Ig this works. parmenides51 wrote: Determine all functions $f : R \to R$ satisfying $f(f(x) + 2y)= 6x + f(f(y) -x)$ for all real numbers $x,y$ As usual let $P(x:y)$ denote the assertion in the above FE. Now $P(f(y):x)\implies f(f(f(y))+2x)=6f(y)+f(0)\implies f$ is Surjective. Let $f(k)=0$ and $f(t)=a+b$ for some $a,b$ such that $f(a)=f(b)$. Clearly $k,t$ exists as $f$ is surjective. So, $$\begin{cases}P(a:t)\implies f(f(a)+2t)=6a+f(b) \\ P(b:t)\implies f(f(b)+2t)=6b+f(a)\implies f(f(a)+2t=6b+f(b)\end{cases}\implies a=b\implies f \text{ is injective }$$Now $P(f(k),k)\implies f(f(0)+2k)=f(0)\implies f(0)=-2k=c(\text{ constant} )$ and $P(0:x)\implies f(f(0)+2x)=f(f(x))\implies f(x)=f(0)+2x\implies \boxed{f(x)=2x+c}$ where $c$ is some constant and this is indeed a solution which clearly satisfies the given equation. $\qquad\blacksquare$

Let $P(x,y)$ be the assertion $f(f(x) + 2y)= 6x + f(f(y) -x)$. $P(f(y),x)\implies f(f(f(y))+2x)=6f(y)+f(0)\implies f$ is surjective Setting $x=0$ and $f(y)=n$, we see that: $$f(f(n))=6n+f(0)\forall n,$$so $f$ is bijective. $P(0,y)\implies f(f(y))=f(f(0)+2y)$, and by injectivity, $\boxed{f(x)=2x+f(0)}$ is the unique solution, which indeed works.

jasperE3 You have mistake here, there should be f(f(x)-f(y)) instead of f(0)

jasperE3 wrote: Let $P(x,y)$ be the assertion $f(f(x) + 2y)= 6x + f(f(y) -x)$. $P(f(y),x)\implies f(f(f(y))+2x)=6f(y)+f(0)\implies f$ is surjective Setting $x=0$ and $f(y)=n$, we see that: $$f(f(n))=6n+f(0)\forall n,$$so $f$ is bijective. $P(0,y)\implies f(f(y))=f(f(0)+2y)$, and by injectivity, $\boxed{f(x)=2x+f(0)}$ is the unique solution, which indeed works. $P(f(y),x)\implies f(f(f(y))+2x)=6f(y)+f(0)$ not should say $ f(f(f(y))+2x)=6f(y)+f(f(x)-f(y))$