Let $\mathbb{Z}^{+}=\{1, 2, 3, …\}$ denote the set of positive integers.
Problem 1. Are there any non-constant functions $f\colon \mathbb{Z}^{+}\to\mathbb{R}$ such that
$$
x f(y) + y f(x) = (x+y) f(x^2+y^2)
$$
for all $x, y\in\mathbb{Z}^{+}$?
Clearly, the constant functions satisfy the functional equation above, and so it is natural to see if there are any non-constant examples. The motivation for this problem comes from a related (and easier problem) from Canadian Mathematical Olympiad (Year 2002):
Problem 2. Find all functions $f\colon \mathbb{Z}^{+}\to\mathbb{Z}^{+}$ such that
$$
x f(y) + y f(x) = (x+y) f(x^2+y^2)
$$
for all $x, y\in\mathbb{Z}^{+}$.
This second problem has a nice solution. I don't want to spoil it for others, so if you want to read it, you can hover over the following solution (I divided it into steps, in case you want to have hints one step at a time):
Solution to Problem 2. We claim that only the constant functions $f:\mathbb{Z}^{+}\to\mathbb{Z}^{+}$ satisfy the functional equation above.
Assume, to the contrary, there is a non-constant function $f$ with this property. Thus, there exist positive integers $a$ and $b$ such that $f(a)<f(b)$. Then using the functional equation, one gets:
$$ (a+b) f(a) < (a+b) f(a^2+b^2) < (a+b) f(b)$$
So, $f(a) < f(a^2+b^2) < f(b)$. We have shown that between any two distinct points in the image of $f$, there is a third point in between. And this process can be repeated forever. However, this is a contradiction since the target of $f$ is the natural numbers $\mathbb{Z}^{+}$. As you can see, this solution does not work when the target is $\mathbb{R}$, hence the reason for asking this question.
Best Answer
Here is a partial result : I show below that $f(k)=f(1)$ for $1\leq k \leq 20$ (and I think it very likely that $f$ is indeed constant).
Let $a=f(1)$ and $A=\lbrace x\in{\mathbb N} | f(x)=a \rbrace$. We see from the functional equation that if any two of $x,y,x^2+y^2$ are in $A$, then the third is in $A$ also. In particular :
Rule $R_1$ : if $x,y\in A$, then $x^2+y^2\in A$. Rule $R_2$ : if $x,y,z\in A$, then $\sqrt{x^2+y^2-z^2} \in A$ (assuming this is an integer).
Since $2=1^2+1^2$, $5=1^2+2^2$, $8=2^2+2^2$, $50=5^2+5^2$, we have by rule $R_1$ that $\lbrace 1,2,5,8,50 \rbrace \subseteq A$. Since $7=\sqrt{5^2+5^2-1^2}$, $4=\sqrt{1^2+8^2-7^2}$, $17=1^2+4^2$, $20=2^2+4^2$ we deduce
$$\lbrace 1,2,4,5,7,8,17,20,50 \rbrace \subseteq A \tag{1}$$
Next, we have $f(10)=f(1^2+3^2)=\frac{3a+f(3)}{4}$ and $f(13)=\frac{3a+2f(3)}{5}$. From $2^2+11^2=5^2+10^2$, we deduce $\frac{2f(11)+11a}{13}=\frac{5f(10)+10a}{15}=\frac{f(10)+2a}{3}=\frac{f(3)+11a}{12}$. Similarly, from $7^2+11^2=1^2+13^2$, we deduce $\frac{7f(11)+11a}{18}=\frac{f(13)+13a}{14}=\frac{f(3)+34a}{35}$. We then have a Cramer $2\times 2$ system in $f(3)$ and $f(11)$ (and parameter $a$), so $f(3)=f(11)=a$. Therefore :
$$\lbrace 1,2,3,4,5,7,8,10,11,13,17,20,50 \rbrace \subseteq A \tag{2}$$
Then, from $9=\sqrt{3^2+11^2-7^2}$, $12=\sqrt{8^2+9^2-1^2}$, $18=3^2+3^2$, $6=\sqrt{2^2+9^2-7^2}$, we further deduce
$$[|1..13|]\cup\lbrace 17,18,20,50 \rbrace \subseteq A \tag{3}$$
Finally, from $14=\sqrt{10^2+10^2-2^2}$, $15=\sqrt{9^2+13^2-5^2}$, $16=\sqrt{8^2+14^2-2^2}$, $19=\sqrt{13^2+14^2-2^2}$, we obtain $[|1..20|] \subseteq A$ as announced.