A functional equation of two variables

6

$begingroup$

Solve the following functional equation :
$f:Bbb Z rightarrow Bbb Z$, $f(f(x)+y)=x+f(y+2017)$

I have no prior experience with solving functional equation but still tried a bit. I set $x=y=0$ to get $f(f(0))=f(2017)$. Can we apply $f^{-1}$ both sides to get $f(0)=2017$? I am unable to carry this on.

functional-equations

share|cite|improve this question

edited Jan 3 at 4:26

Cheerful Parsnip

20.9k23397

asked Jan 2 at 8:32

Epsilon zeroEpsilon zero

34118

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's allowed to take $f^{-1}$ on both sides if a function is monotonic. Can you assume that $f$ is monotonic in this case?
  $endgroup$
  – Matti P.
  Jan 2 at 9:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of $f$ is not injective you can't cancel it out
  $endgroup$
  – Holo
  Jan 2 at 9:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MattiP. Not monotonic, to be able to use $f^{-1}$ you need $f$ to be bijective, and to cancel out $f$ you need it to be injective
  $endgroup$
  – Holo
  Jan 2 at 9:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It holds, because we have that at $y=-2017$, $f(0)=f(f(x)-2017)-x$ and at $f(x)=2017$, $f(0)=f(0)-x|_{f(x)=2017}$ so that $f(0)=2017$.
  $endgroup$
  – TheSimpliFire
  Jan 2 at 9:19
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f(x)=x+2017$ is clearly a solution, and probably the only solution, showing nothing else works is proving to be tricky though.
  $endgroup$
  – Erik Parkinson
  Jan 2 at 9:36
  

  
  
  
  

add a comment | 

6

$begingroup$

Solve the following functional equation :
$f:Bbb Z rightarrow Bbb Z$, $f(f(x)+y)=x+f(y+2017)$

I have no prior experience with solving functional equation but still tried a bit. I set $x=y=0$ to get $f(f(0))=f(2017)$. Can we apply $f^{-1}$ both sides to get $f(0)=2017$? I am unable to carry this on.

functional-equations

share|cite|improve this question

edited Jan 3 at 4:26

Cheerful Parsnip

20.9k23397

asked Jan 2 at 8:32

Epsilon zeroEpsilon zero

34118

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's allowed to take $f^{-1}$ on both sides if a function is monotonic. Can you assume that $f$ is monotonic in this case?
  $endgroup$
  – Matti P.
  Jan 2 at 9:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of $f$ is not injective you can't cancel it out
  $endgroup$
  – Holo
  Jan 2 at 9:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MattiP. Not monotonic, to be able to use $f^{-1}$ you need $f$ to be bijective, and to cancel out $f$ you need it to be injective
  $endgroup$
  – Holo
  Jan 2 at 9:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It holds, because we have that at $y=-2017$, $f(0)=f(f(x)-2017)-x$ and at $f(x)=2017$, $f(0)=f(0)-x|_{f(x)=2017}$ so that $f(0)=2017$.
  $endgroup$
  – TheSimpliFire
  Jan 2 at 9:19
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f(x)=x+2017$ is clearly a solution, and probably the only solution, showing nothing else works is proving to be tricky though.
  $endgroup$
  – Erik Parkinson
  Jan 2 at 9:36
  

  
  
  
  

add a comment | 

6

6

6

3

$begingroup$

Solve the following functional equation :
$f:Bbb Z rightarrow Bbb Z$, $f(f(x)+y)=x+f(y+2017)$

I have no prior experience with solving functional equation but still tried a bit. I set $x=y=0$ to get $f(f(0))=f(2017)$. Can we apply $f^{-1}$ both sides to get $f(0)=2017$? I am unable to carry this on.

functional-equations

share|cite|improve this question

edited Jan 3 at 4:26

Cheerful Parsnip

20.9k23397

asked Jan 2 at 8:32

Epsilon zeroEpsilon zero

34118

$endgroup$

Solve the following functional equation :
$f:Bbb Z rightarrow Bbb Z$, $f(f(x)+y)=x+f(y+2017)$

