IMO Shortlist 2011, Algebra 3

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3647 posts

orl

#1 h Jul 11, 2012, 10:21 PM • 11 Y

Y by Davi-8191, tenplusten, Amir Hossein, itslumi, Jc426, jhu08, megarnie, Adventure10, Mango247, Rounak_iitr, and 1 other user

Determine all pairs $(f,g)$ of functions from the set of real numbers to itself that satisfy $g(f(x+y)) = f(x) + (2x + y)g(y)$ for all real numbers $x$ and $y$ .

Proposed by Japan

This post has been edited 1 time. Last edited by djmathman, Aug 28, 2015, 4:56 AM
Reason: typo

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pco

23450 posts

pco

#2 h Jul 12, 2012, 7:58 AM • 15 Y

Y by Sayan, lovermath, bcp123, eulerou1997, leader, numbertheorist17, Bashy99, Amir Hossein, AlastorMoody, Jc426, jhu08, Adventure10, Mango247, Sedro, and 1 other user

orl wrote:

Determine all pairs $(f,g)$ of functions from the set of real numbers to itsels that satisfy $g(f(x+y)) = f(x) + (2x + y)g(y)$ for all real numbers $x$ and $y.$

Let $P(x,y)$ be the assertion $g(f(x+y))=f(x)+(2x+y)g(y)$

Let $x\ne 0$ :
$P(x,0)$ $\implies$ $g(f(x))=f(x)+2xg(0)$
$P(0,x)$ $\implies$ $g(f(x))=f(0)+xg(x)$
Subtracting, we get $g(x)=\frac{f(x)-f(0)}x+2g(0)$ $\forall x\ne 0$

$P(x,y)$ $\implies$ $g(f(x+y))=f(x)+(2x+y)g(y)$
$P(x+y,0)$ $\implies$ $g(f(x+y))=f(x+y)+(2x+2y)g(0)$
Subtracting, we get $f(x+y)=f(x)+(2x+y)g(y)-(2x+2y)g(0)$

Considering $y\ne 0$ and using previous result, this becomes $f(x+y)=f(x)+(2x+y)\frac{f(y)-f(0)}y+2xg(0)$
Considering $x\ne 0$ and swapping $x,y$ , this becomes $f(x+y)=f(y)+(2y+x)\frac{f(x)-f(0)}x+2yg(0)$

Considering $x,y\ne 0$ and subtracting, we get $f(x)=x^2(\frac{f(y)-f(0)}{y^2}+\frac{g(0)}y)-g(0)x+f(0)$

Setting $y=1$ in the above line, we get $f(x)=x^2(f(1)-f(0)+g(0))-g(0)x+f(0)$ $\forall x\ne 0$

Plugging this in the equality $g(x)=\frac{f(x)-f(0)}x+2g(0)$ $\forall x\ne 0$ we previously got, we get then :
$g(x)=x(f(1)-f(0)+g(0))+g(0)$ $\forall x\ne 0$

Plugging this in original equation, we get two possibilities :
$f(x)=g(x)=0$ $\forall x\ne 0$
$f(x)=x^2+c$ and $g(x)=x$ $\forall x\ne 0$

It's then easy to check that we need the same values for $x=0$ and we get the two families of solutions :
$f(x)=g(x)=0$ $\forall x$
$f(x)=x^2+c$ and $g(x)=x$ $\forall x$

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Orin

22 posts

Orin

#3 h Jul 24, 2012, 2:27 PM • 4 Y

Y by Mehtap, Adventure10, Mango247, and 1 other user

Let $P(x,y)$ be the assertion $g(f(x+y))=f(x)+(2x+y)g(y)$
$P(x,y)-P(y,x) \Rightarrow f(x)-f(y)=(2y+x)g(x)-(2x+y)g(y)$ .
Let $Q(x,y)$ be the assertion $f(x)-f(y)=(2y+x)g(x)-(2x+y)g(y)$
$Q(1,0) \Rightarrow f(1)-f(0)=g(1)-2g(0)$ .....(1)
$Q(x,0)-Q(x,1) \Rightarrow f(1)-f(0)=2xg(1)+g(1)-2xg(0)-2g(x)$ .....(2)
From (1),(2) we get $g(x)=(g(1)-g(0))x+g(0)$
now we have two cases.
Case1 $\Rightarrow g(1)-g(0)=0$
Then $g(x)=g(0)=c \forall x \in \mathbb {R}$ ,where $c$ is a constant.
Plugging $g(x)=c$ in $P(x,y) \Rightarrow f(x)+2cx+cy=c \forall x,y \in \mathbb {R}$
which implies $c=0,f(x)=0$ whcih is indeed a solution.
Case2 $\Rightarrow g(1)-g(0) \neq 0$
We may write $g(x)=ax+b;a \in \mathbb {R}-\{0\},b \in \mathbb {R}$ .
Plugging $g(x)=ax+b$ in $P(x,-x) \Rightarrow af(0)+b=f(x)+xg(-x)$
or, $f(x)=ax^2-bx+af(0)+b$ ,let $af(0)+b=c$
Then $f(x)=ax^2-bx+c$
Plugging these values in $P(0,x)$
$\Rightarrow a^2x^2-abx+ac+b=ax^2+bx+c \forall x \in \mathbb {R}$ ,Let it be $R(x)$ .
$R(0) \Rightarrow ac+b=c$ .....(3)
$R(1)-R(-1) \Rightarrow b(a-1)=0$
Again we have 2 cases.
Case2.1: $a=1$ ,so from (3) we get $b=0$
so $f(x)=x^2+c,g(x)=x$ which indeed satisfy $P(x,y)$ .
Case2.2: $b=0$ ,so from (3) we get $c(a-1)=0$
If $a=1$ it becomes case 2.1
So $c=0$ implying $f(x)=ax^2,g(x)=ax$
Plugging these values in $P(x,0) \Rightarrow a=0$ or $a=1$
if $a=1;f(x)=x^2,g(x)=x$ ,which is a subset of solutions of Case2.1
else if $a=0;f(x)=0,g(x)=0$ .
=============================================
Synthesis of solutions:
1. $f(x)=0,g(x)=0 \forall x \in \mathbb {R}$
2, $f(x)=x^2+c,g(x)=x \forall x \in \mathbb {R}$ .

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omkarkamat

2 posts

omkarkamat

#4 h Apr 9, 2013, 8:22 PM • 2 Y

Y by Adventure10, Mango247

I have another way of solving the problem but don't know if it's right.

