functional equation canada

[ishfaqhaque]

Let $\mathbb N = \{0,1,2,\ldots\}. $Determine all functions $f: \mathbb N \to \mathbb N $such that
$xf(y) + yf(x) = (x+y) f(x^2+y^2)$
for all $x,y \in \mathbb{N} $

[leonardo shawon]

x,y belongs to N
so f(x) belongs to N
f(y) and f(x^2+y^2) belongs to N
now the equation stands
l.H.s= (x+y)N
and r.H.s=(x+y)N

am i wrong?
i am not so good at english
and the model of my phone is Old. Equation mode(full editor) doesnt work well... s o r r y

[Avik Roy]

@leonardo shawon- I guess you should take a look at the latex codes, they are easy to deal with.
regarding the problem, you are to determine all such function $f$ so that the given equality holds. It's not a "STLER" problem
(STLER = Show That Left Equals Right)

[Avik Roy]

The desired functions should be: $f(x) = n$ for some $n \in \mathbb N$
Lets put $y=0, y=x$ and $y=1$ to obtain the following set of equations:
$f(x^2) = f(0)$
$f(x) = f(2x^2)$
$f(x) = (x+1)f(x^2+1) - xf(0)$
where $x$ is assumed to be nonzero.
Now, rearranging the given equation we have-
$x[f(y)-f(x^2+y^2)] = y[f(x^2+y^2) - f(x)]$
This equation reveals that $f(x^2+y^2)$ lies between $min \left (f(x), f(y) \right)$ and $max \left (f(x), f(y) \right)$, inclusive
However, since $f(1) = f(x^2)$ by virtue of the first equation,
$f(x^2+1) = f(0)$

and that reduces the third equation to $f(x) = f(0)$

[Avik Roy]

Ishfaq, was this FE included in your article in Gonit School??

[Masum]

I am a bit confused.
Setting $y=0,f(x^2)=0,$so there is a big possibility for a constant function.If $x=y,f(x)=f(2x^2)=f(8x^4)=f(128x^8)=......$ infinitely times.So combining these two we should have $f(x)=0$

[ishfaqhaque]

@avik vi

no, i didn't give this in my article

[Corei13]

@Avik da : How do we get $f(x^2)=f(1)$?

[Avik Roy]

Just put $x=1$ in $f(x^2)=f(0)$

[Corei13]

Oh! I missed that! :S

I solved it in IMO camp, Here is my solution:
Let, $f$ is not a constant function.
As, range of this function is a subset of $\mathbb{Z}$, The set $S =\{ a | a = |f(n) - f(m)|, n,m \in \mathbb{N} \}/\{0\}$ must have at least one element and so must have a least element. Let $d$ is the least element of the set $S$, clearly $d>0$, and there exists $a,b \in \mathbb{N}$ such that, $d=f(a)-f(b)>0$, But from the given equation we have,
$a[f(b)-f(a^2+b^2)] = b[f(a^2+b^2) - f(a)]$, which implies, $f(a) > f(a^2 + b^2) > f(b)$ so, $|f(a)-f(a^2 + b^2 )| < d$, which contradicts the minimality of $d$. So, $S$ must be an empty set, that is $f$ must be a constant function.