I have no prior experience with solving functional equation but still tried a bit. I set $x=y=0$ to get $f(f(0))=f(2017)$. Can we apply $f^{-1}$ both sides to get $f(0)=2017$? I am unable to carry this on.

functional-equations

functional-equations

share|cite|improve this question

edited Jan 3 at 4:26

Cheerful Parsnip

20.9k23397

asked Jan 2 at 8:32

Epsilon zeroEpsilon zero

34118

share|cite|improve this question

edited Jan 3 at 4:26

Cheerful Parsnip

20.9k23397

asked Jan 2 at 8:32

Epsilon zeroEpsilon zero

34118

share|cite|improve this question

share|cite|improve this question

edited Jan 3 at 4:26

Cheerful Parsnip

20.9k23397

edited Jan 3 at 4:26

Cheerful Parsnip

20.9k23397

edited Jan 3 at 4:26

Cheerful Parsnip

20.9k23397

20.9k23397

asked Jan 2 at 8:32

Epsilon zeroEpsilon zero

34118

asked Jan 2 at 8:32

Epsilon zeroEpsilon zero

34118

asked Jan 2 at 8:32

Epsilon zeroEpsilon zero

34118

34118

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's allowed to take $f^{-1}$ on both sides if a function is monotonic. Can you assume that $f$ is monotonic in this case?
  $endgroup$
  – Matti P.
  Jan 2 at 9:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of $f$ is not injective you can't cancel it out
  $endgroup$
  – Holo
  Jan 2 at 9:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MattiP. Not monotonic, to be able to use $f^{-1}$ you need $f$ to be bijective, and to cancel out $f$ you need it to be injective
  $endgroup$
  – Holo
  Jan 2 at 9:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It holds, because we have that at $y=-2017$, $f(0)=f(f(x)-2017)-x$ and at $f(x)=2017$, $f(0)=f(0)-x|_{f(x)=2017}$ so that $f(0)=2017$.
  $endgroup$
  – TheSimpliFire
  Jan 2 at 9:19
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f(x)=x+2017$ is clearly a solution, and probably the only solution, showing nothing else works is proving to be tricky though.
  $endgroup$
  – Erik Parkinson
  Jan 2 at 9:36
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's allowed to take $f^{-1}$ on both sides if a function is monotonic. Can you assume that $f$ is monotonic in this case?
  $endgroup$
  – Matti P.
  Jan 2 at 9:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of $f$ is not injective you can't cancel it out
  $endgroup$
  – Holo
  Jan 2 at 9:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MattiP. Not monotonic, to be able to use $f^{-1}$ you need $f$ to be bijective, and to cancel out $f$ you need it to be injective
  $endgroup$
  – Holo
  Jan 2 at 9:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It holds, because we have that at $y=-2017$, $f(0)=f(f(x)-2017)-x$ and at $f(x)=2017$, $f(0)=f(0)-x|_{f(x)=2017}$ so that $f(0)=2017$.
  $endgroup$
  – TheSimpliFire
  Jan 2 at 9:19
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f(x)=x+2017$ is clearly a solution, and probably the only solution, showing nothing else works is proving to be tricky though.
  $endgroup$
  – Erik Parkinson
  Jan 2 at 9:36
  

  
  
  
  

1

1

$begingroup$
It's allowed to take $f^{-1}$ on both sides if a function is monotonic. Can you assume that $f$ is monotonic in this case?
$endgroup$
– Matti P.
Jan 2 at 9:01

$begingroup$
It's allowed to take $f^{-1}$ on both sides if a function is monotonic. Can you assume that $f$ is monotonic in this case?
$endgroup$
– Matti P.
Jan 2 at 9:01

$begingroup$
Of $f$ is not injective you can't cancel it out
$endgroup$
– Holo
Jan 2 at 9:02

$begingroup$
Of $f$ is not injective you can't cancel it out
$endgroup$
– Holo
Jan 2 at 9:02

$begingroup$
@MattiP. Not monotonic, to be able to use $f^{-1}$ you need $f$ to be bijective, and to cancel out $f$ you need it to be injective
$endgroup$
– Holo
Jan 2 at 9:04