Plug in (x,y)=(x,-2x) to get g(f(-x))=f(x) for all x in R
And (x,y) = (y,-2y) to get g(f(-y)=f(y) for all y in R

Case 1: f(x) is even.
If this is the case, then g(x) = x for all x in R
The original equation now becomes f(x + y)=f(x)+y(2x+y)
If we take (x,y)=(0,y) we get f(y)=y^2. WLOG let f(0) =c. Where can is a constant,
Then f(y)=y^2 +c for all y in R.

Now we prove that either f(x) is a linear function or a constant function.
We find that f(x)-f(y)=(2y+x)g(x)-(2x+y)g(y)
Plugging in (x,y) = (x,0) ,(1,x),(0,1) we obtain

F(x)-f(0)=xg(x)-2xg(0)
F(1)-f(x)=(2x+1)g(1)-(x+2)g(x)
F(0)-f(1)=2g(0)-g(1)

Taking the sum and dividing by two yields
G(x)=x(g(1)-g(0))+g(0) which proves that it has to a constant or a linear function.

We first examine the case in which it is a constant if such a function exists.
We take f(x)=c1 and this implies that g(x)=c2. Plugging into the equation we see that the only constant that works is 0.

Since this constant works, there is no such linear function that we need to examine and hence the solutions are

F(x)= y^2 +c, f(x)=0 and g(x)=x, g(x)=0

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Legend-crush

199 posts

Legend-crush

#5 h Feb 5, 2015, 9:57 PM • 8 Y

Y by anantmudgal09, Bashy99, A_Math_Lover, Amir Hossein, Jerry122805, Adventure10, Mango247, and 1 other user

Let $P(x,y)\Leftrightarrow g(f(x+y))=f(x)+(2x+y)g(y)$
$P(x,0)\Rightarrow g(f(x))=f(x)+2xg(0)\ (*)$
$P(0,x) \Rightarrow g(f(x))=f(x)+2xg(0) \ (**)$
from (*) and (**) we get $f(x)=f(0)+x(g(x)-2g(0))$
replacing in the functionnal equation we get $P(x,y) \Rightarrow (x+y)g(x+y)=x(g(x)-2g(0))+(2x+y)g(y)$
since the LHS is symetric in x,y , we have $x(g(x)-2g(0))+(2x+y)g(y)=y(g(y)-2g(0))+(2y+x)g(x)$
which yields to $(\forall x,y\neq 0) \ \frac{g(x)-g(0)}{x}=\frac{g(y)-g(0)}{y}$ hence $(\exists a,b\in \mathbb{R}) (\forall x\in \mathbb{R}) g(x)=ax+b$
it also follows that $f(x)=ax^2-bx+f(0)$ forall real x.
replacing in the functionnal equation we get :
1. $f=g=0$
2. $f(x)=x^2+c$ and $g(x)=x \forall x$

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utkarshgupta

2280 posts

utkarshgupta

#6 h Mar 3, 2015, 6:10 AM • 2 Y

Y by Amir Hossein, Adventure10

Solution
Let $P(x,y)$ be the assertion $g(f(x+y)) = f(x) + (2x + y)g(y)$ .
$P(x,0) \implies g(f(x))=f(x)+2xg(0)$
$P(0,x) \implies g(f(x))=f(0)+xg(x)$
$\implies xg(x)+f(0)=f(x)+2xg(0)$
$\implies f(x)=xg(x)+f(0)-2xg(0)$

Replacing this in $P(x,y)$
$g(f(x+y))=xg(x)+f(0)-2xg(0)+(2x+y)g(y)$
Here setting $x=z$ , $y=x+y-z$
$g(f(x+y))=zg(z)+f(0)-2zg(0)+(x+y+z)g(x+y-z)$

Comparing and setting $y=z=1$
$xg(x)-2xg(0)+2xg(1)=2g(0)+(x+2)g(x)$
$\implies g(x)=Ax+B$

Now the question is easy

Thus the solutions are $\boxed {f(x)=0=g(x)}$ or
$\boxed {f(x)=x^2+C}$ and $\boxed {g(x)=x}$

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AMN300

563 posts

AMN300

#7 h Feb 15, 2016, 8:45 PM • 5 Y

Y by MSTang, Amir Hossein, AlastorMoody, THKcandomath, Adventure10

Let $P(x, y)$ be the given assertion.
$P(x, 1) \implies g(f(x+1)) = f(x) + (2x+1) g(1)$ $P(1, x) \implies g(f(x+1)) = f(1) + (x+2) g(x)$ $P(x, 0) \implies g(f(x)) = f(x) + 2x g(0)$ $P(0, x) \implies g(f(x)) = f(0) + x g(x)$ The first two equations mean $f(x) + (2x+1) g(1) = f(1) + (x+2) g(x) = g(f(x+1))$ The second two mean $f(x) + 2x g(0) = f(0) + x g(x) = g(f(x))$ Subtracting the two equations we have $2x(g(1)-g(0)) + g(1) = f(1) - f(0) + 2 g(x)$ So $g(x)$ is linear. Plug in $g(x)=ax+b$ . Then,
$P(x, 0) \implies mf(x) + b = f(x) + 2xb$ If $m=1$ we have $b=0$ . We see that this means $f(x) = x^2+c$ upon plugging in.
If $m \neq 1$ we simplify to $f(x) = \frac{b(2x-1)}{m-1}$ . $P(0, x) \implies mf(x)=f(0)+xg(x)$ so plugging in we see that we need $b=m=0$ .
So the solutions are $\boxed{f(x) \equiv 0, g(x) \equiv 0}$ or $\boxed{f(x) \equiv x^2+c, g(x) \equiv x}$ .

