So here is the Question :-
Find all functions $f$ :- $\mathbb{N}$ $\to$ $\mathbb{N}$ such that :-
$$xf(y) + yf(x) = (x + y)f(x^2 + y^2)$$
I tried substituting values for $x$ and $y$, but I couldn't reach to a possible clue to the solution. Any hints or suggestions will be greatly appreciated!
Best Answer
Suppose there exists an integer $n$ such that $f(1) <f(n)$. With $x=1$ and $y=n$, we have $$ nf(1)+f(n) = (n+1)f(n^2+1) \implies f(1)<f(g(n))<f(n),$$ where $g(n)=n^2+1$. Repeating the same argument using $f(1)<f(g(n))$, we deduce that $$f(1)<f\left(g^{f(n)-f(1)}(n)\right)<f\left(g^{f(n)-f(1)-1}(n)\right)<\ldots<f(g(n)) <f(n),\tag{1}$$ where $g^m(\cdot)$ denotes the composition of the function $g$ repeated $m$ times.
However, there cannot be more than $f(n)-f(1)-1$ integers between $f(n)$ and $f(1)$. Thus, (1) leads to contradiction because $f(x)\in\mathbb{N},\forall x\in\mathbb{N}$. Hence, $f(1)\nless f(n).$
Using similar arguments, we can show that $f(1)\ngtr f(n)$. Therefore, we conclude that $f(x)=f(1)$ for all values of $x$ which trivially satisfies the given relation.