$begingroup$
@MattiP. Not monotonic, to be able to use $f^{-1}$ you need $f$ to be bijective, and to cancel out $f$ you need it to be injective
$endgroup$
– Holo
Jan 2 at 9:04

$begingroup$
It holds, because we have that at $y=-2017$, $f(0)=f(f(x)-2017)-x$ and at $f(x)=2017$, $f(0)=f(0)-x|_{f(x)=2017}$ so that $f(0)=2017$.
$endgroup$
– TheSimpliFire
Jan 2 at 9:19

$begingroup$
It holds, because we have that at $y=-2017$, $f(0)=f(f(x)-2017)-x$ and at $f(x)=2017$, $f(0)=f(0)-x|_{f(x)=2017}$ so that $f(0)=2017$.
$endgroup$
– TheSimpliFire
Jan 2 at 9:19

$begingroup$
$f(x)=x+2017$ is clearly a solution, and probably the only solution, showing nothing else works is proving to be tricky though.
$endgroup$
– Erik Parkinson
Jan 2 at 9:36

$begingroup$
$f(x)=x+2017$ is clearly a solution, and probably the only solution, showing nothing else works is proving to be tricky though.
$endgroup$
– Erik Parkinson
Jan 2 at 9:36

add a comment | 

3 Answers
3

active

oldest

votes

9

$begingroup$

Got it! I probably complicated it more than I had to so if anyone sees any way to simply this I would love to hear feedback.

To solve this, let $y=0$ so that
$f(f(x)) = x+f(2017)$.
Let $c=f(2017)$. Then for all $xinmathbb{Z}$,
$$f(f(x)) = x+c$$

Now plugging $x=y=0$ into the original equation we get
$f(f(0)) = f(2017)$. Taking $f$ of both sides yields
$f(f(f(0))) = f(f(2017))$ which is $f(0)+c=2017+c$ so
$f(0) = 2017$.

Now take $f$ of both sides of the original equation to get
$f(f(f(x)+y)) = f(x+f(y+2017))$
which is
$f(x)+y + c = f(x+f(y+2017))$
Setting $y=-2017$ gives
$$f(x)-2017 + c = f(x+f(0)) = f(x+2017)$$.

Now we return again to the original equation with $y=1$. This gives
$f(f(x)+1) = x+f(2018)$ which by the above formula is
$x+f(1)-2017 + c$. So
$$f(f(x)+1)-f(f(x)) = x+f(1)-2017+c - (x+c) = f(1)-2017$$

Now, as $f(f(x)) = x+c$, $f(f(x))$, and thus $f(x)$, can attain all values in $mathbb{Z}$. So for any $kinmathbb{Z}$, there is an $xinmathbb{Z}$ such that $f(x)=k$, and so the above formula becomes
$$f(k+1)-f(k) = f(1)-2017$$
for all $kinmathbb{Z}$. So for all $kinmathbb{Z}$,
$$f(k) = k+c_2$$
for some $c_2$. So the original equation becomes
$$x+y+2c_2=x+y+2017+c_2$$
so $c_2=2017$. Thus the only solution is
$$f(x) = x+2017$$

share|cite|improve this answer

answered Jan 2 at 10:28

Erik ParkinsonErik Parkinson

8749

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) nice! ${}{}$
  $endgroup$
  – TheSimpliFire
  Jan 2 at 10:38
  

  
  
  
  

add a comment | 

2

$begingroup$

You could take $y=0$ there and get $f(f(x))=x+f(2017)=x+n$ where $n$ is a constant. Thus $f(f(x))$ is a linear function. You can observe (guess?) what $f(x)$ could be like here. If it is so, then you have no problem taking $f^{-1}$ and you can easily find $n$.

The point is how to prove the nature of $f(x)$ if you know $f(f(x))$ is linear. This is an interesting problem. What is the square root (with integer values) of a linear function?