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Ankoganit

3070 posts

Ankoganit

#8 h Oct 18, 2016, 10:56 AM • 8 Y

Y by anantmudgal09, tenplusten, YRNG-BCC168, Amir Hossein, mijail, IAmTheHazard, Adventure10, Mango247

We do not use the notation $P(x,y)$ for the given assertion. Swapping $x,y$ in the given equation and then comparing it with the given yields: $f(x)+(2x+y)g(y)=f(y)+(2y+x)g(x).\qquad (\star)$ Setting $y=0$ in $(\star)$ gives $f(x)=xg(x)-2xg(0)+f(0).\qquad (\smiley )$ Now plug this definition of $f(x)$ into $(\star)$ to get: $-xg(0)+xg(y)=-yg(0)+yg(x).$ For nonzero $x,y$ , this becomes $\frac{g(y)-g(0)}{y}=\frac{g(x)-g(0)}{x}\implies g(x)=kx+g(0)\forall x\in\mathbb R\setminus \{0\}.$ Here $k$ is some constant. Note that this automatically holds for $x=0$ , so no need to worry about that $\mathbb R\setminus \{0\}$ bit. So $g(x)$ is of the form $ax+b$ , and using $(\smiley )$ lets us infer that $f(x)=ax^2-bx+c$ . Now plugging these back into the original equation, we see that only $\boxed{g(x)\equiv f(x)\equiv 0}$ and $\boxed{g(x)\equiv x, \; f(x)\equiv x^2+c}$ work, so these are the only solutions. $\blacksquare$

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john111111

105 posts

john111111

#9 h Nov 1, 2016, 6:20 PM • 2 Y

Y by Amir Hossein, Adventure10

My solution:
Let $P(x,y)$ be the assertion that $g(f(x+y))=f(x)+(2x+y)g(y)$
$P(0,0) \Rightarrow g(f(0))=f(0)$
$P(-x,2x) \Rightarrow g(f(x))=f(-x)$
$P(x,0) \Rightarrow g(f(x))=f(x)+2xg(0)$ (1).
Suppose that $g(0) \neq 0$ .Then (1) easily implies that f is 1-1.
$P(-f(x),f(x)) \Rightarrow g(f(0))=f(-f(x))-f(x)g(f(x)) \Rightarrow f(-f(x))=f(x)f(-x)+f(0)$ .Putting in this $-x$ where $x$ is we obtain
$f(-f(x))=f(-f(-x)$ so from 1-1 $f(x)=f(-x) \Rightarrow x=-x \Rightarrow x=0$ , a contradiction. So $g(0)=0$ and thus from (1) $g(f(x))=f(x)$ (2)
$P(0,x) \Rightarrow g(f(x))=f(0)+xg(x) \Rightarrow f(x)=f(0)+xg(x)$ (3).
Using (2) and (3) in the initial relation we have $f(x+y)=f(x)+(2x+y)g(y) \Rightarrow (x+y)g(x+y)+f(0)=xg(x)+f(0)+(2x+y)g(y) \Rightarrow (x+y)g(x+y)=xg(x)+(2x+y)g(y)$ . Considering here $xy \neq 0$ by swapping $x,y$ we have $xg(x)+(2x+y)g(y)=yg(y)+(2y+x)g(x) \Rightarrow xg(y)=yg(x) \Rightarrow \frac {g(x)} {x}=\frac {g(y)} {y}=c \Rightarrow g(x)=cx$ for every $x \in \mathbb{R}$
(since $g(0)=0$ ) and substituting in (2) we have $f(x)(c-1)=0$ .
If $f(x)=0$ for all real $x$ then (3) gives $g(x)=0$ for all real $x$ (since $g(0)=0$ ) which is indeed a solution.
If $c=1$ we have that $g(x)=x$ and from (3) we obtain $f(x)=f(0)+x^2$ which is a solution,whatever $f(0)$ is.

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ak12sr99

156 posts

ak12sr99

#10 h Oct 8, 2017, 6:28 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Let $P(x,y)$ denote the given functional equation. Then $P(x,0)$ and $P(0,x)$ yield $g(f(x))=f(x)+2xg(0)=f(0)+xg(x) \implies f(x)=f(0)-2xg(0)+xg(x) ....(1).$ Next $P(x,y)$ and $P(x+y,0)$ yield $f(x+y)+2(x+y)g(0)=f(x)+(2x+y)g(y)$ which, in conjunction with $(1)$ , becomes $(f(0)-2(x+y)g(0)+(x+y)g(x+y))+2(x+y)g(0)=(f(0)-2xg(0)+xg(x))+(2x+y)g(y) \iff \frac{g(x)-g(0)}{x}=\frac{g(y)-g(0)}{y}=\lambda$ after which it's just substitution.

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dantaxyz

410 posts

dantaxyz

#11 h Apr 7, 2019, 4:14 AM • 2 Y

Y by AlastorMoody, Adventure10

Let $y=-2x$ to obtain $g(f(-x))=f(x)$ , so $f(-x-y) = f(x)+(2x+y)g(y).$ Let this assertion be $P(x,y)$ .

$P(x,0) \implies f(-x) = f(x) + 2xg(0).$

$P(0,x) \implies f(-x) = f(0) + xg(x)$ .

Compare $P(x,y)$ and $P(y,x)$ to obtain $f(x)+(2x+y)g(y) = f(y) + (2y+x)g(y)$ , and substitute in using the equations above to get $(2x+y)g(y)+f(0)-xg(-x) = (2y+x)g(x) + f(0) - yg(-y)$ . Further substituting gives $(2x+y)g(y)+xg(x)-2xg(0) = (2y+x)g(x) + yg(y) -2yg(0)$ . This simplifies to $2x(g(y)-g(0)) = 2y(g(x)-g(0)) \implies \frac{g(x)-g(0)}{x} = c$ for some constant $c$ . Thus $g$ is linear.

Assume $g(x)=ax+c$ for some reals $a,c$ .

If $a=0$ , we have $f(x) = g(f(-x)) = c$ for all $x$ . Plug this in to obtain $f,g \equiv 0$ .

Else, we have $g(f(x)=af(x)+c=f(-x)$ and $af(-x)+c = f(x)$ , so subtracting these we have $(a+1)(f(x)-f(-x))=0$ . If $a=-1$ , plug this in to obtain no possible solutions. Hence, $af(x)+c=f(-x)=f(x)$ , so either $a=1,c=0$ or $f(x) = \frac{c}{a-1}$ , which is impossible because then $f$ would be constant and $a=0$ .

We know have $g(x)=x$ for all $x$ . Plus this in to $P(0,x)$ to get $f(x)=f(-x) = f(0)+x^2$ .

Hence, our final solutions are $\boxed{g,f \equiv 0}$ and $\boxed{g \equiv x , f \equiv x^2+c}$ .

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FAA2533

65 posts

FAA2533

#12 h Apr 18, 2020, 6:01 AM • 1 Y

Y by THKcandomath

Solution

This post has been edited 14 times. Last edited by FAA2533, Jul 18, 2022, 2:44 PM

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THKcandomath

3 posts

THKcandomath

#14 h Apr 19, 2020, 10:49 AM • 3 Y

Y by Mango247, Mango247, Mango247

FAA2533 wrote:

The answer is ( $f,g$ )=(0,0),( $x^2+c,x$ ).It is easy to check that these solutions satisfy the given equation.We will now show that they are the only solutions.

