IMO ShortList 2002, algebra problem 1

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3647 posts

orl

#1 h Sep 28, 2004, 1:15 PM • 13 Y

Y by jam10307, Bst, mathematicsy, MathLuis, centslordm, junioragd, megarnie, Adventure10, Mango247, ItsBesi, torch, NicoN9, fuzzball2023

Find all functions $f$ from the reals to the reals such that

$f\left(f(x)+y\right)=2x+f\left(f(y)-x\right)$

for all real $x,y$ .

Attachments:

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Philip_Leszczynski

327 posts

Philip_Leszczynski

#2 h Jan 30, 2006, 2:16 AM • 10 Y

Y by Nguyenhuyhoang, AdithyaBhaskar, Vaijan_Mama, centslordm, Adventure10, samrocksnature, megarnie, mathmax12, Mango247, EquationTracker

Yay! My first (first) IMOSL solution!

First we let $f(0) = \alpha$ .

$f(f(x)+y) = 2x + f(f(y)-x)$ $(1)$

Plugging in $x=0$ into (1):

$f(y+\alpha) = f(f(y))$ $(2)$

Plugging in $x=f(y)$ into (1):

$f(f(f(y))+y) = 2f(y) + f(f(y)-f(y)) \Rightarrow f(2y+\alpha) = 2f(y) + \alpha$ $(3)$

Now plugging in $y=-\alpha$ into (3):

$f(-\alpha) = 2f(-\alpha) + \alpha \Rightarrow f(-\alpha) = -\alpha$

Plugging in $y=-\alpha$ into (2):

$\alpha = f(f(-\alpha)) \Rightarrow \alpha = -\alpha \Rightarrow \alpha = 0$

So (2) can be rewritten as: $f(f(x)) = x$ $(4)$

Plugging in $y=0$ into (1):

$f(f(x)) = 2x + f(-x) \Rightarrow f(-x) = -x$

Plugging in $x=-x$ for the last equation gives $f(x)=x$ for all $x$ .

It is easy to see that this function indeed satisfies the initial conditions given in the problem.

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silouan

3952 posts

silouan

#3 h Mar 16, 2006, 1:28 PM • 2 Y

Y by centslordm, Adventure10

Philip_Leszczynski wrote:

$f(f(f(y))+y) = 2f(y) + f(f(y)-f(y)) \Rightarrow f(2y+\alpha) = 2f(y) + \alpha$ $(3)$

You are wrong here sorry you miscalculate I think.
The correct is

$f(f(f(y))+y) = 2f(y) + f(f(y)-f(y)) \Rightarrow f(f(y+\alpha)+y)=2f(y)+\alpha$

from where we can find (plugging where $y=-a$ ) that $f(-\alpha)=0$ and finally
the solutions in not $f(x)=x$ but $f(x)=x+c$

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Jorge Miranda

157 posts

Jorge Miranda

#4 h Jun 2, 2009, 2:07 PM • 23 Y

Y by Patcharapong, Mikasa, Sx763_, TheOneYouWant, Tawan, arshakus, Takeya.O, naw.ngs, Night_Witch123, Lifefunction, Williamgolly, Illuzion, centslordm, Jalil_Huseynov, Chokechoke, mathmax12, Adventure10, Mango247, rstenetbg, ihatemath123, NicoN9, fuzzball2023, and 1 other user

Solution

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dysfunctionalequations

522 posts

dysfunctionalequations

#5 h Dec 8, 2009, 2:52 AM • 10 Y

Y by MSTang, B.J.W.T, Vladimir_Djurica, Adventure10, Mango247, ike.chen, MS_asdfgzxcvb, and 3 other users

Setting $y=-f(x)$ , we have $f(0)=2x+f(f(-x)-x))$ , so $f(0)-2x=f(f(-x)-x)$ , and $f(0)-2x$ is a surjective function of $x$ , so $f$ is surjective as well.

Then, by surjectivity, we can let $c$ be a number for which $f(c)=0$ . Setting $x=c$ , we have $f(y)=2c+f(f(y)-c)$ . Since $f(y)$ can take on all real values $r$ , $r=2c+f(r-c)$ for all real numbers $r$ , implying $f(r-c)=r-2c$ . Letting $s=r-c$ , we find $\boxed{f(s)=s-c}$ for some constant $c$ for all real numbers $s$ . Also, it can be easily verified that all functions of this form satisfy this equation.

