Determine all functions $f:\mathbb{N}\rightarrow\mathbb{N}$ such that $xf(y)+yf(x)=(x+y)f(x^2+y^2)$ for all $x,y\in\mathbb{N}$ (contest question)

Question

The question below is from the 2002 Canada National Olympiad. I have found one family of functions but need help in finding (or proving the non-existence) of others. Suggestions on how to improve the solution method (especially rigour of arguments) would also be appreciated.

Let $\mathbb{N} = {0,1,2,\dots}$. Determine all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that

$xf(y) + yf(x) = (x + y)f(x^2 + y^2) \tag{1}$

for all $x,y \in \mathbb{N}$.

---

Expressing $f$ as a power series:

$f(t) = a_0 + \displaystyle{\sum\limits_{k=1}^{\infty}{a_k x^k}} \tag{2}$

Assuming the power series is convergent, the general coefficient, $a_n$, is the same on both sides of equation (1). [I am not sure of this step].

So for $a_0$

$x(a_0) + y(a_0)=(x+y)(a_0)$

which is satisfied for all values of $a_0$. Hence $\boxed{f(t)=a_0,\:a_0\in\mathbb{N}}$ is a solution.

For any $n > 0$, we have

$\begin{align} x a_n y^n + y a_n x^n &= (x+y)a_n(x^2+y^2)^n \iff \\ a_n(xy^n+yx^n) &= a_n(x+y)(x^2+y^2)^n \end{align}$

and if this is true for fixed $n>0$ and all $x,y\ge0$ then $a_n=0$. So I am unable to find any other functions.

Answer 1

Set $g(x):=f(x)-f(1)$, the we know that $f(x)=f(2x^{2})$ and so the function $g$ (and $f$) can not be a polynomial), because $g$ has infinitely roots. Now we show that the function $g$ must be zero. Assume that there is $x_{0}\in \mathbb{N}$ such that $g(x_{0})\neq 0$, since $\mathbb{N}$ is a countable set, let $x_{0}$ be an element s.t., $g(x_{0})$ has minimum value, but the following relation show that it's impossible and so $g\equiv 0$ i.e., $f(x)=f(1)$ on $\mathbb{N}$. If $f(1)<f(x_{0})$, then $$f(1)<f(1+x_{0}^{2})<f(x_{0}),$$ if $f(x_{0})<f(1)$, then
$$f(x_{0})<f(x_{0}^{2}+1)<f(1).$$ You can obtain the above relations with $f(x^{2}+y^{2})=\dfrac{xf(y)+yf(x)}{x+y}$.

Answer 2

Let $\mathbb{N} = {0,1,2,\dots}$. Determine all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that

$xf(y) + yf(x) = (x + y)f(x^2 + y^2) \tag{OP}$

for all $x,y \in \mathbb{N}$.

Let $k=x=y$, then we get $$ k f(k) + k f(k) = (k+k) f( k^2 + k^2) f(x) = f(2x^2) $$

which implies

$$ f(k) = f( 2 k^2 ) \tag{1} $$

Let $P^n_\mathbb{N}(k)$ be a polynomial $\mathbb{N} \rightarrow \mathbb{N}$, where $k^n$ is the highest power.

When we try $$ f(k) = P^n_\mathbb{N}(k) $$

then we obtain $$ f(k^2) = P^n_\mathbb{N}(k^2) = P^{2n}_\mathbb{N}(k). $$

Whence, due to (1) we get

$$ P^n_\mathbb{N}(k) = P^{2n}_\mathbb{N}(k) \Longrightarrow n = 2n \Longrightarrow n = 0. $$

So the conclusion is that $f(k)$ is just a constant. Thus

$$ f(k) = c \in \mathbb{N}. $$

Answer 3

Set $y=0$. This gives us $$xf(0)=xf(x^2)$$ for all $x$, and so $f(x^2)=f(0)$ for all $x$.

Now take $x=3n^2$ and $y=4n^2$ for some $n$. We get that $$3n^2 f(4n^2) + 4n^2 f(3n^2) = 7n^2 f(25n^4)$$ Since $f(4n^2)=f(25n^4)=f(0)$, we get $$4n^2 f(3n^2) = 4n^2 f(0)$$ and so $f(3n^2)=f(0)$ for all $n$.

Next, let $x=12n^2$ and $y=5n^2$ for some $n$. Then $$12n^2 f(5n^2) + 5n^2 f(12n^2) = 17n^2 f(169n^4)$$ Again, since $f(12n^2)=f(169n^4)=f(0)$, we get that $$12n^2 f(5n^2) = 12n^2 f(0)$$ and so $f(5n^2)=f(0)$ for all $n$.

Now let $x=2n^2$ and $y=n^2$ for some $n$. We get that $$2n^2 f(n^2) + n^2 f(2n^2) = 3n^2 f(5n^2)$$ and since $f(n^2)=f(5n^2)=f(0)$, we get that $$n^2 f(2n^2) = n^2 f(0)$$ giving us that $f(2n^2)=f(0)$ for all $n$.

Finally, let $x=y$. This gives us $$2xf(x) = 2xf(2x^2)=2xf(0)$$ for all $x$, and so $f(x)=f(0)$ for all $x$.

Thus the only functions satisfying the equation are constant functions, and we can check that any constant function does indeed satisfy the equation.