Now here I'm not sure $f(x)$ will end up being a simple function involving only arithmetic operations, or a function also involving the modulo.

share|cite|improve this answer

edited Jan 2 at 9:20

answered Jan 2 at 9:19

FerredFerred

763

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is better posted as a comment. As you don't yet have enough reputation for that, how about answering some other questions and come back? :)
  $endgroup$
  – TheSimpliFire
  Jan 2 at 9:20
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $n$ be an integer. Then at $f(x)=n$ and $y=0$, the equation $f(f(x)+y)=x+f(y+n)$ becomes $f(n+y)=f(y+n)+xBig|_{f(x)=n}$, so $f(0)=n$.

Now at $f(x)=n+1$, we can derive similar relations, at $y=0$ and $y=-1$ respectively, $$xBig|_{f(x)=n+1}=f(n+1)-f(n)=f(n)-f(n-1)implies 2f(n)=f(n-1)+f(n+1).$$

By induction it can be proven that $2f(n)=f(n-x)+f(n+x)$ or $2f(n)=f(x)+f(2n-x)$ for any integer $x$. Assuming differentiability, $f'(x)-f'(2n-x)=0$ so $f$ is either a linear or periodic function. It cannot be periodic because $f(0)=n$ and $xBig|_{f(x)=0}=f(n)-f(0)$.

Thus $f$ is a linear function and putting it in $f(x)=ax+b$, gives us $b=n$ and $a=1$ since $f(f(x))=x+f(n)$.

share|cite|improve this answer

edited Jan 3 at 7:36

answered Jan 2 at 11:08

TheSimpliFireTheSimpliFire

12.6k62360

$endgroup$

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

9

$begingroup$

Got it! I probably complicated it more than I had to so if anyone sees any way to simply this I would love to hear feedback.

To solve this, let $y=0$ so that
$f(f(x)) = x+f(2017)$.
Let $c=f(2017)$. Then for all $xinmathbb{Z}$,
$$f(f(x)) = x+c$$

Now plugging $x=y=0$ into the original equation we get
$f(f(0)) = f(2017)$. Taking $f$ of both sides yields
$f(f(f(0))) = f(f(2017))$ which is $f(0)+c=2017+c$ so
$f(0) = 2017$.

Now take $f$ of both sides of the original equation to get
$f(f(f(x)+y)) = f(x+f(y+2017))$
which is
$f(x)+y + c = f(x+f(y+2017))$
Setting $y=-2017$ gives
$$f(x)-2017 + c = f(x+f(0)) = f(x+2017)$$.

Now we return again to the original equation with $y=1$. This gives
$f(f(x)+1) = x+f(2018)$ which by the above formula is
$x+f(1)-2017 + c$. So
$$f(f(x)+1)-f(f(x)) = x+f(1)-2017+c - (x+c) = f(1)-2017$$

Now, as $f(f(x)) = x+c$, $f(f(x))$, and thus $f(x)$, can attain all values in $mathbb{Z}$. So for any $kinmathbb{Z}$, there is an $xinmathbb{Z}$ such that $f(x)=k$, and so the above formula becomes
$$f(k+1)-f(k) = f(1)-2017$$
for all $kinmathbb{Z}$. So for all $kinmathbb{Z}$,
$$f(k) = k+c_2$$
for some $c_2$. So the original equation becomes
$$x+y+2c_2=x+y+2017+c_2$$
so $c_2=2017$. Thus the only solution is
$$f(x) = x+2017$$

share|cite|improve this answer

answered Jan 2 at 10:28

Erik ParkinsonErik Parkinson

8749

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) nice! ${}{}$
  $endgroup$
  – TheSimpliFire
  Jan 2 at 10:38
  

  
  
  
  

add a comment | 

9

$begingroup$

Got it! I probably complicated it more than I had to so if anyone sees any way to simply this I would love to hear feedback.

To solve this, let $y=0$ so that
$f(f(x)) = x+f(2017)$.
Let $c=f(2017)$. Then for all $xinmathbb{Z}$,
$$f(f(x)) = x+c$$

Now plugging $x=y=0$ into the original equation we get
$f(f(0)) = f(2017)$. Taking $f$ of both sides yields
$f(f(f(0))) = f(f(2017))$ which is $f(0)+c=2017+c$ so
$f(0) = 2017$.