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Rijul saini

904 posts

Rijul saini

#6 h Mar 18, 2011, 5:11 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let $P(x,y)$ denote the assertion $f(f(x)+y) = 2x + f(f(y)-x)$ .
Now,
$P(0,x) \implies f(x+f(0)) = f(f(x))$
$P(f(y),y) \implies f(f(f(y)) + y ) = 2f(y)+f(0) \implies f(f(y+f(0))+y) = 2f(y) + f(0)$
Taking $y=-f(0)$ here, implies $f(0) = 2f(-f(0)) +f(0) \implies f(-f(0)) = 0$ .
Next,
$P(-\frac x2, -f(-\frac x2) - f(0)) \implies x = f(f(-f(-\frac x2) - f(0)) + \frac x2)$
Therefore, $f$ is onto i.e. covers all values of $x$ .
Now,let $f(0) = c$ , and define a function $g : \mathbb{R} \rightarrow \mathbb{R}$ as $g(x) = f(x - c)$ .
Therefore, our original assertion $P(x,y)$ for $f$ changes into the equivalent assertion $Q(x,y)$ for $g$ , as being
$g(g(x+c) + y + c) = 2x + g(g(y+c) - x +c)$
As $f$ is onto, therefore $g$ is also onto. Also, $g(0) = 0$ . And then finally,
$Q(-c,x-c) \implies g(x) = -2c + g(g(x) + 2c) \implies g(g(x) + 2c) = g(x) + 2c$
As $g$ is onto, therefore, $g(x)$ covers all values of $x$ , and therefore, $g(x) + 2c$ also covers all values of $x$ ,
Thus, $g(x) = x$ for every $x \in \mathbb{R}$ .
And therefore, $f(x) = x+f(0)$ for every $x \in \mathbb{R}$ and this is indeed a solution.

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pmtrig

12 posts

pmtrig

#7 h Jul 16, 2013, 4:43 PM • 2 Y

Y by Adventure10, Mango247

Set $y=-f(x)$ then $-2x=f(f(-f(x))-x)$ so $f$ is $surjective$ .
So there exist $a,b$ such that $f(a)=0$ and $f(b)=a$ . Then set $x=a$ and $y=b$ $\Rightarrow$ $a=2a+f(0)$ so $f(-f(0))=0$ .
Now let $x=-f(0)$ $\Rightarrow$ $f(y)=-2f(0)+f(f(y)+f(0))$
let $f(0)=c$ $\Rightarrow$ $(f(y)+c)+c=f(f(y)+c)$ . Since $f$ is $surjective$ we get that $f(y)+c$ is also $surjective$ which implies $f(x)=x+c$ $\forall$ $x$ $\in$ $\mathbb{R}$ where $c=f(0)$ .

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CanVQ

83 posts

CanVQ

#8 h Jan 30, 2014, 5:47 PM • 2 Y

Y by Adventure10, Mango247

orl wrote:

Find all functions $f$ from the reals to the reals such that

$f\left(f(x)+y\right)=2x+f\left(f(y)-x\right)$

for all real $x,y$ .

This is my solution: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3376958#p3376958

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strujabog

91 posts

strujabog

#10 h Jun 20, 2014, 1:08 PM • 2 Y

Y by Adventure10, Mango247

Notice that we can take $x$ such that $x=\frac{1}{2}\cdot f(y)$ .Also it is impossible to be $f(x)>0$ because we can always take $x$ so small to make that absurd.Putting $y=y_1$ ,( $y_1$ is such that $f(y_1)=0$ ),and $x->x_1=\frac{1}{2}\cdot f(y)$ we easily get $f(f(x_1)+y)=2x_1+f(\frac{1}{2}\cdot f(y))$ .Lets assume that there is $x_2$ such that $x_1\neq x_2$ and $f(x_1)=f(x_2)$ .From that it follows: $2x_1+f(\frac{1}{2}\cdot f(y))=f(f(x_1)+y)=f(f(x_2)+y)=2x_2+f(\frac{1}{2}\cdot f(y))$ ,then we get $2x_1=2x_2$ .Clearly the function is an injection.
Putting $x=0$ we get $f(f(0)+y)=f(f(y))$ which is equivalent to $f(x)=x+f(0)$ ,(lets take $f(0)=c$ ).
Now it remains to check if that is a solution: $f(f(x)+y)=f(x+c+y)=x+y+2c$ and $RHS=f(f(y)-x)+2x=f(y-x+c)+2x=y-x+2c+2x=y+x+2c$ .Obviously $LHS=RHS$ .Hence the only solution is $f(x)=x+c$

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strujabog

91 posts

strujabog

#11 h Jun 20, 2014, 1:10 PM • 2 Y

Y by Adventure10, Mango247

please correct me if I am wrong

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strujabog

91 posts

strujabog

#12 h Jun 20, 2014, 5:13 PM • 1 Y

Y by Adventure10

is this solution above ok?

