Let $P(x,y)$ denote the relation $f(x+f(y))+xy=f(x)f(y)+f(x)+y$.

The solutions are $f(x) \equiv x$ and $f(x) \equiv -x$, which work. Let $P(x,y)$ denote the assertion that $$f(x+f(y))+xy=f(x)f(y)+f(x)+y.$$ $P(0,x)$ gives $f(f(x))=f(0)f(x)+f(0)+x$. Notice that $P(f(x),y)$ gives $$f(f(x)+f(y))+yf(x)=f(f(x))f(y)+f(f(x))+y,$$ and subtracting this by its symmetric variant with $x$ and $y$ swapped gives $$yf(x)-xf(y)=f(f(x))f(y)+f(f(x))+y-f(f(y))f(x)-f(f(y))-x.$$ Using $P(0,x)$ to rewrite all the nested $f$'s, the RHS simplifies to $xf(y)-yf(x)$, so we have $xf(y)=yf(x)$. Plugging in $y=1$, we get $f(x)=xf(1)$. If $f(1)=c$, then the functional equation simplifies to $c(x+cy)+xy=c^2xy+cx+y$, which is only true for $c=1,-1$, as desired. $\square$

Solved with Orthogonal.. We claim the only functions are $\boxed{f(x)\equiv x}$ and $\boxed{f(x)\equiv -x}$. It is easy to check that these work. Let $P(x,y)$ denote the given assertion. $P(0,x)$ gives $f(f(x))=f(0)f(x)+f(0)+x$. Comparing $P(f(x),1)$ and $P(f(1),x)$ gives $f(x)\equiv f(1)x$. We can show that $f(1)=\pm 1$, as desired. $\square$

The only solutions are $f(x) = x$ and $f(x) = -x$, which work. Let $P(x,y)$ denote the given assertion. Case 1: $f(0) = 0$ Then $P(0,x): f(f(x)) = x$, so $f$ is an involution. $P(1, f(x)): f(x + 1)= x f(1) + f(1)$, so $f$ is linear, meaning $f(x) = cx$ for some constant $c$. Since $f$ is an involution, either $c = 1$ or $c = -1$. Case 2: $f(0) \ne 0$. Claim: $f$ is injective. Proof: If $f(a)= f(b)$, then $P(0,a)$ compared with $P(0,b)$ gives $a = b$. $\square$ Claim: $x + f(x)$ is injective. Proof: Suppose $a + f(a) = b + f(b)$. Then $P(x,x): f(x + f(x)) - (x + f(x)) = (f(x) - x)(x + f(x))$, so comparing $x = a$ and $x = b$ gives that $(a + f(a)) (f(a) - a)) = (a + f(a)) (f(b) - b)$. If $a + f(a) = 0$, then $P(a,a): f(0) = 0$, absurd. Hence $a + f(a) \ne 0$, so $f(a) - a = f(b) - b$. Hence $$(f(a) - a) + (f(a) + a) = (f(b) - b) + (f(b) + b) \implies f(a) = f(b) \implies a = b$$ $\square$ Now, the equation gives $$f(x + f(y)) - f(x) - y = f(x) f(y) - xy,$$ so swapping $x,y$ here gives $f(x + f(y)) - f(x) - y = f(y + f(x)) - f(y) - x$, so $f(x + f(y)) + (x + f(y)) = f(y + f(x)) + (y + f(x))$, which by our earlier claim implies $x + f(y) = y + f(x)\implies f(x) - x = f(y) - y$, so $f(x) = x + c$ for some constant $c$. $P(0,0): f(f(0)) = f(0)^2 + f(0)$, so $2c = c^2 + c\implies c\in \{0,1\}$. Checking, we see that $c = 1$ fails, so $c = 0$, but this is absurd since $f(0) \ne 0$.