Let P( $x,y$ ) be the assertion $g(f(x+y))=f(x)+(2x+y)g(y)$ .

Let $f(0)=c$ and $g(0)=d$ .Now
P(0, $x$ ): $g(f(x))=f(0)+xg(x)$
P( $x$ ,0): $g(f(x))=f(x)+2xg(0)$

From the last two equations,we get
$f(x)=c-2xd+xg(x)$

Using the 2nd and the last equation,we get
$c+(x+y)g(x+y)=c-2xd+xg(x)+(2x+y)g(y)$

Exploiting asymmetry,we find
$x(g(y)-d)=y(g(x)-d)$
or $g(x)=hx+d$
Thus plugging value,we get $f(x)=hx^2-dx+c$
Plugging back into the given equation we find the only two solutions ( $f,g$ )=(0,0),( $x^2+c,x$ ).

Nice

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niyu

830 posts

niyu

#16 h Apr 20, 2020, 11:36 PM

Y by

We claim that the only solutions are $(f(x), g(x)) \equiv (0, 0)$ , and $(f(x), g(x)) \equiv (x^2 + e, x)$ , where $e \in \mathbb{R}$ is a real constant. It is easy to check that these are indeed solutions. We now show that these are all.

Let $P(x, y)$ denote the given assertion. From $P(x, -2x)$ , we find that
$\begin{align*} g(f(-x)) &= f(x). \end{align*}$ Hence, $P(x, y)$ rewrites as
$\begin{align*} f(-x - y) &= f(x) + (2x + y)g(y). \end{align*}$ Now, from $P(x, 0)$ and $P(0, x)$ we respectively have
$\begin{align*} f(-x) &= f(x) + 2xg(0) \\ f(-x) &= f(0) + xg(x). \end{align*}$ In particular, we have
$\begin{align*} f(x) + 2xg(0) &= f(0) + xg(x) \\ f(x) - f(0) &= x(g(x) - 2g(0)). \end{align*}$ Hence, we further rewrite $P(x, y)$ as
$\begin{align*} (f(-x - y) - f(0)) &= (f(x) - f(0)) + (2x + y)g(y) \\ (-x - y)(g(-x - y) - 2g(0)) &= x(g(x) - 2g(0)) + (2x + y)g(y) \\ (-x - y)(g(-x - y) - g(0)) + (x + y)g(0) &= x(g(x) - g(0)) - xg(0) + (2x + y)g(y) \\ (-x - y)(g(-x - y) - g(0)) &= x(g(x) - g(0)) + (2x + y)(g(y) - g(0)). \end{align*}$
Now, let $h(x) = x(g(x) - g(0))$ . Rewriting the previous equation in terms of $h$ , we have
$\begin{align*} h(-x - y) &= h(x) + \frac{2x + y}{y}h(y). \end{align*}$ Noting that $h(0) = 0$ , by putting $y = -x$ in the previous equation we obtain
$\begin{align*} h(0) &= h(x) - h(-x) \\ h(x) &= h(-x), \end{align*}$ so $h$ is even. Therefore, we have
$\begin{align*} h(x + y) &= h(x) + \frac{2x + y}{y}h(y). \end{align*}$ Denote this assertion by $Q(x, y)$ . Let $c = h(1)$ . From $Q(x, 1)$ , we obtain
$\begin{align*} h(x + 1) &= h(x) + c(2x + 1). \end{align*}$ From $Q(1, y)$ , we have
$\begin{align*} h(1 + y) &= c + \frac{y + 2}{y}h(y) \\ h(y) + c(2y + 1) &= c + \frac{y + 2}{y}h(y) \\ \frac{2}{y}h(y) &= 2cy \\ h(y) &= cy^2. \end{align*}$ Hence, recalling that $h(x) = x(g(x) - g(0)) = cx^2$ , we find that $g(x) = cx + d$ (where $d = g(0)$ ). Recalling that $f(x) - f(0) = x(g(x) - 2g(0))$ , we have $f(x) = cx^2 - dx + e$ , where $e = f(0)$ .

Now, from $g(f(-x)) = f(x)$ , we have
$\begin{align*} c(cx^2 + dx + e) + d &= cx^2 - dx + e. \end{align*}$ Equating the quadratic coefficients, we have $c^2 = c$ , so $c = 0, 1$ . Suppose $c = 0$ . By equating the linear coefficients, we find that $cd = -d$ , so $d = 0$ . Equating constant coefficients then yields $e = ce + d = 0$ . Hence, in this case, $(f(x), g(x)) \equiv (0, 0)$ . Otherwise, suppose $c = 1$ . Equating the linear coefficients yields $cd = d = -d$ , so $d = 0$ . Therefore, we have $(f(x), g(x)) \equiv (x^2 + e, x)$ .

Thus, as we have exhausted all cases, the only solutions to the functional equation are $(f(x), g(x)) \equiv (0, 0)$ , and $(f(x), g(x)) \equiv (x^2 + e, x)$ for $e$ a real constant, as claimed. $\Box$

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pad

1671 posts

pad

#17 h Jun 4, 2020, 11:38 PM • 1 Y

Y by Mango247

Plug in $y=-2x$ ; we get $g(f(-x)) = f(x)$ . So the equation becomes $f(-x-y) = f(X)+(2x+y)g(y)$ . Call this $P(x,y)$ . $P(x,0)$ gives $f(-x) = f(x)+2xg(0)$ , and $P(0,y)$ gives $f(-y)=f(0)+yg(y)$ , i.e. $f(y)=f(0)-yg(-y)$ . So $P(x,y)$ becomes
$\begin{align*} &f(0)+(x+y)g(x+y) = [f(0)-xg(-x)] + (2x+y)g(y) \\ \implies &(2x+y)[g(x+y) - g(y)] = x[g(x+y)-g(-x)]. \end{align*}$ Call the last equation $Q(x,y)$ . $Q(x,0)$ for $x\not = 0$ gives
$2[g(x)-g(0)] = g(x)-g(-x) \implies g(-x)=2g(0)-g(x).$ In fact, the above also holds for $x=0$ . Let $h(x)=g(x)-g(0)$ . Then $h(-x)=-h(x)$ . Now, $Q(x,y)$ becomes
$\begin{align*} &(2x+y)[g(x+y)-g(y)] = x[g(x+y) + g(x)-2g(0)] \\ \implies &(2x+y)[h(x+y) - h(y)] = x[h(x+y)+h(x)] \\ \implies &(x+y)h(x+y) = xh(x) + (2x+y)h(y). \end{align*}$ Let $k(x)=xh(x)$ . Then the above becomes
$k(x+y) = k(x) + k(y) + \tfrac{2x}{y}k(y).$ Switching $x$ and $y$ above gives $k(x+y) = k(x)+k(y) + \tfrac{2y}{x}k(x)$ . Therefore, $\tfrac{2y}{x}k(x) = \tfrac{2x}{y}k(y)$ , i.e. $k(x) = \tfrac{x^2}{y^2}k(y)$ . Fixing $y$ to be a constant, this tells us that $k(x)=Cx^2$ for some constant $C$ . Now we backtrack; $h(x)=Cx$ , so $g(x)=Cx+D$ for some constant $D$ . And $f(x)=f(0)-xg(-x)=Cx^2-Dx+E$ for some constant $E$ . Plugging back into the original equation and checking gives the solutions as $(f,g)=(0,0),(x^2+E,x)$ for a constant $E$ .

