Find all functions f with f(x+y)+f(x)f(y)=f(xy)+(y+1)f(x)+(x+1)f(y)

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Find all functions f with f(x+y)+f(x)f(y)=f(xy)+(y+1)f(x)+(x+1)f(y)

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Source: (4th Middle European Mathematical Olympiad, Individual Competition, Problem 1)

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#1 h Sep 11, 2010, 12:34 PM • 2 Y

Y by Adventure10, Mango247

Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that for all $x, y\in\mathbb{R}$ , we have
$f(x+y)+f(x)f(y)=f(xy)+(y+1)f(x)+(x+1)f(y).$

This post has been edited 1 time. Last edited by Amir Hossein, May 3, 2018, 8:45 AM

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binaj

#2 h Sep 11, 2010, 2:56 PM • 2 Y

Y by Adventure10, Mango247

do you know anybody who solved this problem?

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#3 h Sep 11, 2010, 3:00 PM • 2 Y

Y by Adventure10, Mango247

binaj wrote:

do you know anybody who solved this problem?

I'm sure someone solved it, but I didn't and neither did the rest of the Austrian team...

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pco

#4 h Sep 11, 2010, 7:44 PM • 7 Y

Y by Tawan, Jerry37284, lazizbek42, Adventure10, Mango247, Blackbeam999, and 1 other user

Martin N. wrote:

Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that for all $x\mbox{, }y\in\mathbb{R}$ , we have
$f(x+y)+f(x)f(y)=f(xy)+(y+1)f(x)+(x+1)f(y)\mbox{.}$

(4th Middle European Mathematical Olympiad, Individual Competition, Problem 1)

Let $P(x,y)$ be the assertion $f(x+y)+f(x)f(y)=f(xy)+(y+1)f(x)+(x+1)f(y)$

$P(0,0)$ $\implies$ $f(0)^2=2f(0)$ and so $f(0)=0$ or $f(0)=2$

If $f(0)=2$ , $P(x,0)$ $\implies$ $f(x)=x+2$ which is not a solution. So $f(0)=0$

$P(1,-1)$ $\implies$ $f(1)f(-1)=3f(-1)$ and so $f(-1)=0$ or $f(1)=3$

If $f(1)=3$ , $P(x-1,1)$ $\implies$ $\boxed{f(x)=3x}$ which indeed is a solution.

If $f(1)\ne 3$ , then $f(-1)=0$

$P(-x,1)$ $\implies$ $f(-x+1)=(3-f(1))f(-x)+(-x+1)f(1)$
$P(x,-1)$ $\implies$ $f(x-1)=f(-x)$
$P(x-1,-1)$ $\implies$ $f(x-2)=f(-x+1)$

And so $f(x-2)=(3-f(1))f(x-1)+(-x+1)f(1)$
But $P(x-2,1)$ $\implies$ $f(x-1)=(3-f(1))f(x-2)+(x-1)f(1)$
Adding these two lines gives $(2-f(1))(f(x-2)+f(x-1))=0$

If $f(1)\ne 2$ , this implies $f(x-2)=-f(x-1)$ and so $f(x+1)=-f(x)$ and $f(1)=0$
But then $P(x,1)$ $\implies$ $f(x+1)=3f(x)$ and so $-f(x)=3f(x)$ and $\boxed{f(x)=0}$ $\forall x$ which indeed is a solution.

