functional equation australia

[the arrivals]

determine all real valued function f define on non negative integers satisfying
\[ f(x+y)+f(x-y)=f(3x) \]

a question i like to be answered.
determine all functions define on R+ to R satisfying the condition
\[f(x+y)=f(x^2+y^2)\]
working a few time you may rather find the relations
$f(x)=f(x^2)$
$f(x+y)=f(x)+f(y)=f(x^2+y^2)=f(x^2)+f(y^2)$
and also $f(nx)=nf(x)$
seeing these relations what can i claim about its charactaristics? in other words i d like to be given the answers with full description please cuz i m novice

[Moon]

Why don't you use LaTeX? It is quite easy to use, and in this post what you just need to add dollar signs at the start and end of the equation.

[Cryptic.shohag]

Solution for both of the functions should be $f(x)=0$

[Corei13]

For $f(x+y) = f(x^2) + f(y^2)$:
Note that, for any $b, a\in \mathbb{R}$ and $ |b| \geq |a|$, we can find $r \in \mathbb{R}$ such that,
$b^2 = r^2 + (a-r)^2$ or $r^2 - a r - \frac{b^2 - a^2}{2}=0$ because, $a^2 + 2(b^2 - a^2) \geq 0$
So, $f(b)=f(b^2)=f(r^2 +(a-r)^2)=f([r] + [a-r])=f(a)$, so, f is a constant function.

[sourav das]

I find a nice solution to 1st F.E. It is easy to show that $f(0)=0, 2f(x)=f(2x)=f(3x)$. Now plug in $x=3x$ and $y=x$ and find out that $6f(x)=4f(x)$ So$f(x)=0$

[*Mahi*]

$f(3x)=f(2x)$ turns it into the classic Cauchy equation, whose solution is $f(x)=xf(1)$. Setting this in the given equation, we get $f(1)=0$ and so $f(x)=0$ for all $x$.

[nayel]

...the classic Cauchy equation, whose solution is $f(x)=xf(1)$.

This isn't true in general.

[*Mahi*]

...the classic Cauchy equation, whose solution is $f(x)=xf(1)$.

This isn't true in general.

Hmm I know I was just too lazy to write that it's for the functions $f:\mathbb{Q} \to \mathbb{Q}$

[nayel]

But still you can't use it in your solution, because the function in the question is defined on the non-negative integers. So replacing $x-y$ and $x+y$ with, say, $u$ and $v$ our functional equation becomes $f(u)+f(v)=f(u+v)$, only for non-negative integers $u,v$ with the same parity. Your solution would work if this was true for all non-negative integers $u,v$, but you would probably have to derive it.

[*Mahi*]

Hmmm...in my solution I used induction and then thought Cauchy equation is worth giving a try but now it seems it blows everything up...