Goutham wrote: Let $f:\mathbb{R}\longrightarrow \mathbb{R}$ be a function such that $f(x+y+xy)=f(x)+f(y)+f(xy)$ for all $x, y\in\mathbb{R}$. Prove that $f$ satisfies $f(x+y)=f(x)+f(y)$ for all $x, y\in\mathbb{R}$. Let $P(x,y)$ be the assertion $f(x+y+xy)=f(x)+f(y)+f(xy)$ $P(0,0)$ $\implies$ $f(0)=0$ $P(x,-1)$ $\implies$ $f(-x)=-f(x)$ Let $x,y\ne -1$ $P(x,\frac y{x+1})$ $\implies$ $f(x+y)=f(x)+f(\frac y{x+1})+f(\frac {xy}{x+1})$ $P(x,-\frac y{x+1})$ $\implies$ $f(x-y)=f(x)-f(\frac y{x+1})-f(\frac {xy}{x+1})$ $P(y,\frac x{y+1})$ $\implies$ $f(x+y)=f(y)+f(\frac x{y+1})+f(\frac {xy}{y+1})$ $P(y,-\frac x{y+1})$ $\implies$ $-f(x-y)=f(y)-f(\frac x{y+1})-f(\frac {xy}{y+1})$ Adding these four lines, we get $f(x+y)=f(x)+f(y)$ $\forall x,y\ne -1$ It remains to get rid of the constraint $x,y\ne -1$ : Let $x\ne 0$ : $x-1\ne -1$ and $1\ne -1$ and so $f((x-1)+1)=f(x-1)+f(1)$ and so $f(x-1)=f(x)+f(-1)$, still true when $x=0$. So $f(x+y)=f(x)+f(y)$ $\forall x\ne -1,\forall y$ So $f(x+y)=f(x)+f(y)$ $\forall x,\forall y\ne -1$ And $f(1+1)=f(1)+f(1)$ and so $f(-1-1)=f(-1)+f(-1)$ and so $f(x+y)=f(x)+f(y)$ when $x=y=-1$ So $f(x+y)=f(x)+f(y)$ $\forall x,y$ Q.E.D.

Generalization 1 Let $a\in {{\mathbb{R}}^{*}}$ and $f:\mathbb{R}\to \mathbb{R}$ be a function such that $f(x+y+axy)=f(x)+f(y)+f(axy)$ for all $x, y\in\mathbb{R}$. Prove that $f$ satisfies $f(x+y)=f(x)+f(y)$ for all $x, y\in\mathbb{R}$.

Generalization 2 Let $a,b\in {{\mathbb{R}}^{*}},a+b\ne 0$ and $f:\mathbb{R}\to \mathbb{R}$ be a function such that $f(ax+by+xy)=af(x)+bf(y)+f(xy)$ for all $x, y\in\mathbb{R}$. Prove that $f$ satisfies $f(x+y)=f(x)+f(y)$ for all $x, y\in\mathbb{R}$.

Ah, it is in fact from LL.

Isn't it very simple It can be clearly seen that $f(x)=0$ $f(x+y+xy+0)=f(x+y)+f(xy)+f(0)$ or $f(x+y+xy)=f(x+y)+f(xy)$ Subtracting the above and given equation we get $f(x+y)=f(x)+f(y)$

div5252 wrote: $f(x+y+xy+0)=f(x+y)+f(xy)+f(0)$ Why?

oh i thought we could divide it into any 3 sums

My solution : We have : $f(x+y+xy) = f(x)+f(y)+f(xy)$ $(1)$ Plug in $x=y=0$ $\rightarrow$ $f(0) = 0$ Plug in $ y = -1$ $\rightarrow$ $f(-x) = -f(x)$ Plug in $ y = 1 $ $\rightarrow$ $f(2x+1) = 2f(x) + 1$ $(2)$ From $(1)$ and $(2)$ $\rightarrow$ $f(2(xy+x+y)+1) = 2f(xy+x+y) + f(1) = 2[f(x)+f(y)+f(xy)] +f(1)$ On the other hand we have : $f(2(xy+x+y)+1) = f(2x+1+y(2x+1) +y) \\ = f(2x+1) + f(y(2x+1)) + f(y) \\ =2f(x) + f(1) + f(y) + f(2xy + y)$ $\rightarrow$ $f(2xy + y) = f(y) + 2f(xy)$ $(3)$ Plug in $x = \frac{-1}{2} $ to $(3)$ we have : $f(y) = -2f(\frac{-y}{2}) = 2f(\frac{y}{2})$ Combine this with $(3)$ , it follows that $f(2xy+y) = f(y) + f(2xy)$ With $u,v \in\mathbb{R} , v = 0$ we have $f(u+v) = f(u) + f(v)$ With $u,v \in\mathbb{R} ; u , v \ne 0$ let $x = \frac{u}{2v} ; y = v$ $\rightarrow$ $f(u+v) = f(u) + f(v)$ So $f(x+y) = f(x) + f(y)$ $\forall x,y$

sorta involved...

f odd

f(x+n)=f(x)+nf(1) for integers n

f(nx)=nf(x) for all integers n

Going back to the original problem, plugging in $(x,y)=(u+\sqrt{u^2-2v},u-\sqrt{u^2-2v})$ for nonnegative $u$ and $v$ such that $u\ge \sqrt{2v}$ implies $$f(u+\sqrt{u^2-2v})+f(u-\sqrt{u^2-2v})+f(2v)=f(2u+2v)=2f(u+v).$$Now taking $(x,y)=(2u+2\sqrt{u^2-2v},2u-2\sqrt{u^2-2v})$, $$f(2u+2\sqrt{u^2-2v})+f(2u-2\sqrt{u^2-2v})+f(8v)=f(4u+8v)=4f(u+2v).$$Pulling out the integers from the arguments and using the previous relation, $$2(2f(u+v)-2f(v))=4f(u+2v)-8f(v)\implies f(u+2v)=f(u+v)+f(v)$$Now letting $b=v$ and $a=u+v$, we see that for all $a\ge b+\sqrt{2b} \ge 0$, we have $f(a+b)=f(a)+f(b)$. To establish this for $a < b+\sqrt{2b}$ as well, pick a large integer $N$ such that $x+N \ge b+\sqrt{2b}$, and note that $$f(x+N+b)=f(x+N)+f(b)\iff f(x+b)+Nf(1)=f(x)+Nf(1)+f(b).$$In particular this implies that $f(a+b)=f(a)+f(b)$ holds for all $b\ge 0$ and any $a$. But since $f$ is odd, it holds for all $b$ as well, and we are done.