IAmTheHazard wrote: Find all functions $f : \mathbb{R}\to\mathbb{R}$ such that for all real numbers $x$ and $y$, $$f(x+f(y))+xy=f(x)f(y)+f(x)+y.$$ Andrew Carratu $f(x)f(y)f(z)=f(x+f(y))f(z)+xyf(z)-f(z)f(x)-yf(z)=f(x+f(y)+f(z))+(x+f(y))f(z)-f(x+f(y))-z+xyf(z)-f(x+f(z))-xz+f(x)+z-yf(z)$ Now by the symmetry of $y,z$ we get that: $f(x+f(y)+f(z))+(x+f(y))f(z)-f(x+f(y))-z+xyf(z)-f(x+f(z))-xz+f(x)+z-yf(z)=$ $f(x+f(z)+f(y))+(x+f(z))f(y)-f(x+f(z))-y+xzf(y)-f(x+f(y))-xy+f(x)+y-zf(y)$ $\Rightarrow xf(z)+xyf(z)-xz-yf(z)=xf(y)+xzf(y)-xy-zf(y)$ In this for $x=0$ we get that: $-yf(z)=-zf(y)\Rightarrow \frac{f(z)}{z}=\frac{f(y)}{y}\Rightarrow f(x)=cx$ Now in the start we have that: $cx+c^2y+xy=c^2xy+cx+y\Rightarrow c^2y=y\Rightarrow c=+-1$ So we get that: $\boxed{f(x)\equiv x}$ and $\boxed{f(x)\equiv -x}$. In the same way we can solve the problem if $f : \mathbb{R^+}\to\mathbb{R^+}$

Cute one. $P(0,x)$ -> $f(f(x))=f(x)f(0)+f(0)+x$ $(1)$ By looking at $P(f(x),y)$ and $P(f(y),x)$ we have $f(f(x)+f(y))=f(f(x))f(y)+f(f(x))+y-f(x)y=f(f(y))f(x)+f(f(y))+x-f(y)x$, and by using $(1)$ we observe that $xf(y)=yf(x)$, which means that $f(x)=cx$. And by plugging it into equation one can prove that only solutions are $\boxed{f(x)\equiv x}$ and $\boxed{f(x)\equiv -x}$.

I am not sure about the surjectivety part $\textbf{Answer:}$ $f(x)=\pm x \forall x \in \mathbb{R}$ $\textbf{Solution:}$ Let $P(x,y)$-denote the given assertion. $\textbf{Claim:}$ $f$-bijective $\textbf{Proof:}$ $P(0,x) \implies f(f(x))=f(x)f(0)+f(0)+x \implies f$-is surjective

Since $f$ is both injective and surjective we get that $f$-is bijective. $\square$ $\textbf{Claim:}$ $f(0)=0$ $\textbf{Proof:}$ Since $f$-is surjective there exists an $\alpha$ such that $f(\alpha)=0 \iff \exists \alpha \in \mathbb{R} : f(\alpha)=0$ $P(\alpha,0) \implies f(\alpha+f(0))=0=f(\alpha) \implies f(\alpha+f(0))=f(\alpha) \stackrel{f-injective}{\implies} \alpha+f(0)=\alpha \implies f(0)=0.$ $\square$ $\textbf{Claim:}$ $f(x)= \pm x \forall \in \mathbb{R}$ $\textbf{Proof:}$ $P(0,x) \implies f(f(x))=x$ $...(*)$ $P(x,x) \implies f(x+f(x))+x^2=f(x)^2+f(x)+x \implies f(x+f(x))-x-f(x)=f(x)^2-x^2$ $...(3)$ $P(f(x),f(x)) \stackrel{(*)}{\implies} f(x+f(x))+f(x)^2=x^2+x+f(x) \implies f(x+f(x))-x-f(x)=x^2-f(x)^2$ $...(4)$ $(3)-(4) \implies f(x)^2-x^2=x^2-f(x)^2 \implies f(x)^2=x^2 \implies f(x)= \pm x \forall \in \mathbb{R}$ $\blacksquare$ $\textbf{POINTWISE TRAP!}$

This solution might be one of the lamest ever : $P(x,y):f(x+f(y))+xy=f(x)f(y)+f(x)+y$ We will just simplify $P(f(y),1)$ to find the answer : $\boxed{f(f(y)+f(1)) +f(y)-1= f(f(y))(f(1)+1)}$ Now : $P(0,y)$ , gives $f(f(y))=f(0)(f(y)+1)+y$ So for the right side we get that : $\boxed{f(f(y))(f(1)+1)=f(y).f(0)(f(1)+1)+f(0)(f(1)+1)+y(f(1)+1)}$ For the left side using $P(f(1),y)$ , we get that : $f(f(y)+f(1))=f(f(1))(f(y)+1)+y(1-f(1))=[f(0)(f(1)+1)+1](f(y)+1)+y(1-f(1))$ So :$\boxed{f(f(y)+f(1))+f(y)-1=f(0)(f(1)+1).f(y)+f(0)(f(1)+1)+f(y)+1+f(y)-1+y(1-f(1))}$ Identifying both sides will be left with : $2f(y)+y(1-f(1)) = y(f(1)+1)$ , thus $f(y)=yf(1)$ , the rest is obvious .

