Not true. Let f(x) be x^2. Then f(4)=16 but f(2)+f(2)=2*4=8.

mathwarrior557 wrote: Not true. Let f(x) be x^2. Then f(4)=16 but f(2)+f(2)=2*4=8. Are you sure ? If $f(x)=x^2$ then $f(x+y)=(x+y)^2=x^2+y^2+2xy$ and $f(x)=x^2, f(y)=y^2.$ But we had $f(x+y)=f(x)+f(y)$, this implies $f(x) = x^2$ is not true for this function. Remark. $f(x)=cx$ is the only solution of the functional equation $f(x+y)=f(x)+f(y)$ where $c$ is a constant real number.

Assume f(a) + f(b) = f(a+b) but f(ab) + f(a) + f(b) =\= f(a+b+ab) Then: f(a+b+ab) =\= f(ab) + f(a+b) lets define ab=k and a+b= n f(k+n) =\= f(k) + f(n) contradiction!

Note: A moderator should move this to the Olympiad Algebra forum. We only prove the nontrivial direction. Let $P(x,y)\implies f(xy+x+y)=f(xy)+f(x)+f(y)$. First \[P(x,0)\implies f(0)=0,\\ P(x,1)\implies f(2x+1)=2f(x)+1,\\ P(x,-1)\implies f(x)=-f(-x).\]Thus we find \begin{align*} [2f(x)+1]-[2f(x-1)+1] &= f(2x+1)-f(2x-1)\\ &=f(2x+1)+f(-2x+1)\\ &=[2f(x)+f(1)]+[2f(-x)+f(1)]=2f(1), \end{align*}so \[f(x)-f(x-1)=f(1)\]and \[2f(x)+f(1)=f(2x+1)=f(2x)+f(1)\implies f(2x)=2f(x).\]Now \[P(x,y)+P(x,-y)\implies f(xy+x+y)+f(-xy+x-y)=2f(x)=f(2x),\]whence \[f(2x)+f(xy-x+y)=f(xy+x+y).\]Letting $u=2x$ and $v=xy-x+y$ (so that we cover all $(u,v)$ with $x=u/2$ and $y=(u+2v)/(u+2)$ when $u\ne-2$), \[f(u)+f(v)=f(u+v).\]If $u=-2$, then \[f(-2)+f(v)=f(v)-f(2)=f(v)-2f(1)=f(v-2)=f(u+v),\]so we're done.

Amir Hossein wrote: . Remark. $f(x)=cx$ is the only solution of the functional equation $f(x+y)=f(x)+f(y)$ where $c$ is a constant real number. Well obviously, it's the Cauchy FE.

Case 1: $f(x+y)=f(x)+f(y) ~~~\forall x,y\in \mathbb{R}.$ $\text{So, }f(x+y+xy)=f(x)+f(y)+f(xy). \text{ Q.E.D. } \blacksquare$ Case 2: $f(x+y+xy)=f(x)+f(y)+f(xy) ~~~\forall x,y \in \mathbb{R}.$ $P(0,0)\implies f(0)=0.$ $P(x,-1)\implies f(x)=-f(-x).$ $P(x,1)\implies f(2x+1)=2f(x)+f(1).$ $x\to a+b+ab \implies f(2(a+b+ab)+1)$ $=2f(a+b+ab)+f(1)=2f(ab)+2f(a)+2f(b)+f(1).$ \begin{align*} P(a,2b+1)\implies f(2(a+b+ab)+1)\\ =f(b+(2b+1)+a(2b+1))\\ =f(a)+2f(b)+f(1)+f(2ab+a) \\ \implies f(2ab+a)=2f(ab)+f(a) ~~~~~(*).\end{align*}$b\to -1/2 \implies 0=2f(-a/2)+f(a)=-2f(a/2)+f(a) \implies f(a)=2f(a/2) \implies f(2x)=2f(x).$ Now from (*) we have $f(2ab+a)=f(2ab)+f(a),$ which fulfills the claim by setting variables to $x,y \in \mathbb{R}.$ Note $x,y$ can take $0$, since $f(0)=0.$ $\blacksquare$

Amir Hossein wrote: Remark. $f(x)=cx$ is the only solution of the functional equation $f(x+y)=f(x)+f(y)$ where $c$ is a constant real number. Not necessarily, that's only true under more conditions. https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation#Existence_of_nonlinear_solutions_over_the_real_numbers

3333 wrote: Assume f(a) + f(b) = f(a+b) but f(ab) + f(a) + f(b) =\= f(a+b+ab) Then: f(a+b+ab) =\= f(ab) + f(a+b) lets define ab=k and a+b= n f(k+n) =\= f(k) + f(n) contradiction! Note that they have just done in "one direction."

darshpatel wrote: Amir Hossein wrote: Remark. $f(x)=cx$ is the only solution of the functional equation $f(x+y)=f(x)+f(y)$ where $c$ is a constant real number. Not necessarily, that's only true under more conditions. https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation#Existence_of_nonlinear_solutions_over_the_real_numbers True, I'm sure Amir Hossein had that in mind.

Note $f(x+y)=f(x)+f(y)$ trivially implies $f(x+y+xy)=f(x)+f(y)+f(xy)$. So we will only prove the other direction. Let $P(x,y)$ denote the assertion that $f(x+y+xy)=f(x)+f(y)+f(xy)$. $P(0,0): f(0)=3f(0) \implies f(0)=0$ $P(x,-1): f(-1)=f(x)+f(-1)+f(-x) \implies f(x)+f(-x)=0 \implies f(-x)=-f(x)$ $P(x,1): f(2x+1)=2f(x)+f(1)$. Now, to make use of this, $P(2x+1,2y+1)$. $LHS=f(4xy+4x+4y+3)=2f(2xy+2x+2y+1)+f(1)=2(2f(xy+x+y)+f(1))+f(1)=4f(xy+x+y)+3f(1)$ $RHS=f(2x+1)+f(2y+1)+f(4xy+2x+2y+1)=2f(x)+2f(y)+2f(2xy+x+y)+3f(1)$ Hence, $4f(xy+x+y)=2f(x)+2f(y)+2f(2xy+x+y)$. $4f(xy+x+y)=2f(xy+x+y)+2f(x)+2f(y)+2f(xy)$. So, $$2f(xy+x+y)+2f(x)+2f(y)+2f(xy)=2f(x)+2f(y)+2f(2xy+x+y) \implies f(xy+x+y)+f(xy)=f(2xy+x+y)$$We therefore would wish for $xy=a$ and $xy+x+y=b$ to take on all real values. $a,b$ only work if $xy=a,x+y=b-a$ has a solution, which is true if and only if $$D=(b-a)^2-4a=a^2+b^2-2ab-4a\geq 0$$Note we can also always swap $(a,b)$ to see if that works. Hence if one of $a,b\leq0$ then $f(a)+f(b)=f(a+b)$. For $a,b>0$, $$f(a)+f(b)=f(a+b) \iff -f(-a)-f(-b)=-f(-a-b) \iff f(-a)+f(-b)=f(-a-b)$$which is true. So we are done.