Solve over $\mathbb{R}$:
$f(x^2 + y) = f(x^{27} + 2y) + f(x^4)$
from Chan's handout, "Intro to functional equations", (https://web.evanchen.cc/handouts/FuncEq-Intro/FuncEq-Intro.pdf)
My solution differs from the one given, and I was just wondering whether mine had some flaw or not. Please point out mistakes/corrections/improvements.
Plugging in $x,y=0$ we get $f(0)=0$,Plugging in $x=1,y=0$ we get $f(1)=0$ Now, for $x=1,y=q$, we get $f(1+q)=f(1+2q)$ replacing $1+q$ by $p$, $f(p)=f(p+q)$ and by setting $p=0$ we get $f(q)=0$ for all $q \in \mathbb{R}$.
My main doubt is whether I can set $p=0$ (or anything) as I feel that could be a mistake because that would fix $q$.
Another thought I had was for $x=0,y=y$, we get $f(y)=f(2y)$. Can this be put to any good use?
Thanks a lot for your help!
Well, you're right about fixing $q$. What you did was assume $p=1+q$, so the value of $p$ is dependent on $q$ and vice versa, so that's illegal. Setting $p=0$ is the same as setting $q=-1$, which only works for this $q$, in other words, you've proved only that $f(-1)=0$
I think if you want another solution you can try $y=x^4-x^2$, this gets you: $$f(x^{27}+2x^4-2x^2)=0$$ Since the expression inside can attain any real value, then $f(y)=0$ for all $y \in \mathbb{R}$