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fclvbfm934

759 posts

fclvbfm934

#13 h Aug 17, 2014, 9:54 PM • 1 Y

Y by Adventure10

Let $f(0) = a$ . We will first prove that $f$ is surjective i.e. the range of $f$ is $\mathbb{R}$ . Let $y = -f(x)$ and we get
$a - 2x = f(f(-f(x)) - x)$
and because $a-2x$ can range over all the reals, there must exist some real $z$ such that $f(z) = a-2x$ .

Now let $r$ be some real constant such that $f(r) = 0$ . Let $x = r$ and we have $f(y) = 2r + f(f(y) - r)$ . If we let $y$ be some number such that $f(y) = r$ , then we have $r = 2r + a$ so $r = -a$ , so $f(-a) = 0$ . Hence, we can replace $r$ with $-a$ and get $f(y) = -2a + f(f(y)+a)$ , and since $f(y)$ can take on any real value, let's just replace $f(y)$ with $z$ so we have $f(z+a) = z+2a$ . Letting $x = 0$ and $y = z$ in the original FE gives us
$f(z+a) = f(f(z))$
so we have $f(f(z)) = z+2a$ . Now we show that $f$ is injective. Suppose we have $f(x) = f(y)$ for some $x, y$ , so then $f(f(x)) = x+2a = f(f(y)) = y+2a$ which implies that $x = y$ . So now, since $f(z+a) = f(f(z))$ , we have $f(z) = z+a$ .

Therefore, our solutions are $f(x) = x+a$ for some $a \in \mathbb{R}$ . Checking that these work is trivial.

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Nanas

44 posts

Nanas

#14 h Oct 8, 2014, 9:49 PM • 2 Y

Y by fclvbfm934, Adventure10

fclvbfm934 wrote:

Hence, we can replace $r$ with $-a$ and get $f(y) = -2a + f(f(y)+a)$ , and since $f(y)$ can take on any real value, let's just replace $f(y)$ with $z$ so we have $f(z+a) = z+2a$ . .

I think you could have stopped here, as $a$ is constant and $z$ can be chosen arbitrarily, so for any $x$ , we may let $z=x-a$ , and we get the solution .. Am I messing something ?

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MSTang

6012 posts

MSTang

#15 h Nov 16, 2014, 4:40 AM • 3 Y

Y by Nanas, Adventure10, Mango247

I think you are right @Nanas.

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Mukhammadiev

79 posts

Mukhammadiev

#16 h Dec 20, 2014, 8:24 AM • 2 Y

Y by Adventure10 and 1 other user

By putting $P(x;0)$ we have $f(f(x))=2x+f(f(0)-x)$ which begets $f$ is injective and by $P(0,y)$ we have $f(f(0)+y)=f(f(y))$ . Obviously $f(y)=y+f(0)$ . It is easy to verify that it is indeed a solution.

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Jndd

1417 posts

Jndd

#130 h Jul 28, 2024, 3:44 AM • 1 Y

Y by radian_51

Setting $y=-f(x)$ , we get $f(f(x)-f(f(x))=f(0)=2x+f(f(-f(x))-x),$ so by varying $x$ across all reals, we get that $f$ is surjective. Now, using that $f$ is surjective, we must have some $t\in\mathbb{R}$ such that $f(t)=0$ . Then, plugging in $x=t$ gives us $f(f(t)+y)=f(y)=2t+f(f(y)-t),$ and letting $x=f(y)$ , we have $x-2t=f(x-t)$ , which can be written as $f(x)=x-t$ . Replacing the constant $-t$ with $c$ and plugging it back into the original equation, we see that all such equations work, so our answer is $f(x)=x+c$ for all $x$ .

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Markas

150 posts

Markas

#131 h Aug 20, 2024, 8:07 PM • 1 Y

Y by radian_51

Let y = -f(x). We get $f(0) = 2x + f(f(-f(x) - x)$ $\Rightarrow$ $f(f(-f(x) - x) = f(0) - 2x$ and now f(0) is fixed and we can vary x however we want $\Rightarrow$ f goes trough all values $\Rightarrow$ f is surjective. Now since f is surjective there $\exists$ t, for which $f(t) = 0$ . Let x = t. We get $f(y) = 2t + f(f(y) - t)$ . Let f(y) - t = u which goes trough all values $\Rightarrow$ $u - t = f(u)$ $\Rightarrow$ we get that $f(x) = x - t$ , for every t. Now we have to check it so plugging in we get x - t + y - t = 2x + y - t - x - t, which is correct $\Rightarrow$ we are ready.