For above that equation doesn't imply that f is surjective as changing x to some value will also change f(x)

For above that equation doesn't imply that f is surjective as changing x to some value will also change f(x) an example is $f(x)=(x+1)^2$ ,will get on the right : $(x+1)^2+(x+1)$ , which doesn't take all values in $R$ such as $-4$

The solutions are $f(x) = x$ and $f(x)=-x$, which are both easy to verify. Claim: $f(0) = 0$. Proof: Assume that $f(0) \neq 0$ FTSOC. Taking $P(x,0)$ gives us $$f(x+f(0)) = f(x) \cdot (1 + f(0)),$$ so incrementing $x$ by $f(0)$ gives some geometric sequence. But the original equation can be rewritten as $$f(x+f(y)) - f(x)(1+f(y)) = y(1-x),$$ and if we increment $x$ by $f(0)$ here, the LHS will multiply by $1+f(0)$. However, the RHS cannot grow exponentially since it is negative for all $x>1$ and positive for all $x < 1$, something no exponential function does. This gives us a contradiction. Now, taking $P(0,x)$ gives us $f(f(x)) = x$, implying surjectivity, and $P(1,x)$ gives us $$f(1+f(x)) = f(1)(1+f(x)),$$ which is enough to imply that $f$ is linear. Plugging this into our original equation gives the claimed solutions.

