Iveela wrote: Find all functions $f : \mathbb{R} \to \mathbb{R}$ and $h : \mathbb{R}^2 \to \mathbb{R}$ such that \[f(x+y-z)^2=f(xy)+h(x+y+z, xy+yz+zx)\]for all real numbers $x,y,z$. Let $P(x,y,z)$ be the assertion $f(x+y-z)^2=f(xy)+h(x+y+z,xy+yz+zx)$ Note that $(x+y+z)^2\ge 3(xy+yz+zx)$ and so we have no information on $h(x,y)$ when $x^2<3y$ Let $a=f(0)$ Let $x\ne 0$. Subtracting $P(x,\frac yx-x,\frac yx)$ from $P(x,\frac yx,\frac yx-x)$, we get New assertion $Q(x,y)$ : $f(y-x^2)=f(y)-f(2x)^2+a^2$ $\forall y$ and $\forall x\ne 0$, still true when $x=0$ Subtracting $Q(x,0)$ from $Q(x,x^2)$, we get $f(x^2)+f(-x^2)=2a$ and so $f(x)+f(-x)=2a$ $\forall x$ Subtracting $Q(x,0)$ from $Q(x,y)$, we get $f(y-x^2)=f(y)+f(-x^2)-a=f(y)-f(x^2)+a$ and so $f(x)-a$ is additive. Subtracting $P(x,y,0)$ from $P(x,0,y)$, we get $f(x+y)^2-f(x-y)^2=f(xy)-a$ Which is, using additivity of $f-a$ : $(f(x)+f(y)-a)^2-(f(x)-f(y)+a)^2=f(xy)-a$ And so $f(xy)=4f(x)f(y)-4af(x)+a$ Swapping $x,y$ and subtracting, we get $af(x)=af(y)$ and so Either $f(x)=a$ $\forall x$, either $a=0$ and $f(xy)=4f(x)f(y)$ and so $f(x)$ is linear and we have $f(x)=\frac x4$ $\forall x$ 1) If $f(x)=a$ constant, $P(x,y)$ is $h(x+y+z,xy+yz+zx)=a^2-a$ and solution : $\boxed{\text{S1 : }f(x)=a\quad\forall x\text{ and }h(x,y)=a^2-a\quad\forall x^2\ge 3y\text{ and }h(x,y)\text{ is anything we want }\forall x^2<3y}$, which indeed fits 2) If $f(x)=\frac x4$, $P(x,y)$ is $h(x+y+z,xy+yz+zx)=\left(\frac{x+y-z}4\right)^2-\frac {xy}4$ $=\frac{x^2+y^2+z^2-2xy-2yz-2zx}{16}$ $=\frac{(x+y+z)^2-4(xy+yz+zx)}{16}$ And solution : $\boxed{\text{S2 : }f(x)=\frac x4\quad\forall x\text{ and }h(x,y)=\frac{x^2-4y}{16}\quad\forall x^2\ge 3y\text{ and }h(x,y)\text{ is anything we want }\forall x^2<3y}$, which indeed fits