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ezpotd

1301 posts

ezpotd

#132 h Aug 24, 2024, 8:00 PM • 1 Y

Y by radian_51

We claim the answer is all functions of the form $f(x) = x + c$ , it is easy to verify all of these functions work.

Set $y =-f(x)$ , then we get $f(0) - 2x = f(f(y) - x)$ , varying $x$ shows that $f$ is surjective. Now we show $f$ is injective, assume otherwise and we have $f(a) = f(b)$ for unequal $a,b$ , then set $y = c$ such that $f(c) = a + b$ , then we get $2a + f(b)= f(f(b) + c)= f(f(a) + c)=2b + f(a) = 2b + f(b)$ , giving $a = b$ , contradiction. Now set $x = 0$ , and undo a layer of $f$ via bijectivity, we get $f(0) + y = f(y)$ , so $f(y) = y + c$ for some $c$ .

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balllightning37

389 posts

balllightning37

#133 h Aug 31, 2024, 4:22 PM • 1 Y

Y by radian_51

This is nice

The answer is $f(x)=x-c$ , which trivially works.

If $y=-f(x)$ , we get that $f(0)=2x+f(f(-f(x)-x))$ , so $f$ is surjective.

Now, we take a constant $c$ such that $f(c)=0$ . $P(c,y)$ gives $f(y)=2c+f(f(y)-c)$ which rearranges to $f(y)-c=f(f(y)-c)+c$ . As $f$ is surjective, for all $x$ we can find $y$ such that $x=f(y)-c$ . Thus, for all $x$ , $f(x)=x-c$ . It is easily verified that $c$ can be any real number, so we are done.

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alexanderhamilton124

400 posts

alexanderhamilton124

#134 h Oct 14, 2024, 3:51 PM • 1 Y

Y by radian_51

$P(x, -f(x))$ gives $f(0) = 2x + f(f(-f(x)) - x)$ which means $f$ is surjective. Take $y$ such that $f(y) = x$ , and we get $f(f(x) + y) = 2x + f(0)$ . Say $f(a) = f(b)$ , plugging this in into the new equation, we get $a = b$ , so $f$ is injective. $P(0, y)$ gives $f(y + f(0)) = f(f(y)) \implies f(y) = y + f(0)$ , so $f(x) = x + c$ , which clearly works.

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Anshu_Singh_Anahu

69 posts

Anshu_Singh_Anahu

#135 h Nov 21, 2024, 3:52 PM • 2 Y

Y by alexanderhamilton124, radian_51

alexanderhamilton124 wrote:

. Take $y$ such that $f(y) = x$ , and we get $f(f(x) + y) = 2x + f(0)$ . Say $f(a) = f(b)$ , plugging this in into the new equation, we get $a = b$ ,

You cant say a=b becoz y is not fixed you taken y such that f(y) = x ,

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Saucepan_man02

1360 posts

Saucepan_man02

#138 h Jan 22, 2025, 7:31 AM • 2 Y

Y by Sreepranad, radian_51

Note that $f$ is surjective. Let $f(a)=0$ .
$P(a, y)$ gives: $f(y) = 2a+f(f(y)-a)$ . Since $f$ is surjective, let $t=f(y)-a \in \mathbb R$ . Thus: $f(t)=t-a$ for all $t \in \mathbb R$ for some constant $a$ .

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alexanderhamilton124

400 posts

alexanderhamilton124

#139 h Jan 22, 2025, 7:41 AM • 2 Y

Y by Nobitasolvesproblems1979, radian_51

Anshu_Singh_Anahu wrote:

alexanderhamilton124 wrote:

. Take $y$ such that $f(y) = x$ , and we get $f(f(x) + y) = 2x + f(0)$ . Say $f(a) = f(b)$ , plugging this in into the new equation, we get $a = b$ ,

You cant say a=b becoz y is not fixed you taken y such that f(y) = x ,

Thank you for spotting the error! Will try to fix today.

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Maximilian113

575 posts

Maximilian113

#140 h Feb 24, 2025, 4:41 AM • 1 Y

Y by radian_51

$y=-f(x)$ yields $f(x)$ is surjective. Thus let $k$ be such that $f(k)=0,$ then $x=k \implies f(y)=2k+f(f(y)-k) \implies m-2k=f(m-k)$ for all real $m.$ Hence $f(x)=x-k,$ which after substitution into the assertion simply works.