Now take $f$ of both sides of the original equation to get
$f(f(f(x)+y)) = f(x+f(y+2017))$
which is
$f(x)+y + c = f(x+f(y+2017))$
Setting $y=-2017$ gives
$$f(x)-2017 + c = f(x+f(0)) = f(x+2017)$$.

Now we return again to the original equation with $y=1$. This gives
$f(f(x)+1) = x+f(2018)$ which by the above formula is
$x+f(1)-2017 + c$. So
$$f(f(x)+1)-f(f(x)) = x+f(1)-2017+c - (x+c) = f(1)-2017$$

Now, as $f(f(x)) = x+c$, $f(f(x))$, and thus $f(x)$, can attain all values in $mathbb{Z}$. So for any $kinmathbb{Z}$, there is an $xinmathbb{Z}$ such that $f(x)=k$, and so the above formula becomes
$$f(k+1)-f(k) = f(1)-2017$$
for all $kinmathbb{Z}$. So for all $kinmathbb{Z}$,
$$f(k) = k+c_2$$
for some $c_2$. So the original equation becomes
$$x+y+2c_2=x+y+2017+c_2$$
so $c_2=2017$. Thus the only solution is
$$f(x) = x+2017$$

share|cite|improve this answer

answered Jan 2 at 10:28

Erik ParkinsonErik Parkinson

8749

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) nice! ${}{}$
  $endgroup$
  – TheSimpliFire
  Jan 2 at 10:38
  

  
  
  
  

add a comment | 

9

9

9

$begingroup$

Got it! I probably complicated it more than I had to so if anyone sees any way to simply this I would love to hear feedback.

To solve this, let $y=0$ so that
$f(f(x)) = x+f(2017)$.
Let $c=f(2017)$. Then for all $xinmathbb{Z}$,
$$f(f(x)) = x+c$$

Now plugging $x=y=0$ into the original equation we get
$f(f(0)) = f(2017)$. Taking $f$ of both sides yields
$f(f(f(0))) = f(f(2017))$ which is $f(0)+c=2017+c$ so
$f(0) = 2017$.

Now take $f$ of both sides of the original equation to get
$f(f(f(x)+y)) = f(x+f(y+2017))$
which is
$f(x)+y + c = f(x+f(y+2017))$
Setting $y=-2017$ gives
$$f(x)-2017 + c = f(x+f(0)) = f(x+2017)$$.

Now we return again to the original equation with $y=1$. This gives
$f(f(x)+1) = x+f(2018)$ which by the above formula is
$x+f(1)-2017 + c$. So
$$f(f(x)+1)-f(f(x)) = x+f(1)-2017+c - (x+c) = f(1)-2017$$

Now, as $f(f(x)) = x+c$, $f(f(x))$, and thus $f(x)$, can attain all values in $mathbb{Z}$. So for any $kinmathbb{Z}$, there is an $xinmathbb{Z}$ such that $f(x)=k$, and so the above formula becomes
$$f(k+1)-f(k) = f(1)-2017$$
for all $kinmathbb{Z}$. So for all $kinmathbb{Z}$,
$$f(k) = k+c_2$$
for some $c_2$. So the original equation becomes
$$x+y+2c_2=x+y+2017+c_2$$
so $c_2=2017$. Thus the only solution is
$$f(x) = x+2017$$

share|cite|improve this answer

answered Jan 2 at 10:28

Erik ParkinsonErik Parkinson

8749

$endgroup$

Got it! I probably complicated it more than I had to so if anyone sees any way to simply this I would love to hear feedback.

To solve this, let $y=0$ so that
$f(f(x)) = x+f(2017)$.
Let $c=f(2017)$. Then for all $xinmathbb{Z}$,
$$f(f(x)) = x+c$$

Now plugging $x=y=0$ into the original equation we get
$f(f(0)) = f(2017)$. Taking $f$ of both sides yields
$f(f(f(0))) = f(f(2017))$ which is $f(0)+c=2017+c$ so
$f(0) = 2017$.