f(x+f(y))+x.y=f(x).f(y)+f(x)+y Let' s call this P(x,y) P(0,0) gives us f(f(0))=f(0)^2+f(0) this call * P(x, 0) gives us f(x+f(0))=f(x).f(0)+f(x) x->-f(0) gives us f(0)=f(-f(0)).f(0)+f(-f(0)) from this, we can get that: f(0)/f(f(0))=f(0)+1 from this, * gives us f(0)^2=f(-f(0)).f(f(0))=A P(-f(0), f(0)) f(-f(0)+f(f(0)))-f(0)^2=A+f(-f(0))+f(0) f(-f(0)+f(0)^2+f(0))=2.f(0)^2+f(0)+f(-f(0)) this give f(f(0)^2)=f(0)^2+f(f(0))+f(-f(0)) P(f(0), f(0)^2) f(f(0)+f(f(0)^2))+f(0)^3=f(f(0)).f(f(0)^2)+f(f(0))+f(0)^2 f(f(0)+f(0)^2+f(f(0))+f(-f(0)))+f(0)^3=f(f(0)).f(f(0)^2)+f(f(0))+f(0)^2 f(2f(f(0))+f(-f(0)))+f(0)^3=f(f(0)).f(f(0)^2)+f(f(0))+f(0)^2 Now, P(2f(f(0)), -f(0)), which give us f(2f(f(0))+f(-f(0)))-2.f(f(0)).f(0)=f(2f(f(0))).f(-f(0))+f(2f(f(0)))-f(0) Now P(x, f(0)) W f(x+f(f(0)))+f(0).x=f(f(0)).f(x)+f(x)+f(0) x->f(f(0)), I don' t know why I didn't put this here W f(2f(f(0)))+f(0).f(f(0))=f(f(0)).f(f(f(0)))+f(f(f(0)))+f(0) Now I want to find f(f(f(0))) if we take this P(0, y) we will get this f(f(y))=f(0).f(y)+f(0)+y y->f(0), f(f(f(0)))=f(f(0)).f(0)+2.f(0) if we call f(0) to be (a) and f(f(0)) to be (b), where b=a^2+a( from the beginning) Then, f(2f(f(0)))+f(0).f(f(0))=f(f(0)).(f(f(0)).f(0)+2.f(0))+f(f(0)).f(0)+2.f(0)+f(0) we will get this f(2f(f(0)))+a.b=b.(b.a+2.a)+b.a+2.a+a, which is equivalent to f(2f(f(0)))=b(b.a+2.a)+3.a f(2f(f(0))+f(-f(0)))-2.f(f(0)).f(0)=f(2f(f(0))).f(-f(0))+f(2f(f(0)))-f(0), Now, from this, we have: f(2f(f(0))+f(-f(0)))-2.a.b=(b.(b.a+2.a)+3.a).f(-f(0))+b.(b.a+2.a)+3.a-a, call this T From this, f(2f(f(0))+f(-f(0)))+f(0)^3=f(f(0)).f(f(0)^2)+f(f(0))+f(0)^2, we have that the left side is equal to, also I want to say that f(f(0)^2)=f(0)^2+f(f(0))+f(-f(0))=a^2+b+f(-f(0)) Now, nightmare is coming The left side=b(a^2+b+f(-f(0)))+b+a^2-a^3 But we have that: f(0)^2=f(f(0)).f(-f(0)), which is equal to a^2=b.f(-f(0)), i.e the left side=a^2.b+b^2+a^2+b+a^2-a^3=a^2.b+b^2+2.a^2+b-a^3, call D Now-if we multiply whole equation with b, we will get this (The left side).b=2.b^2.a+(b^2.a+2.a.b+3.a).a^2+b^3.a+2.a.b^2+2.a.b From D multiply with (b), we get this a^2.b^2+b^3+2.a^2.b+b^2-a^3.b to be equal to 2.b^2.a+b^2.a^3+2.a^3.b+3.a^3+b^3.a+2.a.b^2+2.a.b Now, if this above wasn' t nightmare, after we will replaceable b with a^2+a- this will be a real bad nightmare, let's start (teeth) a^2.(a^2+a)^2+(a^2+a)^3+2.a^2.(a^2+a)+(a^2+a)^2-a^3.(a^2+a)=2.(a^2+a)^2.a+(a^2+a)^2.a^3+2.a^3.(a^2+a)+3.a^3+(a^2+a)^3.a+2.a.(a^2+a)^2+2.a.(a^2+a) After we use Wolfram Alpha, oops, just a joke, I figured it all myself (sad) , we will get this +2.a^7+3.a^6+6.a^5+5.a^4+4.a^3+a^2=0, the solution of this two in Real numbers, the one of them is approximately to some digit-negative and stranger, the other solution is 0 Then we have a=0=f(0) Then, if we put this in this: f(f(y)=f(0).f(y)+f(0)+y=y, i.e the function is involution Now, let P(f(x), f(y)) gives us f(f(x)+y)+f(x).f(y)=x.y+f(y)+x But P(y,x) gives us f(f(x)+y)+x.y=f(x).f(y)+f(y)+x, call G Let sum this two, then this gives us f(f(x)+y)=f(y)+x Put this in G gives x.y=f(x).f(y), y->x gives f(x)^2=x^2 f(x)=+x, f(x)=-x

P(f(x),y) ,change x,y after this and y=0 implies that f(f(x))×(f(0)+1)=f(x)×f(f(0))+f(f(0))+x-x×f(0). Plug x=0 we get f(f(y))=f(0)×f(y)+f(0)+y we get f is injective. Also we use this equation above and we get f(x)×(f(0)²+f(0)-f(f(0))) =-2x×f(0)-(f(0)²+f(0)-f((0))) if f(0)²+f(0)-f(f(0)) isn't equal 0 then f(x)=ax+b we check and we get f(x)=x or f(x)=-x. Other case implies that f(0)=0.In common equation we plug x=0 and we get f(f(x))=x.In next steps P(x,f(y)) and we change x,y we get x×f(y)=y×f(x) y=1 and we get f(x)=cx.It is easy to check and find f(x)=x and f(x)=-x answers.We are done.

$P(0,x)\implies f(f(x))=f(0)(f(x)+1)$ $P(f(x), y)\implies f(f(x)+f(y))+yf(x)=f(0)(f(x)+1)(f(y)+1)+xf(y)+x+y$ $P(f(x),y)-P(x, f(y))\implies xf(y)=yf(x)\implies \boxed{f(x)=\pm x}$