Hence $f(x) \equiv x+c$ for some constant $c \in \mathbb R$ are the only solutions.

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ali123456

52 posts

ali123456

#141 h Feb 24, 2025, 10:58 PM • 1 Y

Y by radian_51

$\textbf{Claim:}$ $f$ is surjective
$\textbf{Proof:}$ we notice that $P(x;-f(x)) \implies f(f(-x)-x)=f(0)-2x$
$\textbf{Claim:}$ $f$ is injective
$\textbf{Proof:}$ Let $x$ and $y$ such that $f(x)=f(y)$ Since f is surjective there exist $a$ such that $f(a)=x+y$ and so we conclude that $x=y$ by $P(x;a)$ and $P(y;a)$
$\textbf{Last step:}$ : $P(0;y) \implies f(f(0)+y)=f(f(y) \implies f(y)=y+f(0)$ Hence $f(y)=y+c$ with $c$ a real constant

This post has been edited 2 times. Last edited by ali123456, Feb 24, 2025, 11:07 PM

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Tony_stark0094

69 posts

Tony_stark0094

#143 h Mar 3, 2025, 11:56 AM • 1 Y

Y by radian_51

dysfunctionalequations wrote:

Setting $y=-f(x)$ , we have $f(0)=2x+f(f(-x)-x))$ , so $f(0)-2x=f(f(-x)-x)$ , and $f(0)-2x$ is a surjective function of $x$ , so $f$ is surjective as well.

Then, by surjectivity, we can let $c$ be a number for which $f(c)=0$ . Setting $x=c$ , we have $f(y)=2c+f(f(y)-c)$ . Since $f(y)$ can take on all real values $r$ , $r=2c+f(r-c)$ for all real numbers $r$ , implying $f(r-c)=r-2c$ . Letting $s=r-c$ , we find $\boxed{f(s)=s-c}$ for some constant $c$ for all real numbers $s$ . Also, it can be easily verified that all functions of this form satisfy this equation.

how can you let f(r)=r?

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Tony_stark0094

69 posts

Tony_stark0094

#144 h Mar 3, 2025, 11:59 AM • 1 Y

Y by radian_51

djmathman wrote:

Here's a solution that I believe avoids injectivity and surjectivity altogether.

Let $x=0$ to get $f(f(0)+y)=f(f(y))$ and let $y=0$ to get $f(f(x))=2x+f(f(0)-x)$ . Substituting our expression for $f(f(x))$ into the first equation yields $f(f(0)+y)=f(f(y))=2y+f(f(0)-y).$ Now make the change of variables $m=f(0)+y$ , $n=f(0)-y$ . Then $2y=m-n$ , so $f(m)=m-n+f(n)\implies f(m)-m=f(n)-n.$ Let $g(x)=f(x)-x$ . This implies $g(m)=g(n)$ for all pairs of real numbers $(m,n)$ , so $g$ is constant. Thus $f(x)-x=c\implies f(x)=x+c$ , where $c$ is a constant. Checking yields that all such functions work.

too good

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Marcus_Zhang

980 posts

Marcus_Zhang

#145 h Mar 16, 2025, 2:13 PM • 1 Y

Y by radian_51

Surjectivity.

This post has been edited 2 times. Last edited by Marcus_Zhang, Mar 16, 2025, 2:14 PM

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lpieleanu

3007 posts

lpieleanu

#146 h Mar 27, 2025, 4:03 PM • 2 Y

Y by megarnie, radian_51

Solution

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Ilikeminecraft

665 posts

Ilikeminecraft

#147 h Apr 25, 2025, 8:01 AM

Y by

I claim that $f(x) = x + c.$

Plug in $y = -f(x)$ to get that $f(f(-f(x)) - x) = f(0) - 2x,$ and thus $f$ is surjective. Let $P(x, y)$ denote our assertion.

Since $f$ is surjective, let us assume that $f(k) = 0.$ With $P(0, k),$ we get that $f(f(0) + k) = 0.$ With $P(-k, 0),$ we have that $f(f(-k)) = -2k.$ With $P(0, -k),$ we see that $f(f(0) - k) = -2k.$ With $P(k, 0),$ we get that $f(0) = 2k.$ Thus, we get that $f(k) = -2k,$ but by assumption, $f(k) = 0.$ Thus, $k = 0,$ and hence $f(0) = 0.$

With $P(0, x),$ we see that $f(x) = f(f(x)).$ Since $f(x)$ is surjective, we have that $f(x) = x.$

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