Now take $f$ of both sides of the original equation to get
$f(f(f(x)+y)) = f(x+f(y+2017))$
which is
$f(x)+y + c = f(x+f(y+2017))$
Setting $y=-2017$ gives
$$f(x)-2017 + c = f(x+f(0)) = f(x+2017)$$.

Now we return again to the original equation with $y=1$. This gives
$f(f(x)+1) = x+f(2018)$ which by the above formula is
$x+f(1)-2017 + c$. So
$$f(f(x)+1)-f(f(x)) = x+f(1)-2017+c - (x+c) = f(1)-2017$$

Now, as $f(f(x)) = x+c$, $f(f(x))$, and thus $f(x)$, can attain all values in $mathbb{Z}$. So for any $kinmathbb{Z}$, there is an $xinmathbb{Z}$ such that $f(x)=k$, and so the above formula becomes
$$f(k+1)-f(k) = f(1)-2017$$
for all $kinmathbb{Z}$. So for all $kinmathbb{Z}$,
$$f(k) = k+c_2$$
for some $c_2$. So the original equation becomes
$$x+y+2c_2=x+y+2017+c_2$$
so $c_2=2017$. Thus the only solution is
$$f(x) = x+2017$$

share|cite|improve this answer

answered Jan 2 at 10:28

Erik ParkinsonErik Parkinson

8749

share|cite|improve this answer

share|cite|improve this answer

answered Jan 2 at 10:28

Erik ParkinsonErik Parkinson

8749

answered Jan 2 at 10:28

Erik ParkinsonErik Parkinson

8749

answered Jan 2 at 10:28

Erik ParkinsonErik Parkinson

8749

8749

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) nice! ${}{}$
  $endgroup$
  – TheSimpliFire
  Jan 2 at 10:38
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) nice! ${}{}$
  $endgroup$
  – TheSimpliFire
  Jan 2 at 10:38
  

  
  
  
  

$begingroup$
(+1) nice! ${}{}$
$endgroup$
– TheSimpliFire
Jan 2 at 10:38

$begingroup$
(+1) nice! ${}{}$
$endgroup$
– TheSimpliFire
Jan 2 at 10:38

add a comment | 

2

$begingroup$

You could take $y=0$ there and get $f(f(x))=x+f(2017)=x+n$ where $n$ is a constant. Thus $f(f(x))$ is a linear function. You can observe (guess?) what $f(x)$ could be like here. If it is so, then you have no problem taking $f^{-1}$ and you can easily find $n$.

The point is how to prove the nature of $f(x)$ if you know $f(f(x))$ is linear. This is an interesting problem. What is the square root (with integer values) of a linear function?

Now here I'm not sure $f(x)$ will end up being a simple function involving only arithmetic operations, or a function also involving the modulo.

share|cite|improve this answer

edited Jan 2 at 9:20

answered Jan 2 at 9:19

FerredFerred

763

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is better posted as a comment. As you don't yet have enough reputation for that, how about answering some other questions and come back? :)
  $endgroup$
  – TheSimpliFire
  Jan 2 at 9:20
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

You could take $y=0$ there and get $f(f(x))=x+f(2017)=x+n$ where $n$ is a constant. Thus $f(f(x))$ is a linear function. You can observe (guess?) what $f(x)$ could be like here. If it is so, then you have no problem taking $f^{-1}$ and you can easily find $n$.

The point is how to prove the nature of $f(x)$ if you know $f(f(x))$ is linear. This is an interesting problem. What is the square root (with integer values) of a linear function?

Now here I'm not sure $f(x)$ will end up being a simple function involving only arithmetic operations, or a function also involving the modulo.

share|cite|improve this answer

edited Jan 2 at 9:20

answered Jan 2 at 9:19

FerredFerred

763

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is better posted as a comment. As you don't yet have enough reputation for that, how about answering some other questions and come back? :)
  $endgroup$
  – TheSimpliFire
  Jan 2 at 9:20
  
  
  

  
  
  
  

add a comment | 

2

2

2

$begingroup$

You could take $y=0$ there and get $f(f(x))=x+f(2017)=x+n$ where $n$ is a constant. Thus $f(f(x))$ is a linear function. You can observe (guess?) what $f(x)$ could be like here. If it is so, then you have no problem taking $f^{-1}$ and you can easily find $n$.