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AliAlavi

27 posts

AliAlavi

#35 h Jul 21, 2022, 9:15 AM • 1 Y

Y by Mango247

Let $P(x,y)$ denote the assertion
$P(x,0) , P(0,x) \Rightarrow f(x) = f(0) - 2xg(0) + xg(x)$
Case 1: $g(0) = 0$
$f(x) = f(0) + xg(x) \Rightarrow g(f(x+y)) = xg(x) + yg(y) + 2xg(y) + f(0) [Q(x,y)]$
$Q(x,y),Q(y,x) \Rightarrow 2xg(y) = 2yg(x) \Rightarrow g(x) = ax+b$
Case 2: $g(0) != 0$
$P(x,-2x) \Rightarrow g(f(-x)) = f(x) *$
$P(x,f(y)),* \Rightarrow f(-x-f(y)) = f(x) + (2x+y)f(-y) [H(x,y)]$
$g(0,0) \Rightarrow f(-f(0)) = f(0) \Rightarrow -f(0) = 0 \Rightarrow g(0) = 0$ but we knew $g(0)$ is not equal to $0$

Solutions : $f(x)=g(x)=0$ and $g(x)=x+b,f(x)=x^2+c$

This post has been edited 5 times. Last edited by AliAlavi, Jan 22, 2023, 10:23 AM

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megarnie

5537 posts

megarnie

#36 h Aug 26, 2022, 5:39 PM

Y by

Either both $f$ and $g$ are the zero function, or $f(x) = x^2 + c$ and $g(x) = x$ , for any constant $c$ . These work.

Let $P(x,y)$ denote the given assertion.

$P(x,y)$ and $P(y,x)$ gives $f(x) + (2x+y)g(y) = f(y) + (2y+x)g(x)$
Let $Q(x,y)$ be this assertion.

$P(0,0): g(f(0)) = f(0)$ .

$P(x,-x): f(0) = f(x) + xg(-x)$ .

$Q(x,0): f(x) + 2xg(0) = f(0) + xg(x)$ .

This implies $f(x) = f(0) + xg(x) - 2xg(0) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$
So $\begin{align*} f(x) + xg(-x) + xg(x) - 2xg(0) = f(x) \\ \implies x(g(-x) + g(x)) = 2xg(0) \\ \implies g(-x) + g(x) = 2g(0)\\ \end{align*}$
By $(1)$ , $Q(x,y)$ can be written as $f(0) + xg(x) - 2xg(0) + (2x+y)g(y) = f(0) + yg(y) - 2yg(0) + (2y+x)g(x)$ Simplifying gives $x(g(y) - g(0)) = y(g(x) - g(0))$
Let $h(x) = g(x) - g(0)$ . Then $xh(y) = yh(x)$ , so $h(x) = ax$ for all $x$ , and some constant $a$ . Thus $g(x) = ax+b$ for constants $a,b$ .

Set $c=f(0)$ .

Then by $(1)$ $\begin{align*} f(x) \\ = f(0) + x(ax+b) - 2xb \\ = ax^2 - bx + c \\ \end{align*}$
$P(0,x): g(f(x)) = f(0) + xg(x)$ .

Thus $af(x) = ax^2 + bx + c - b$ . So $a^2x^2 - (ab)x + ac = ax^2 + bx + (c-b)$
So $a=a^2\implies a=0$ or $a=1$ .

Case 1: $a=0$ .
Then $b$ and $c-b$ are both $0$ , so $b=c=0$ , which implies both $f$ and $g$ are identically zero.

Case 2: $a=1$ .
Then $bx = -bx$ , so $b = -b\implies b=0$ .

In this case, we have $f(x) = x^2 + c$ and $g(x) = x$ , as desired.

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Ibrahim_K

62 posts

Ibrahim_K

#37 h Jan 15, 2023, 8:30 PM

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Let $P(x,y)$ denote the given assertion.

$P(0,1)-P(1,0) \implies f(0)+g(1)=f(1)+2g(0) \ \ \ \ (1)$ $P(0,x)-P(x,0) \implies f(0)+xg(x)=f(x)+2xg(0) \ \ \ \ (2)$ $P(x,1)-P(1,x) \implies f(x)+g(1)(2x+1)=g(x)(x+2)+f(1) \ \ \ \ (3)$
Expanding equation $3$ and replacing $f(x)-g(x)x$ with $f(0)-2xg(0)$ yields

$f(0)+2xg(1)+g(1)=2xg(0)+2g(x)+f(1)$
Replacing $f(0)+g(1)$ with $f(1)+2g(0)$ implies

$xg(1)+g(0)=xg(0)+g(x) \implies g(x)=x(g(1)-g(0))+g(0) \implies g(x)=kx+b$
Using this in equation $2$ yields

$f(x)=x^2k-xb+f(0) \iff f(x)=x^2k-xb+c$
By putting $f(x)=x^2k-xb+c$ and $g(x)=kx+b$ and using $c(k-1)=b$ (by $P(0,0)$ ) we get

$(x+y)(k(k-1)(x+y)-b(k+1))=0 \implies \text{ true for all real x,y hence } x+y \ne 0 \implies k(k-1)(x+y)-b(k+1)=0$
By $x=0,y=0$ and $x=1,y=0$ we get

$k(k-1)=0 \ \ , \ \ b(k+1)=0 \implies k=1,b=0 \text { or } k=0,b=0$
$k=0,b=0 \implies \boxed{ f \equiv g \equiv 0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k=1,b=0 \implies \boxed{ g(x)=x , f(x)=x^2+c}$

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mathscrazy

113 posts

mathscrazy

#38 h Mar 6, 2023, 11:26 AM

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Answer is $f(x)=x^2+c$ , $g(x)=x$ for any constant $c$ , and $f\equiv g \equiv 0$ . Let $P(x,y)$ be the assertion in given equation.