If $f(1)=2$ , then $P(x,1)$ $\implies$ $f(x+1)=f(x)+2x+2$ and so :

$f(x+n)=f(x)+2nx+n^2+n$
So $f(n)=n^2+n$

Then $P(\frac pq,q)$ $\implies$ $f(\frac pq)=(\frac pq)^2+\frac pq$ and we got $f(x)=x^2+x$ $\forall x\in\mathbb Q$

Let then $g(x)=f(x)-x^2-x$ such that $g(x)=0$ $\forall x\in\mathbb Q$ . The original equation becomes :

New assertion $Q(x,y)$ : $g(x+y)+g(x)g(y)+g(x)(y^2-1)+g(y)(x^2-1))=g(xy)$

$Q(x,1)$ $\implies$ $g(x+1)=g(x)$

Using this property and subtracting $Q(x,y)$ from $Q(x,y+1)$ , we get $g(x)(2y+1)=g(xy+x)-g(xy)$

So $g(x+y)=g(x)+g(y)(\frac{2x}y+1)$ $\forall x,y\ne 0$

So $g(x)+g(y)(\frac{2x}y+1)=g(y)+g(x)(\frac{2y}x+1)$ and so $\frac{g(y)}{y^2}=\frac{g(x)}{x^2}$ $\forall x,y\ne 0$
So $g(x)=ax^2$ and so $g(x)=0$ and $\boxed{f(x)=x^2+x}$ which indeed is a solution.

Hence the three solutions of the equation :

$f(x)=0$
$f(x)=3x$
$f(x)=x^2+x$

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#5 h Sep 11, 2010, 9:54 PM • 2 Y

Y by Tawan, Adventure10

A bit different solution
$f(x+y)+f(x)f(y)=f(xy)+(y+1)f(x)+(x+1)f(y)$
$P(0,0) \implies {f(0)}^2=2f(0)$
if $f(0)=2$ from $P(x,0) \implies f(x)=x+2$ which is not a solution
so $f(0)=0$
Put $P(1,-1) \implies f(1)f(-1)=3f(-1)$
if $f(1)=3$
$P(x-1,1) \implies f(x)=3x$ which is a solution
so we have $f(-1)=0$
$P(-1,-1) \implies f(-2)=f(1)$
$P(-2,1) \implies f(-1)=f(-2)(3-f(1))+(-1)f(1)$
so $0=f(1)(2-f(1))$
if f(1)=0 we get
$P(x,-1) \implies f(x-1)=f(-x)$
$P(x-1,1) \implies f(x)=3f(x-1)=3f(-x)$
so from $f(x)=3f(-x)$ we get $f(x)=3f(-x)=3(3f(-(-x)))=9f(x) \implies f(x)=0$ which is a solution
if f(1)=2 we get
$P(x,-1) \implies f(x-1)=f(-x)$
$P(x-1,1) \implies f(x)=f(x-1)+2x=f(-x)+2x$
so $f(x)=f(-x)+2x$
$P(x,-x) \implies f(0)+f(x)f(-x)=f(-x^2)+(x+1)f(-x)+(-x+1)f(x)$
$0+f(x)(f(x)-2x)=(f(x^2)-2x^2)+(x+1)(f(x)-2x)+(-x+1)f(x) \implies f(x)^2-2xf(x)=f(x^2)-2x^2+xf(x)-2x^2+f(x)-2x-xf(x)+f(x)$
$f(x)^2=f(x^2)-4x^2+2f(x)+2xf(x)-2x$
$P(x,x) \implies f(2x)+ {f(x)}^2=f(x^2)+2(x+1)f(x)$
$f(2x)+(f(x^2)-4x^2+2f(x)+2xf(x)-2x)=f(x^2)+2(x+1)f(x)$
$f(2x)=4x^2+2x$ or
$f(x)=x^2+x$ which also shows to be a solution

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#6 h Oct 31, 2010, 3:35 PM • 2 Y

Y by Tawan, Adventure10

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Zhero

#7 h Nov 1, 2010, 10:51 PM • 4 Y

Y by LMat, Tawan, Adventure10, Mango247

FantasyLover wrote:

Rewriting the functional equation in terms of $g$ , we have $g(x)g(y)=g(xy)+g(x)+g(y)-g(x+y)+2xy$ .

Just a comment: if we let $h(x) = g(x) - 1$ , then $(h(x) + 1)(h(y) + 1) + h(x+y) + 1 = h(xy) + 1 + h(x) + 1 + h(y) + 1 + 2xy$ , or $h(x)h(y) + h(x+y) = h(xy) + 2xy + 1$ , which is exactly 2005 A4.