The point is how to prove the nature of $f(x)$ if you know $f(f(x))$ is linear. This is an interesting problem. What is the square root (with integer values) of a linear function?

Now here I'm not sure $f(x)$ will end up being a simple function involving only arithmetic operations, or a function also involving the modulo.

share|cite|improve this answer

edited Jan 2 at 9:20

answered Jan 2 at 9:19

FerredFerred

763

$endgroup$

You could take $y=0$ there and get $f(f(x))=x+f(2017)=x+n$ where $n$ is a constant. Thus $f(f(x))$ is a linear function. You can observe (guess?) what $f(x)$ could be like here. If it is so, then you have no problem taking $f^{-1}$ and you can easily find $n$.

The point is how to prove the nature of $f(x)$ if you know $f(f(x))$ is linear. This is an interesting problem. What is the square root (with integer values) of a linear function?

Now here I'm not sure $f(x)$ will end up being a simple function involving only arithmetic operations, or a function also involving the modulo.

share|cite|improve this answer

edited Jan 2 at 9:20

answered Jan 2 at 9:19

FerredFerred

763

share|cite|improve this answer

share|cite|improve this answer

edited Jan 2 at 9:20

edited Jan 2 at 9:20

edited Jan 2 at 9:20

answered Jan 2 at 9:19

FerredFerred

763

answered Jan 2 at 9:19

FerredFerred

763

answered Jan 2 at 9:19

FerredFerred

763

763

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is better posted as a comment. As you don't yet have enough reputation for that, how about answering some other questions and come back? :)
  $endgroup$
  – TheSimpliFire
  Jan 2 at 9:20
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is better posted as a comment. As you don't yet have enough reputation for that, how about answering some other questions and come back? :)
  $endgroup$
  – TheSimpliFire
  Jan 2 at 9:20
  
  
  

  
  
  
  

1

1

$begingroup$
This is better posted as a comment. As you don't yet have enough reputation for that, how about answering some other questions and come back? :)
$endgroup$
– TheSimpliFire
Jan 2 at 9:20

$begingroup$
This is better posted as a comment. As you don't yet have enough reputation for that, how about answering some other questions and come back? :)
$endgroup$
– TheSimpliFire
Jan 2 at 9:20

add a comment | 

2

$begingroup$

Let $n$ be an integer. Then at $f(x)=n$ and $y=0$, the equation $f(f(x)+y)=x+f(y+n)$ becomes $f(n+y)=f(y+n)+xBig|_{f(x)=n}$, so $f(0)=n$.

Now at $f(x)=n+1$, we can derive similar relations, at $y=0$ and $y=-1$ respectively, $$xBig|_{f(x)=n+1}=f(n+1)-f(n)=f(n)-f(n-1)implies 2f(n)=f(n-1)+f(n+1).$$

By induction it can be proven that $2f(n)=f(n-x)+f(n+x)$ or $2f(n)=f(x)+f(2n-x)$ for any integer $x$. Assuming differentiability, $f'(x)-f'(2n-x)=0$ so $f$ is either a linear or periodic function. It cannot be periodic because $f(0)=n$ and $xBig|_{f(x)=0}=f(n)-f(0)$.

Thus $f$ is a linear function and putting it in $f(x)=ax+b$, gives us $b=n$ and $a=1$ since $f(f(x))=x+f(n)$.

share|cite|improve this answer

edited Jan 3 at 7:36

answered Jan 2 at 11:08

TheSimpliFireTheSimpliFire

12.6k62360

$endgroup$

add a comment | 

2

$begingroup$

Let $n$ be an integer. Then at $f(x)=n$ and $y=0$, the equation $f(f(x)+y)=x+f(y+n)$ becomes $f(n+y)=f(y+n)+xBig|_{f(x)=n}$, so $f(0)=n$.

