]from which it follows that $f(0)=0$. Next, $P(x,0)$ yields, \[f(x^2)=f(x^2)=2f(0)=xf(x)\]Further, $P(x,x)$ gives \begin{align*} f(x^2)+2f(x^2) &= xf(2x)+xf(x)\\ 3xf(x) &= xf(2x) + xf(x)\\ 2xf(x) &= xf(2x) \end{align*}from which it follows that $f(2x)=f(x)$ for all $x\ne 0$. Now, consider $x\ne y$. Comparing $P(x,y)$ and $P(y,x)$ will give us, \begin{align*} P(x^2) - P(y^2) &= (x-y)f(x+y) + yf(x)-xf(y)\\ xf(x) - yf(y) &= (x-y)f(x+y) + yf(x)-xf(y)\\ (x-y)(f(x+y)-f(x)-f(y)) &= 0 \end{align*}Since $x\ne y$, it thus follows that $f(x+y)=f(x)+f(y)$. In conjunction with our previous observation that $f(2x)=2f(x)$, we can conclude that $f$ is additive, \[f(x+y)=f(x)+f(y) \text{ for all }x,y\in \mathbb{R}\]Now, let $f(1)=c \in \mathbb{R}$. Then, $P(x,1)$ gives, \begin{align*} f(x^2)+2f(x) &= xf(x+1) + f(x)\\ xf(x) + f(x) &= x(f(x)+f(1))\\ f(x) &= xf(1) = xc \end{align*}for all $x\in \mathbb{R}$, which implies that all solutions to the given functional equation indeed take on the claimed forms.

$$f(x^2)+2f(xy)=xf(x+y)+yf(x)$$Asserting, $x=y=0$ $$f(0)=0$$Asserting $y=0$ $$f(x^2)=xf(x)$$ Asserting $y=-x$ $$f(x^2) = -f(-x^2)$$Hence, $f$ is an even function. Asserting $y=x$ $$2f(x) = f(2x)$$Replacing $x$ with $y$ $$f(y^2)+2f(xy)=yf(x+y)+xf(y)$$$$f(x^2)-f(y^2)=(x-y)f(x+y)+yf(x)-xf(y)$$$$(x-y)(f(x)+f(y)) = (x-y)f(x+y)$$For $x\neq y \rightarrow f(x) + f(y) = f(x + y)$ This is an additive Cauchy functional equation. Asserting $x = 1$ $$f(1)+2f(y) = f(y+1) + yf(1)$$$$f(y) = yf(1)$$Let $f(1) = k$, for some $k\in\mathbb{Z^+}$. Hence $f(x) = kx$ for all $x\in\mathbb{R}$

The idea is as follows: $P(0, 0) \implies f(0) = 0, P(x, 0)\implies f(x^2) = xf(x)$. Now we can use $P(x, 1)$ and $P(1, x)$ to get a system of equations in $f(x)$ and $f(x+1)$, from where we eliminate $f(x+1)$ to see $f(x) = xf(1) \forall x \ne 1$, but it holds for $x = 1$, so $f(x) = cx$. This works. $\square$

Let $P(x, y)$ denote the assertion. . $P(0, 0)$ gives $f(0) = 0$. $P(x, 0)$ gives $f(x^2) = xf(x)$. $P(x, 1)$ gives $f(x^2) + f(x) = xf(x + 1) \implies (x + 1)f(x) = xf(x + 1)$. $P(1, y)$ gives $f(1) + 2f(y) - yf(1) = f(y + 1) \implies x(f(1) + 2f(x) - xf(1)) = (x + 1)f(x) \implies 2xf(x) - x^2f(1) + xf(1) = (x + 1)f(x) \implies (x - 1)f(x) = f(1)x(x - 1) \implies f(x) = cx$ for $x \neq 1$. Plugging $P(1, y)$ for sufficiently large $y$ gives $f(1) = c$, and $f(x) = cx$. This clearly works, and we're done.