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#8 h Jan 2, 2019, 1:10 AM • 2 Y

Y by Adventure10, Mango247

Let $P(x,y)$ be the assertion into the problem statement.
$P(0,0) \implies f(0)^2=2f(0)$ so either $f(0)=0$ or $f(0)=2.$

Case 1: $f(0)=2.$
$P(x,0)\implies 3f(x)=2+f(x)+2x+2\implies f(x)=x+2$ . Plugging this into the problem statement yields that it is not a solution.

Case 2: $f(0)=0.$
$P(-1,1)\implies f(1)f(-1)=3f(-1)$ so $(f(1)-3)f(-1)=0$ . Either $f(1)=3$ or $f(-1)=0.$

Case 2.1: $f(1)=3.$
$P(x-1,1)\implies \boxed{f(x)=3x}$ and plugging this in indeed works.

Case 2.2: $f(-1)=0.$
$P(2,-1)\implies f(1)=f(-2).$
$P(-2,1)\implies f(-2)f(1)=3f(-2)-f(1).$ From above, $f(1)=f(-2)$ so it becomes $f(1)^2=2f(1)$ so either $f(1)=0$ or $f(1)=2$ .

Case 2.2.1: $f(1)=0$ .
$P(x,-1)\implies f(x-1)=f(-x). P(x-1,1)\implies f(x)=3f(x-1)$ . Substituting $f(x-1)$ from first into the second yields $f(x)=3f(-x).$ Let $x=x$ and $x=-x$ into the past function to get $f(x)=3f(-x)$ and $f(-x)=3f(x).$ Substituting the $f(-x)$ from the second into the first yields $f(x)=3f(-x)=3(3f(x))=9f(x)\implies \boxed{f(x)=0}.$ We can easily check that this function works.

Case 2.2.2: $f(1)=2.$
$P(x-1,1)\implies f(x)=f(x-1)+2x$ .
$P(x,-1)\implies f(x-1)=f(-x)$ . Combining these yields $f(x)=f(-x)+2x.$ We can easily compute $f(2)=6$ by $P(1,1)$ .
$P(x,2)=f(2x)=f(x+2)+3f(x)-6x-6=f(x+1)+2(x+2)+3f(x)-6x-6=f(x)+2(x+1)+2(x+2)+3f(x)-6x-6=4f(x)-2x$ , where we used $f(x)=f(x-1)+2x$ twice.
$P(x,x)=f(2x)+(f(x))^2=f(x^2)+2xf(x)+2f(x)=4f(x)-2x+(f(x))^2+2xf(x)+2f(x)$ , where we simplified $f(2x).$
$P(x,-x)=f(x)f(-x)=f(-x^2)+(x+1)f(-x)+(-x+1)f(x)$ . Using $f(x)=f(-x)+2x,$ we have $f(-x^2)=f(x^2)-2x^2$ and $f(-x)=f(x)-2x.$ Substituting these in yields $f(x)^2-2xf(x)=f(x^2)-4x^2-2x+2f(x).$ Subtracting this from the $P(x,x)$ line gives $\boxed{f(x)=x^2+x}$ and we can easily check that this function works.

Hence, the three solutions are $f(x)=0, f(x)=3x,$ and $f(x)=x^2+x$ .