Claim : $g(f(x))=f(x)=f(-x)$
Proof :
$P(x,0) : g(f(x))=f(x)+2xg(0) \dots [\spadesuit]$ .
Assume possible FTSOC $g(0) \neq 0$ . Then $[\spadesuit] \implies f$ injective.
$P(-x,2x) : g(f(x))=f(-x) \dots [\clubsuit]$ .
$P(0,y) : f(-y)=f(0)+yg(y)$ .
Now, $P(0,f(y)) : f(-f(y))=f(0)+f(y)f(-y)$ . And, $P(0,f(-y)) : f(-f(-y))=f(0)+f(-y)f(y)$ . Hence $f(-f(y))=f(-f(-y)) \implies -f(y)=-f(-y)$ by injectivity. But this itself contradicts injectivity.
Hence our assumption that $g(0) \neq 0$ was wrong. Hence $g(0)=0$ .
$[\spadesuit]$ and $[\clubsuit]$ imply $g(f(x))=f(x)=f(-x)$ , as claimed. $\square$

$P(0,x) : f(x) = f(0) + xg(x) \dots [\diamondsuit]$ Using this, comparing $P(x,y)$ and $P(y,x)$ : $f(x)+(2x+y)g(y)=f(y)+(2y+x)g(x) \implies 2xg(y)=2yg(x)$ . Hence, $xg(y)=yg(x) \dots [\heartsuit]$ If $f\equiv 0$ , from $[\diamondsuit]$ we imply that $g \equiv 0$ .
Assume $f \not\equiv 0$ . Hence there exists $t$ such that $f(t) \neq 0$ . $y=t$ in $[\heartsuit]$ and using claim, we get $g(x)=x \text{ for all } x \overset{[\diamondsuit]}{\implies} f(x) = c+x^2$ On verifying, these two solutions indeed work. Hence we get the claimed answers.

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HamstPan38825

8855 posts

HamstPan38825

#39 h Mar 25, 2023, 3:27 PM

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The only solutions are $f(x) = x^2+c$ and $g(x) = x$ , or $f(x) = 0$ and $g(x) = 0$ .

Using symmetry, we have $f(x) + (2x+y)g(y) = f(y) + (2y+x)g(x).$ It turns out that this is the only equation we need.

By setting $x=0$ , $f(0) + yg(y)=f(y) + 2yg(0),$ and by setting $y=-x$ , $f(x) + xg(-x) = f(-x) - xg(x).$ Substituting for $f(x)$ and $f(-x)$ in both expressions, we obtain $g(-x) + g(x) = g(0)x.$ In particular, letting $x=0$ , $g(0) = 0$ , so $g$ is odd. As a result, plugging this back into $P(x, -x)$ , $f$ is even too.

Now, compare the two equations $P(x, qx)$ and $P(x, -qx)$ for any real number $q$ , given below:
$\begin{align*} f(x) + (q+2) x g(qx) &=f(qx) + (2q+1) x g(x) \\ f(x) + (2-q) x g(-qx) &= f(-qx) + (1-2q) g(x). \end{align*}$ Using $g$ odd and $f$ even, we can combine these to read $g(qx) = qg(x)$ . Hence $g(x) = cx$ for constants $c$ , and $f(x) = cx^2+c_1$ . Plugging this back into the original equation, we must have $c = 1$ , which yields the desired solution set.

This post has been edited 1 time. Last edited by HamstPan38825, Apr 2, 2023, 5:27 PM

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pco

23450 posts

pco

#40 h Mar 25, 2023, 3:38 PM

Y by

HamstPan38825 wrote:

The only solutions are $f(x) = x^2+c$ and $g(x) = x$ .

You should read the 38 previous posts. They show that there is another solution.

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Taco12

1757 posts

Taco12

#41 h Apr 3, 2023, 3:48 PM • 2 Y

Y by Jndd, megarnie

Let $P(x,y)$ denote the assertion.

$\begin{align*} P(0,1) \rightarrow g(f(1))&=f(0)+g(1) \\ P(1,0) \rightarrow g(f(1)) &= f(1)+2g(0). \\ \end{align*}$
Thus, $f(0)+g(1)=f(1)+2g(0) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1).$
We also have
$\begin{align*} P(x,1) \rightarrow g(f(x+1))&=f(x)+(2x+1)g(1) \\ P(1,x) \rightarrow g(f(x+1)) &= f(1)+(x+2)g(x). \\ \end{align*}$
Thus, $f(x)+(2x+1)g(1)=f(1)+(x+2)g(x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2).$ Next,
$\begin{align*} P(x,0) \rightarrow g(f(x))&=f(x)+2xg(0) \\ P(0,x) \rightarrow g(f(x)) &= f(0)+xg(x). \\ \end{align*}$
Thus, $f(x)+2xg(0)=f(0)+xg(x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3).$
By doing $(2)-(3)$ and using $(1)$ , we get $(2x+1)g(1)-2xg(0)=g(1)-2g(0)+2g(x),$ implying $g$ is linear. Set $g(x)=ax+b$ . Plugging that into $(3)$ , we get $f(x)+2xb=f(0)+ax^2+bx \rightarrow f(x)=ax^2-bx+c$ . Plugging $f$ and $g$ back into the original equation now gives $f(x)=x^2+c$ and $g(x)=x$ or $f\equiv g \equiv 0$ .

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WinterSecret

7 posts

WinterSecret

#42 h May 13, 2023, 2:57 PM

Y by

Does this work?

