1) $f(y)+f(z)=f(y+z)+f(0)$. $f^2(x+y+z)=f^2(x+y)+2f(xz+yz)+f^2(z)\\=f^2(x)+f^2(y)+f^2(z)+2f(xy)+f(xz+yz)$. Obviously we have $f(xz+yz)+f(xy)=f(xy+yz)+f(xz)$. Let $y=-x$ above, we get $f(-x^2)+f(0)=f(-x^2-xz)+f(xz)$, or $f(y)+f(z)=f(y+z)+f(0),y<-z$. Let $y<0$, $f(-y)+f(2y)=f(y)+f(0)$ and $2f(y)=f(2y)+f(0)$ imply $f(-y)+f(y)=2f(0),\forall y$. For the case $y>-z$, we have $-y<z$, then $f(-y)+f(-z)=f(-y-z)+f(0)$ implies $f(y)+f(z)=f(y+z)+f(0)$ 2) $(f(x)-f(0))(f(y)-f(0))=f(xy)-f(0)$. $P(0,0)$ implies $f(0)=0,-2$. Plug 1) into $P(y,z)$, we will get 2) Now we see $g(x)=f(x)-f(0)$ satisfies Cauchy equation and $g(xy)=g(x)g(y)$. So $f(x)=ax+b$. The solutions are $f(x)=x$, $f(x)=x-2$, $f(x)=0$ or $f(x)=-2$.

xxp2000 wrote: Let $y<0$, $f(-y)+f(2y)=f(y)+f(0)$ and $\color{red} 2f(y)=f(2y)+f(0)$ imply $f(-y)+f(y)=2f(0),\forall y$. . I don't get this step. Can you explain?

socrates wrote: xxp2000 wrote: Let $y<0$, $f(-y)+f(2y)=f(y)+f(0)$ and $\color{red} 2f(y)=f(2y)+f(0)$ imply $f(-y)+f(y)=2f(0),\forall y$. . I don't get this step. Can you explain? Since $y<0$, $-y<-2y$ implies $f(-y)+f(2y)=f(y)+f(0)$ $y<-y$ implies $f(y)+f(y)=f(2y)+f(0)$

socrates wrote: Determine all functions $f:\Bbb{R}\to\Bbb{R}$ such that \[ f^2(x+y)=f^2(x)+2f(xy)+f^2(y), \]for all $x,y\in \Bbb{R}.$ Nice problem! My Solution $P(x,-x)\implies f^2(x)+f^2(-x)=f^2(0)-2f(-x^2)$ $P(x+y,-x)+P(x,y)\implies f^2(y)+f^2(x+y)=f^2(x+y)+2f(-x^2-xy)+f^2(-x)+f^2(x)+f^2(y)+2f(xy)\implies$ $\implies 2f(-x^2-xy)+2f(xy)=-(f^2(-x)+f^2(x))=2f(-x^2)+f^2(0)\implies f(-x^2-xy)+f(xy)=f(-x^2)-\frac{1}{2}f^2(0)$ $h(x)=f(x)+\frac{1}{2}f^2(0)$ Then $h(-x^2)=h(xy)+h(-x^2-xy)$ $xy=a,-x^2-xy=b$ $a+b<0$ $a+b<0\implies h(a+b)=h(a)+h(b)$ Claim.$h-additive$ Proof: $a,c\in R$ $b<min[-a,-(a+c)]\implies a+b<0,a+b+c<0$ $a+b<0 \implies h(a+b)=h(a)+h(b)$ $a+b+c<0\implies h(a+b)+h(c)=h(a+b+c)$ $h(a)+h(b)+h(c)=h(a+b)+h(c)=h(a+b+c)=h(a+c)+h(b)\implies h(a+c)=h(a)+h(c)$ $h-additive\implies f-additive$ $f^2(x)+f^2(y)+2f(xy)=(f(x+y))^2=(f(x)+f(y))^2=f^2(x)+f^2(y)+2f(x)f(y)\implies f(x)f(y)=f(xy)$ Then $f-multiplicative$ $f(x+y)=f(x)+f(y),f(xy)=f(x)f(y)\implies f(x)=0,f(x)=x$

Functional_equation wrote: socrates wrote: Determine all functions $f:\Bbb{R}\to\Bbb{R}$ such that \[ f^2(x+y)=f^2(x)+2f(xy)+f^2(y), \]for all $x,y\in \Bbb{R}.$ Nice problem! My Solution $P(x,-x)\implies f^2(x)+f^2(-x)=f^2(0)-2f(-x^2)$ $P(x+y,-x)+P(x,y)\implies f^2(y)+f^2(x+y)=f^2(x+y)+2f(-x^2-xy)+f^2(-x)+f^2(x)+f^2(y)+2f(xy)\implies$ $\implies 2f(-x^2-xy)+2f(xy)=-(f^2(-x)+f^2(x))=2f(-x^2)+f^2(0)\implies f(-x^2-xy)+f(xy)=f(-x^2)-\frac{1}{2}f^2(0)$ $h(x)=f(x)+\frac{1}{2}f^2(0)$ Then $h(-x^2)=h(xy)+h(-x^2-xy)$ $xy=a,-x^2-xy=b$ $a+b<0$ $a+b<0\implies h(a+b)=h(a)+h(b)$ Claim.$h-additive$ Proof: $a,c\in R$ $b<min[-a,-(a+c)]\implies a+b<0,a+b+c<0$ $a+b<0 \implies h(a+b)=h(a)+h(b)$ $a+b+c<0\implies h(a+b)+h(c)=h(a+b+c)$ $h(a)+h(b)+h(c)=h(a+b)+h(c)=h(a+b+c)=h(a+c)+h(b)\implies h(a+c)=h(a)+h(c)$ $h-additive\implies f-additive$ $f^2(x)+f^2(y)+2f(xy)=(f(x+y))^2=(f(x)+f(y))^2=f^2(x)+f^2(y)+2f(x)f(y)\implies f(x)f(y)=f(xy)$ Then $f-multiplicative$ $f(x+y)=f(x)+f(y),f(xy)=f(x)f(y)\implies f(x)=0,f(x)=x$ This solution is not correct, $f=-2$ is also solution.