This post has been edited 1 time. Last edited by Jzhang21, Jan 2, 2019, 1:11 AM
Reason: typo

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#10 h Jan 16, 2022, 5:10 AM

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Answer:Click to reveal hidden text

This post has been edited 1 time. Last edited by AMichael, Jan 17, 2022, 1:05 AM
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#11 h Apr 28, 2025, 4:38 PM

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Let $P(x,y)$ denote the given assertion.
$P(0,0): f(0) + f(0)^2 = 3f(0) \to f(0)^2 = 2f(0) \to f(0) = 0$ or $2$ .
If $f(0) = 2$ then $P(x,0): 3f(x) = 2 + f(x) + 2x + 2 \to 2f(x) = 2x + 4 \to f(x) = x + 2$ which we can check and see is not a solution.
So $f(0) = 0$ . $P(-1, 1): f(-1)f(1) = f(-1) + 2f(-1) \to f(-1)f(1) = 3f(-1)$ .
Case 1: $f(-1) \neq 0 \to f(1) = 3$ , then $P(x, 1): f(x+1) + 3f(x) = f(x) + 2f(x) + 3(x+1) \to f(x+1) = 3(x+1) \to \underline{f(x) = 3x}$ which is indeed a solution.
Case 2: $f(-1) = 0$ . $P(x, -1): f(x-1) = f(-x) \to f(1) = f(-2)$
$P(1, -2): f(1)^2 = f(1) - f(1) + 2f(1) = 2f(1)$ , which means either $f(1)=0$ or $f(1) = 2$
If $f(1)=0$ then $P(x-1, 1): f(x) = f(x-1) + 2f(x-1) \to f(x) = 3f(-x) \quad x:= -x \to f(-x) = 3f(x) \to f(x) = 9f(x) \to \underline{f(x) = 0}$ which is a solution.
And if $f(1) = 2$ then $P(x-1, 1): f(x) + 2f(x-1) = f(x-1) + 2f(x-1) + 2x \to f(x) = f(x-1) + 2x = f(-x) + 2x$ , so $f(2) = f(-2) + 4 = f(1) + 4 = 6$
$P(x, 2): f(x+2) + 6f(x) = f(2x) + 3f(x) + 6x + 6, f(x+2) = f(x+1) + 2(x+2) = f(x) + 2(x+2) + 2(x+1) = f(x) + 4x + 6 \to f(x) + 4x + 6 + 6f(x) = f(2x) + 3f(x) + 6x + 6 \to 4f(x) = f(2x) + 2x$
So $f(2x) = 4f(x) - 2x$ , now $P(x, x): 4f(x) - 2x + f(x)^2 = f(x^2) + 2xf(x) + 2f(x) \to f(x)^2 + 2f(x)= f(x^2) + 2xf(x) + 2x$
$P(x, -x): f(x)f(-x) = f(-x^2) -xf(x) + f(x) + xf(-x) + f(-x) \to f(x)(f(x) - 2x) = f(x^2) - 2x^2 - xf(x) + f(x) + x(f(x) - 2x) + f(x) - 2x \to f(x)^2 - 2xf(x) = f(x^2) - 4x^2 + 2f(x) -2x$
$\to f(x)^2 + 4x^2 + 2x= f(x^2) + 2xf(x) + 2f(x)$
Now we have:
$f(x)^2 + 2f(x)= f(x^2) + 2xf(x) + 2x$
$f(x)^2 + 4x^2 + 2x= f(x^2) + 2xf(x) + 2f(x)$
Minusing these equations, we have:
$2f(x) - 4x^2 - 2x = 2x - 2f(x) \to 4f(x) = 4x^2 + 4x \to \underline{f(x) = x^2 + x}$ , plugging this into the problem equation, we can check that this is a solution.
So finally, we proved the answers are $\boxed{f(x) = 3x \quad \forall x \in \mathbb{R}}$ , $\boxed{f(x) = 0 \quad \forall x \in \mathbb{R}}$ , and $\boxed{f(x) = x^2 + x \quad \forall x \in \mathbb{R}}$ . $\blacksquare$

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#12 h Apr 28, 2025, 4:46 PM

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The vast majority of these types of problems are bashable by doing $P(x,1)$ to get an expression of $f(x+1)$ in terms of $f(x)$ and then doing $P(x,y+1)$ to get an expression of $f(x+y)$ in terms of $x,y,f(x),f(y)$ .

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