Firstly let $P(x,y)$ be the assertion to the function.
$P(0,0) \rightarrow g(f(0))=f(0)$ $P(x,-x) \rightarrow g(f(0))=f(x)+xg(-x)$ So, we have that $f(x)=f(0)-xg(-x)$ . Now, subtitude all the $f$ 's into $g$ ' and we have
$g(f(0)-(x+y)g(-x-y))=f(0)-xg(-x)+(2x+y)g(y)$ Subtituding $x\rightarrow -x$ and $y\rightarrow -y$ and also letting $f(0)=c$ we have
$g(c+(x+y)g(x+y))=c+xg(x)-(2x+y)g(-y)$ $P(0,x) \rightarrow g(c+xg(x))=c-xg(-x)$ $P(x,0) \rightarrow g(c+xg(x))=c+xg(x)-2xg(0)$ So, we have that $g(x)+g(-x)=2g(0)$ for all $x\in \mathbb{R}$ , subtitude $g(-y)=2g(0)-g(y)$ we have
$g(c+(x+y)g(x+y))=c+xg(x)+(2x+y)g(y)-(4x+2y)g(0)\cdots \bigstar$ Also, swapping $x$ and $y$ implies
$g(c+(x+y)g(x+y))=c+yg(y)+(2y+x)g(x)-(2x+4y)g(0)\cdots \square$ Taking the average of $\bigstar$ and $\square$ we have
$g(c+(x+y)g(x+y))=c+(x+y)(g(x)+g(y))-(3x+3y)g(0)$ Hence, $P(x+y,0)$ concludes that $g(x+y)+g(0)=g(x)+g(y)$ if $x+y\ne 0.$ But, when $x+y=0$ it is obvious that $g(x+y)+g(0)=g(x)+g(y)$ .
Now, the functional equation becomes
$g(c)+g((x+y)g(x+y))-g(0)=c+(x+y)(g(x+y)+g(0))-(3x+3y)g(0)$ Thus,
$g((x+y)g(x+y))=(x+y)g(x+y)+(1-2x-2y)g(0)$ And so,
$g(xg(x))=xg(x)+(1-2x)g(0)\cdots \clubsuit$
Now, let $h(x)=g(x)-x$ so, $g(x)=h(x)+x$ . Note that $h(0)=g(0)$ and $h(x+y)+h(0)=h(x)+h(y)$ . So, the functional equation becomes
$h(xg(x))+xg(x)=xg(x)+(1-2x)h(0)$ $h(xh(x)+x^2)=(1-2x)h(0)$ $h(xh(x))+h(x^2)-h(0)=(1-2x)h(0)$ Taking $x=1$ yields
$h(h(1))+h(1)=0 \rightarrow g(h(1))=0 \rightarrow g(g(1)-1)=0.$ Now, let $t=g(1)-1$ so, $g(t)=0$ . Also subtituding $x=t$ in $\clubsuit$ gives $g(0)=(1-2t)g(0)\rightarrow t=0$ or $g(0)=0.$
But, either way we have $g(0)=0$ . Comparing the equivalency in $\bigstar$ and $\square$ we have $xg(y)=yg(x)\cdots \diamondsuit$ .

Observe that:
i.) If $t=0$ then $g(1)=1$ and subtituding $y=1$ in $\diamondsuit$ we have that $\boxed{g(x)=x}$ for all $x$ and $\boxed{f(x)=f(0)-xg(-x)=x^2+c}$
ii.) If $t\ne 0$ then subtituding $y=t$ in $\diamondsuit$ we have that $\boxed{g(x)=0}$ for all $x$ and $f(x)=f(0)-xg(-x)=c$ and it is easy to check that $c=0$ so, $\boxed{f(x)=0}$
Hence, there are 2 pairs of function that satisfies the problem that is: $(f,g)=(x^2+c,x),(0,0)$

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IAmTheHazard

5000 posts

IAmTheHazard

#43 h Aug 13, 2023, 5:24 PM • 1 Y

Y by centslordm

no brain solution

The answer is $f \equiv g \equiv 0$ , as well as $f(x)=x^2+c$ and $g(x)=x$ , which clearly work. Let $P_1(x,y)$ denote the given assertion.

From $P_1(-x,2x)$ , we get $g(f(x))=f(-x)$ , hence we can rewrite the functional equation as $f(-x-y)=f(x)+(2x+y)g(y)$ . Let $P_2(x,y)$ denote this assertion: we will first ignore the original equation and solve this one instead.

Since $f$ works iff $f+c$ does for any fixed constant $c$ , WLOG let $f(0)=0$ . Then from $P_2(x,-x)$ we find that $f(x)+xg(-x)=0 \implies f(x)=-xg(-x)$ . Thus we can once again rewrite this equation as $(x+y)g(x+y)=-xg(-x)+(2x+y)g(y)$ . From plugging in $y=0$ , we get $xg(-x)=-xg(x)$ , hence we can rewrite this equation as $(x+y)g(x+y)=xg(x)+(2x+y)g(y)$ . Call this equation $P_3(x,y)$ ; we will solve it first.

It is clear that any linear functions work, and that the set of solutions forms a $\mathbb{R}$ -vector space, hence we can shift so $g(0)=g(1)=0$ , which additionally implies that $g(-1)=0$ . Suppose that there exists some $a$ with $g(a) \neq 0$ , so $a \not \in \{-1,0,1\}$ . From $P(1,a-1)$ we find that $ag(a)=(a+1)g(a-1)$ . From $P(-1,a)$ we find that $(a-2)g(a)=(a-1)g(a-1)$ . Since $(a+1)g(a-1)=ag(a) \neq 0$ , we can divide to obtain
$\frac{a-2}{a}=\frac{a-1}{a+1} \implies (a-2)(a+1)=a(a-1) \implies -2=0,$ which is absurd. Hence $g \equiv 0$ , so the only solutions to this equation are linear.

Suppose that $g(x)=ax+b$ . Then $f(x)=-x(-ax+b)=ax^2-bx+c$ , since we shifted $f$ so that $f(0)=0$ . Instead of directly plugging this back into the original solution, we instead inspect $g(f(x))=f(-x)$ , which implies that $a=0$ or $a=1$ by comparing the leading coefficients. If $a=0$ , then $g(x)=b$ , and $f(x)=-bx+c$ , but it is clear that $f$ must be constant, hence $g \equiv 0$ , from which $f \equiv 0$ is clear. If $a=1$ , then since $g(f(0))=f(0) \implies g(c)=c$ , we find that $b=0$ , hence $g(x)=x$ and $f(x)=x^2+c$ , which is our other solution. $\blacksquare$

This post has been edited 1 time. Last edited by IAmTheHazard, Aug 13, 2023, 9:11 PM

