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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.

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    $\begingroup$ 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
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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 $x\in\mathbb{Z}$, $$f(f(x)) = x+c$$

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 $k\in\mathbb{Z}$, there is an $x\in\mathbb{Z}$ such that $f(x)=k$, and so the above formula becomes $$f(k+1)-f(k) = f(1)-2017$$ for all $k\in\mathbb{Z}$. So for all $k\in\mathbb{Z}$, $$f(k) = k+c_2$$ 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 $$f(x) = x+2017$$

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

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

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

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 $x\Big|_{f(x)=0}=f(n)-f(0)$.

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

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