Can $(x+y)^2$ be written in the form $f(x) g(y) + g(x)f(y)$? Are there functions $f$ and $g$ such that $f(x)g(y) + g(x)f(y) = (x+y)^2$. I'm asking this because in quantum mechanics, the symmetric wavefunction of two identical particles is written as $\psi(x_1, x_2) =\frac{1}{\sqrt{2}}[ f(x_1) g(x_2) + g(x_1) f(x_2) ]$.
 A: The answer is NO.
Suppose that $f,g : {\mathbb R} \to {\mathbb R}$  satisfy your condition. Then clearly $f$ cannot be zero everywhere, so we have an $x_1$ such that $f(x_1)\neq 0$, and then
$$
g(y)=\frac{(x_1+y)^2-g(x_1)f(y)}{f(x_1)} \tag{1}
$$
Reinjecting (1) into the original equation, we obtain
$$
(x+y)^2=f(x)\bigg(\frac{(x_1+y)^2-g(x_1)f(y)}{f(x_1)}\bigg)+
f(y)\bigg(\frac{(x_1+x)^2-g(x_1)f(x)}{f(x_1)}\bigg) \tag{2}
$$
which can be rewritten as
$$
\bigg((x_1+x)^2-2g(x_1)f(x)\bigg)f(y)=f(x_1)(x+y)^2-f(x)(x_1+y)^2\tag{3}
$$
Now I claim that $f$ must necessarily be a polynomial of degree at most $2$. To see why, consider $h(x)=(x_1+x)^2-2g(x_1)f(x)$. If $h$ iz zero everywhere, then $f(x)=\frac{(x_1+x)^2}{2g(x_1)}$ is a polynomial of degree at most $2$. Otherwise, we have a $x_2$ such that $h(x_2)\neq 0$, and then by (3) we have that $f(y)=\frac{f(x_1)(x_2+y)^2-f(x_2)(x_1+y)^2}{h(x_2)}$ is a polynomial of degree at most $2$ again.
Since the roles of $f$ and $g$ are symmetric, $g$ must also be a polynomial of degree at most $2$. So there are constants $a_1,b_1,c_1,a_2,b_2,c_2$ with
$$
f(x)=a_1x^2+b_1x+c_1, \ g(x)=a_2x^2+b_2x+c_2 \tag{4}
$$
Then $f(x)g(y)+f(y)g(x)=2a_1a_2(xy)^2$ plus other monomials. So one of $a_1$ or $a_2$, say $a_2$, must be zero.
Then $f(x)g(y)+f(y)g(x)=a_1b_2x^2y$ plus other monomials, so one of $a_1$or $b_2$ must be zero.
If $b_2=0$, then $g$ is constant and this is easily seen to be impossible. So we must have $a_1=0$.
But then $f(x)g(y)+f(y)g(x)$ has no monomial in $x^2$ or $y^2$, and this is a contradiction.
A: There are no functions $f,g$ defined on $\{0,1,2\}$ so that $f(x)g(y) + g(x)f(y) = (x+y)^2$ for all $x,y \in \{0,1,2\}$.
Proof.  Assume such $f,g$ exist.
Take $x=y=0$,
$$
2f(0)g(0) = 0 .
$$
At least one of $f(0), g(0)$ is $0$.  WLOG assume $f(0)=0$.  Write
$a = g(0)$.
Take $x=0$,
$$
f(0)g(y)+g(0)f(y) = y^2\\
af(y) = y^2
$$
Now $y \ne 0$ sometimes, so $a \ne 0$ and $f(y) = y^2/a$ for all $y$.
Take $x=1,y=1$,
$$
2f(1)g(1) = 4\\
\frac{2g(1)}{a} = 4\\
g(1)=2a
$$
Take $x=1,y=2$,
$$
f(1)g(2)+g(1)f(2)=9
\\
\frac{g(2)+8a}{a} = 9
\\
g(2) = a
$$
Take $x=2, y=2$,
$$
2f(2)g(2) = 16\\
\frac{8a}{a}=16\\
8=16
$$
Contradiction.