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Math4Life7

1703 posts

Math4Life7

#44 h Nov 4, 2023, 3:52 AM

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We can see that $f(y) + (2y+x)g(x) = g(f(y+x)) = g(f(x+y)) = f(x) + (2x+y)g(y)$ . Let $P(x, y)$ denote the assertion $f(y) + (2y+x)g(x) = f(x) + (2x+y)g(y)$ We can see from $P(x, 0)$ that we have $f(0) + xg(x) = f(x) + 2xg(0) \Rightarrow f(x) = xg(x) + f(0) - 2xg(0)$ We can plug this back into our $P(x, y)$ thingy to get $yg(y) + f(0) - 2yg(0) + (2y+x)g(x) = xg(x) + f(0) - 2xg(0) + (2x+y)g(y) \Rightarrow \frac{g(x) - g(0)}{x} = \frac{g(y) - g(0)}{x}$ Thus $g(x) = ax + c$ where $c = g(0)$ . Going back to $P(x, 0)$ we have $f(x) = xg(x) + f(0) - 2xg(0) = ax^2 + cx + b - 2xc = ax^2 -cx + b$ . We can plug our values for $f$ and $g$ back into our original equation to find that we get $ax^2 - cx + b +2ax^2 + 2cx + axy + cy = a(ax^2 + 2axy + by^2 -cx - cy + b)+c$ . This obviously gives that $a = 1, 0$ and $c = 0$ . Checking we can find that $\boxed{f(x) = x^2 + c, g(x) = x}$ and $\boxed{f = 0, g = 0}$ work for a constant $c$ . $\blacksquare$

This post has been edited 1 time. Last edited by Math4Life7, Nov 4, 2023, 3:55 AM

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lelouchvigeo

172 posts

lelouchvigeo

#45 h Jan 10, 2024, 10:29 AM

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The solutions are $\boxed{f(x) \equiv 0, g(x) \equiv 0}$ or $\boxed{f(x) \equiv x^2+c, g(x) \equiv x}$ .
Solution which I will fully upload later

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P2nisic

406 posts

P2nisic

#47 h Jun 24, 2024, 2:29 PM

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orl wrote:

Determine all pairs $(f,g)$ of functions from the set of real numbers to itself that satisfy $g(f(x+y)) = f(x) + (2x + y)g(y)$ for all real numbers $x$ and $y$ .

Proposed by Japan

Let $P(x,y)\Leftrightarrow g(f(x+y))=f(x)+(2x+y)g(y)$
$P(x,0)\Rightarrow g(f(x))=f(x)+2xg(0)\$
$P(0,x) \Rightarrow g(f(x))=f(x)+2xg(0) \$
from this two we get $f(x)=f(0)+x(g(x)-2g(0))$ (1)

$P(x,-2x)$ gives $g(f(-x))=f(x)$
Using this the first became: $f(-x-y)=f(x)+(2x+y)g(y)$

Now we have that:
$f(-x-y-z)=f(x+y)+(2x+2y+z)g(z)=f(-x)+(-2x-y)g(-y)+(2x+2y+z)g(z)$
By symmetry on $y,z$ we get that:
$(-2x-y)g(-y)+(2x+2y+z)g(z)=(-2x-z)g(-z)+(2x+2z+y)g(y)$
For $x=y=0$ we have that $-g(-z)=g(z)-2g(0)$
so the last one became:
$(2x+y)g(y)-2g(0)g(y)+(2x+2y+z)g(z)=(2x+z)g(z)-2g(0)g(z)+(2x+2z+y)g(y)\Rightarrow (2z+2g(0))g(y)= (2y+2g(0))g(z)\Rightarrow g(x)=ax+b$
And from (1) we have the form of $f(x)$
Now esily we get that: the only solutions are $(f(x), g(x)) \equiv (0, 0)$ , and $(f(x), g(x)) \equiv (x^2 + e, x)$ , where $e \in \mathbb{R}$ is a real constant.

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Mathandski

715 posts

Mathandski

#48 h Dec 9, 2024, 7:57 PM

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$\text{[math]}$

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alba_tross1867

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alba_tross1867

#49 h Jan 24, 2025, 2:10 PM

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orl wrote:

Determine all pairs $(f,g)$ of functions from the set of real numbers to itself that satisfy $g(f(x+y)) = f(x) + (2x + y)g(y)$ for all real numbers $x$ and $y$ .

Proposed by Japan

$P(-x,2x) : g(f(x))=f(-x)$ .
$P(x,0) : g(f(x))=f(x)+2xg(0) \Rightarrow f(-x)=f(x)+2xg(0)$
$P(0,x) : f(-x)=f(0)+xg(x)$
Thus, $f(x)=f(0)+xg(x)-2xg(0)$
$P(x,y)$ and $P(y,x)$ gives us : $f(x)+(2x+y)g(y)=f(y)+(2y+x)g(x)$
Replacing $f$ we get : $f(0)+xg(x)-2xg(0)+2xg(y)+yg(y)=f(0)+yg(y)+2yg(x)+xg(x)-2yg(0)$
Then : $x(g(y)-g(0))=y(g(x)-g(0))$ which yields $g(x)=mx+p\ \forall x\neq 0$ with $p=g(0)$ which means $g$ is linear.
We get $f(x)=f(0)+ax^2+bx$ replacing in the fe ends the prob.
Answer is both functions are $0$ or $(f,g)=(x^2+c,x)$

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Ilikeminecraft

300 posts

Ilikeminecraft

#50 h Mar 18, 2025, 5:34 PM

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bruh. the answer is $f\equiv x^2 + c, g\equiv x$

$(-x, 2x+y)$ gives $f(x) + (2x +y)g(y) = f(0) + (x + y)g(x + y).$

$f(x) + (2x + y)g(y) = f(y) + (x + 2y) g(x)$ follows from symmetry trick
we will spam this
$(x, 0)$ gives $f(x) + 2xg(0) = f(0) + xg(x),$ which gives $g(x) = \frac{f(x) - f(0)}{x} + 2g(0)$
$(1, x)$ gives $f(1) + (2 + x)g(x) = f(x) + (2x + 1)g(1),$ or $g(x) = \frac{f(x) - f(1) + 2xg(1) + g(1)}{x + 2}$

by setting these two to be equal, we get $xf(x) - xf(1) + 2x^2g(1) + xg(1) = (f(x) - f(0) + 2xg(0))(x + 2) = xf(x) - xf(0) + 2x^2g(0) + 2f(x) - 2f(0) + 4xg(0)$
simplifying, we observe that we can write $f\equiv ax^2 + bx + c$
thus, $g\equiv ax + g(0)$
plugging into the given, we can tell that $a = 1$
thus, $g\equiv x +g(0), f\equiv x^2 + g(0)x + c$
simple plugging in tells us $g(0) = 0$

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