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Source: ELMO Shortlist 2024/A3

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#1 h Jun 22, 2024, 3:40 PM • 3 Y

Y by ItsBesi, ihatemath123, pho1234

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

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#2 h Jun 22, 2024, 3:42 PM

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Solution

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#3 h Jun 22, 2024, 7:37 PM

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Solution

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#4 h Jun 22, 2024, 7:50 PM

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

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CyclicISLscelesTrapezoid

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CyclicISLscelesTrapezoid

#5 h Jun 22, 2024, 11:33 PM • 1 Y

Y by pho1234

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. $\square$

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#6 h Jun 23, 2024, 2:23 AM

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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. $\square$

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#7 h Jun 23, 2024, 2:35 AM • 1 Y

Y by KevinYang2.71

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 $f(0) \ne 0$ .

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#8 h Jun 23, 2024, 11:12 AM

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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 $f : \mathbb{R^+}\to\mathbb{R^+}$

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#9 h Jun 23, 2024, 12:44 PM

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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 $\boxed{f(x)\equiv -x}$ .

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#13 h Jul 2, 2024, 8:41 PM

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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

f-injective

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 $f(\alpha)=0 \iff \exists \alpha \in \mathbb{R} : f(\alpha)=0$

$P(\alpha,0) \implies f(\alpha+f(0))=0=f(\alpha) \implies f(\alpha+f(0))=f(\alpha) \stackrel{f-injective}{\implies} \alpha+f(0)=\alpha \implies f(0)=0.$ $\square$

$\textbf{Claim:}$ $f(x)= \pm x \forall \in \mathbb{R}$

$\textbf{Proof:}$
$P(0,x) \implies f(f(x))=x$ $...(*)$

$P(x,x) \implies f(x+f(x))+x^2=f(x)^2+f(x)+x \implies f(x+f(x))-x-f(x)=f(x)^2-x^2$
$...(3)$

$P(f(x),f(x)) \stackrel{(*)}{\implies} f(x+f(x))+f(x)^2=x^2+x+f(x) \implies f(x+f(x))-x-f(x)=x^2-f(x)^2$ $...(4)$

$(3)-(4) \implies f(x)^2-x^2=x^2-f(x)^2 \implies f(x)^2=x^2 \implies f(x)= \pm x \forall \in \mathbb{R}$ $\blacksquare$

$\textbf{POINTWISE TRAP!}$

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#14 h Aug 1, 2024, 5:13 PM

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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 .

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#15 h Aug 1, 2024, 5:26 PM

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

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#16 h Aug 1, 2024, 5:39 PM

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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 $-4$

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#17 h Aug 9, 2024, 3:42 PM

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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.

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bo18

#18 h Sep 18, 2024, 12:23 AM

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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

This post has been edited 1 time. Last edited by bo18, Sep 18, 2024, 5:58 PM
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#19 h Jan 2, 2025, 1:35 PM

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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.

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lksb

#20 h Jan 2, 2025, 1:49 PM

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

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awesomeming327.

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awesomeming327.

#21 h Jan 29, 2025, 1:08 AM

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The answer is $f(x)=x$ or $f(x)=-x$ , which clearly work. Now we prove that this is the only solution. Let $P(x,y)$ denote the assertion, and let $f(0)=c$ .

We have
$\begin{align*} P(0,0) &\implies f(c) = c^2+c \\ P(0,x) &\implies f(f(x)) = cf(x) + c + x \\ P(x,0) &\implies f(x+c) = (c+1)f(x) \end{align*}$
Claim 1: $f(0)=0$ .

Take $P(c,1)$ which gives us on one hand,
$f(c+f(1))+c=(c^2+c)(f(1)+1)+1$ But on the other hand, we have
$\begin{align*} f(f(1)+c)+c &= (c+1)f(f(1)) + c \\ &= (c+1)(cf(1) + c + 1) + c \\ &=(c^2+c)(f(1)+1)+2c+1 \end{align*}$ which proves our claim.

Now, $P(0,x)$ gives $f(f(x))=x$ . Thus when we use $P(f(x),y)$ we get
$f(f(x)+f(y))+f(x)y=xy+x+y$ Since $xy+x+y-f(f(x)+f(y))$ is symmetric, we must have $f(x)y=f(y)x$ , implying that $f(x)/x$ is a constant. Since $f$ is an involution, this constant is $\pm 1$ .

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#22 h Feb 28, 2025, 4:54 AM

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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

Let $P(x,y)$ be the assertion $f(x+f(y))+xy=f(x)f(y)+f(x)+y$ .
$P(0,x)\Rightarrow f(f(x))=f(0)(f(x)+1)+x$
$P(f(x),y)\Rightarrow f(f(x)+f(y))-f(0)(f(x)+1)(f(y)+1)=xf(y)-yf(x)+x+y$
Swapping $x,y$ gives $yf(x)-xf(y)=xf(y)-yf(x)$ , then setting $y=1$ , we have $f(x)=xf(1)$ . Testing, only $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$ work.

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#23 h Apr 1, 2025, 4:32 PM

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Probably similar solutions are already mentioned. Anyway, I'll go on with mine.
Let $P(x, y)$ denote the given assertion.
$P(0, x): \underline{f(f(x)) = f(0)(f(x) + 1) + x}$
$P(f(x), y): f(f(x) + f(y)) + f(x)y = f(f(x))(f(y) + 1) + y = (f(0)(f(x) + 1) + x)(f(y) + 1) + y = f(0)f(x)f(y) + f(0)f(y) + xf(y) + f(0)f(x) + f(0) + x + y$
$\rightarrow f(f(x) + f(y)) = (x + y) + f(0)f(x)f(y) + f(0)(f(x)f(y) + f(y) + f(x) + 1) + xf(y) - f(x)y$
Now, if we swap $x,y$ , the left side stays the same, and all the parts in the right side stay the same, except the $xf(y) - f(x)y$ , so:
$xf(y) - f(x)y = yf(x) - f(y)x \rightarrow 2xf(y) = 2yf(x) \rightarrow \frac{f(x)}{x} = \frac{f(y)}{y} \rightarrow \underline{f(x) = kx}$ And by checking this format in the equation, we find that only $k = 1, -1$ are the solutions, therefore the only answers are $\boxed{f(x) = x \quad \forall x \in \mathbb{R}}$ and $\boxed{f(x) = -x \quad \forall x \in \mathbb{R}}$ . $\blacksquare$

This post has been edited 1 time. Last edited by Bardia7003, Apr 1, 2025, 4:34 PM
Reason: typo

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#24 h Yesterday at 3:28 PM

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One linear:
$P(x+f(z),y)$ and swap $y,z$ you can find that $\frac{f(x)}{x}$ is a constant

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