Functional Equation

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Source: 2006 Red MOP Homework Algebra 1.2

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4216 posts

#1 h Mar 17, 2013, 5:47 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfying
$f(x+f(y))=x+f(f(y))$
for all real numbers $x$ and $y$ , with the additional constraint $f(2004)=2005$ .

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139 posts

goo

#2 h Mar 17, 2013, 7:38 PM • 3 Y

Y by JSGandora, ssilwa, Adventure10

Setting $x=a-2005$ and $y=2004$ gives us $f(a) = a-2005 + f(2005)$ .
Setting $a=2004$ yields $2005=(-1) + f(2005)$ or, equivalently, $f(2005)=2006$ .
Thus, $f(a) = a+1$ .

.
Since $f(a)=a+1$ is clearly a solution, we're done.

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152 posts

rahman

#3 h Mar 17, 2013, 8:21 PM • 2 Y

Y by Adventure10, Mango247

let x=1 and f(y)=k than we have : $f(1+k)=1+f(k)$
since f is defined on R we can have $f(1+k)=1+f(k),\Rightarrow f(1)=1+f(0),f(2)=1+f(1)=2+f(0)...f(2004)=2004+f(0)$
now since $f(2004)=2005\wedge f(2004)=2004+f(0) \Rightarrow f(0)=1$
the functoin i $f(x)=1+x$
maybe im wrong but if someone can tell me i;ll be thankfull

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2932 posts

#4 h Mar 18, 2013, 11:26 AM • 3 Y

Y by rahman, Adventure10, Mango247

@rahman: You need to prove that $f$ is surjective before you can assume $k = f(y)$ . (Otherwise, there are some values of $k$ that you cannot find a $y$ for it, which means you cannot plug in that value of $k$ .)

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rahman

#5 h Mar 19, 2013, 10:03 AM • 2 Y

Y by Adventure10, Mango247

But how to prove that a function is surjective

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2932 posts

#6 h Mar 19, 2013, 10:18 AM • 2 Y

Y by Adventure10, Mango247

Prove that for any $x$ , there exists $y$ such that $f(y )=x$ .

In this case, it is actually easy to prove:
Fix $y$ and let $x = x-f(f(y))$ . We have $f( \text{something} ) = x$ . Since $x$ can take all values, we have proven that $f$ is surjective.

In other functional equations, this might be harder or even impossible (if $f$ is indeed not surjective, obviously you cannot "prove" that it is surjective).

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A-Thought-Of-God

454 posts

A-Thought-Of-God

#7 h May 23, 2020, 6:32 PM • 1 Y

Y by AmirKhusrau

JSGandora wrote:

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfying
$f(x+f(y))=x+f(f(y))$ for all real numbers $x$ and $y$ , with the additional constraint $f(2004)=2005$ .

Here is my solution

We claim that without the constraint $f(x)=x+c$ is the only solution.
Let $P(x,y)$ denotes the assertion. $P(-f(y),y) \implies f(0)=-f(y)+f(f(y)) \implies f(f(y)) = f(y) +c$ where $c=f(0)$ . Substituting this back to original equation, we get that $f(x+f(y)) = x+f(y)+c$ . Now then $P(x-f(y),y) \implies \boxed{f(x)=x+c}$ .

With Constraint : Using the given constraint, we get $a=1$ . So $f(x)=x+1$ is the only solution atlast.
$\blacksquare$

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#8 h May 23, 2020, 8:16 PM • 2 Y

Y by Muaaz.SY, Williamgolly

it's very easy

We have clarity $f$ is surjective hence :
$P(f(x),y)-P(f(y),x)\implies f(x)=x+c$
put $x=2004\implies c=1\implies f(x)=x+1$

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11317 posts

#9 h Jun 6, 2021, 12:48 AM • 1 Y

Y by Mango247

$P(x-2005,2004)\Rightarrow f(x)=x+c$ , testing gives $\boxed{f(x)=x+1}$ .

This post has been edited 1 time. Last edited by jasperE3, Jun 6, 2021, 1:00 AM

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1524 posts

#10 h Jun 6, 2021, 3:51 AM

Y by

JSGandora wrote:

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfying
$f(x+f(y))=x+f(f(y))$ for all real numbers $x$ and $y$ , with the additional constraint $f(2004)=2005$ .

Let $P(x,y)$ the assertion of the given FE:
$P(x-f(f(y)),y)$
$f(x+f(y)-f(f(y)))=x \implies f \; \text{surjective}$ Since $f$ is surjective set $f(y)=t$ and $P(-t,y)$
$f(t)=t+f(0)$ So now set $t=2004$
$f(2004)=2004+f(0) \implies f(0)=1$ Thus the only solution is:

$\boxed{f(x)=x+1 \; \forall x \in \mathbb R}$

Thus we are done :blush:

:blush:

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5606 posts

#11 h Sep 21, 2021, 10:50 PM

Y by

As usual, let $P(x,y)$ denote the given assertion.
$P(x-f(f(y)),y): f(f(x-f(f(y))+f(y))=x$ . From here, we know $f$ is surjective.

Let $k$ be a real number such that $f(k)=2004$ .

$P(-2004,k): f(0)=f(2004)-2004=1$ .

Let $f(y)=-x$ . Now, we have $1=x+f(-x)\implies f(-x)=-x+1$ , so $\boxed{f(x)=x+1}$ is the only solution, and it clearly works.

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ZETA_in_olympiad

2211 posts

ZETA_in_olympiad

#12 h Jul 9, 2022, 11:32 AM

Y by

Setting $x=x-f(0)$ and $y=0$ gives $f(x)=x+k.$ But only $f(x)=x+1$ works.

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148 posts

NicoN9

#13 h Apr 29, 2025, 5:49 AM

Y by

Came from the FE handout of Pang-Chung Wu!

We solve for the FE first. The answer is $f(x)\equiv 0, x + c$ , which works.

Putting $f(x)$ in $x$ , we see that $f(x)+f(f(y))=f(f(x)+f(y))=f(y)+f(f(x))$ , thus $f(f(x))=f(x)+c$ for some $c$ . Putting $y=0$ , we easily see that $f$ is linear. the rest is easy. (In particular, the answer to original problem is $f(x)=x+1$ .)

This post has been edited 1 time. Last edited by NicoN9, Apr 29, 2025, 5:50 AM

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ray66

#14 h Apr 29, 2025, 6:08 AM

Y by

Let $P(x,y)$ be the assertion that the FE is true. Plug in $y=2024$ to get $f(x+2025)=x+f(2025)$ , so $f$ is a linear function $f(x)=x+c$ . But $f(2024)=2025$ , so $c=1$ and $f(x)=x+1$

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