Now at $f(x)=n+1$, we can derive similar relations, at $y=0$ and $y=-1$ respectively, $$xBig|_{f(x)=n+1}=f(n+1)-f(n)=f(n)-f(n-1)implies 2f(n)=f(n-1)+f(n+1).$$

By induction it can be proven that $2f(n)=f(n-x)+f(n+x)$ or $2f(n)=f(x)+f(2n-x)$ for any integer $x$. Assuming differentiability, $f'(x)-f'(2n-x)=0$ so $f$ is either a linear or periodic function. It cannot be periodic because $f(0)=n$ and $xBig|_{f(x)=0}=f(n)-f(0)$.

Thus $f$ is a linear function and putting it in $f(x)=ax+b$, gives us $b=n$ and $a=1$ since $f(f(x))=x+f(n)$.

share|cite|improve this answer

edited Jan 3 at 7:36

answered Jan 2 at 11:08

TheSimpliFireTheSimpliFire

12.6k62360

$endgroup$

add a comment | 

2

2

2

$begingroup$

Let $n$ be an integer. Then at $f(x)=n$ and $y=0$, the equation $f(f(x)+y)=x+f(y+n)$ becomes $f(n+y)=f(y+n)+xBig|_{f(x)=n}$, so $f(0)=n$.

Now at $f(x)=n+1$, we can derive similar relations, at $y=0$ and $y=-1$ respectively, $$xBig|_{f(x)=n+1}=f(n+1)-f(n)=f(n)-f(n-1)implies 2f(n)=f(n-1)+f(n+1).$$

By induction it can be proven that $2f(n)=f(n-x)+f(n+x)$ or $2f(n)=f(x)+f(2n-x)$ for any integer $x$. Assuming differentiability, $f'(x)-f'(2n-x)=0$ so $f$ is either a linear or periodic function. It cannot be periodic because $f(0)=n$ and $xBig|_{f(x)=0}=f(n)-f(0)$.

Thus $f$ is a linear function and putting it in $f(x)=ax+b$, gives us $b=n$ and $a=1$ since $f(f(x))=x+f(n)$.

share|cite|improve this answer

edited Jan 3 at 7:36

answered Jan 2 at 11:08

TheSimpliFireTheSimpliFire

12.6k62360

$endgroup$

Let $n$ be an integer. Then at $f(x)=n$ and $y=0$, the equation $f(f(x)+y)=x+f(y+n)$ becomes $f(n+y)=f(y+n)+xBig|_{f(x)=n}$, so $f(0)=n$.

Now at $f(x)=n+1$, we can derive similar relations, at $y=0$ and $y=-1$ respectively, $$xBig|_{f(x)=n+1}=f(n+1)-f(n)=f(n)-f(n-1)implies 2f(n)=f(n-1)+f(n+1).$$

By induction it can be proven that $2f(n)=f(n-x)+f(n+x)$ or $2f(n)=f(x)+f(2n-x)$ for any integer $x$. Assuming differentiability, $f'(x)-f'(2n-x)=0$ so $f$ is either a linear or periodic function. It cannot be periodic because $f(0)=n$ and $xBig|_{f(x)=0}=f(n)-f(0)$.

Thus $f$ is a linear function and putting it in $f(x)=ax+b$, gives us $b=n$ and $a=1$ since $f(f(x))=x+f(n)$.

share|cite|improve this answer

edited Jan 3 at 7:36

answered Jan 2 at 11:08

TheSimpliFireTheSimpliFire

12.6k62360

share|cite|improve this answer

share|cite|improve this answer

edited Jan 3 at 7:36

edited Jan 3 at 7:36

edited Jan 3 at 7:36

answered Jan 2 at 11:08

TheSimpliFireTheSimpliFire

12.6k62360

answered Jan 2 at 11:08

TheSimpliFireTheSimpliFire

12.6k62360

answered Jan 2 at 11:08

TheSimpliFireTheSimpliFire

12.6k62360

12.6k62360

add a comment | 

add a comment | 

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