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This was Problem 1 from the 2019 International Mathematics Olympiad.

Find all functions $f$ : $\mathbb{Z}$ $\rightarrow$ $\mathbb{Z}$ that satisfy $f(2x) + 2f(y) = f(f(x+y))$ whenever $x , y$ $\in$ $\mathbb{Z}$.

And this is my progress...

Let $x=y$

then $f(2x) + 2f(y) = f(f(x+y))$

then $f(2x) + 2f(x) = f(f(2x))$ (substitute $x$ with $y$)

then $2f(x) + 2f(x) = f(2f(x))$ (from Cauchy's theory $f(nx) = nf(x)$ where $n > 0$)

then $4f(x) = 2f(f(x))$ (from Cauchy's theory $f(nx) =nf(x)$ where $n > 0$)

then $2f(x) = f(f(x))$ (divide both sides by $2$)

then $f(x) = 2x$ which is one of the satisfied fuctions.

Let $x = -y$

then $f(2x) + 2f(y) = f(f(x+y))$

then $f(-2y) + 2f(y) = f(f(-y+y))$ (substitute $x$ with $-y$)

then $2f(-y) + 2f(y) = f(f(-y+y))$ (from Cauchy's theory $f(nx)=nf(x)$ where $n>0$ )

then $2f(-y) + 2f(y) = f(f(0))$ (from additional inverse property of any integers)

then $2f(-y) + 2f(y) = 0$ (from Cauchy's theory $f(0)=0$ )

then $f(-y) = -f(y)$ (from additional inverse property of any integers, since $f(y)$ $\in$ $\mathbb{Z}$)

then all functions $f$ : $\mathbb{Z}$ $\rightarrow$ $\mathbb{Z}$ that satisfy $f(2x) + 2f(y) = f(f(x+y))$ whereas $x,y$ $\in$ $\mathbb{Z}$ are odd.

I can't continue because I'd almost lost my dinner when I solve this, so I decided to take some Clorazepate and sleep.

Can you continue or check my progress? Is it right or not? Can you help me please?

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    $\begingroup$ Where do you get $f(nx)=nf(x)$? $\endgroup$ – Lord Shark the Unknown Jul 18 at 2:30
  • $\begingroup$ @LordSharktheUnknown By induction. Thankfully, I found it on my Olympiad training book $\endgroup$ – Supakorn Srisawat Jul 18 at 2:36
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    $\begingroup$ Induction? There are solutions $f$ to this problem where $f(2x)\ne 2f(x)$. $\endgroup$ – Lord Shark the Unknown Jul 18 at 2:37
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    $\begingroup$ @SupakornSrisawat Please use the tag [contest-math] when asking about problems relating to Olympiads. $\endgroup$ – user574848 Jul 18 at 6:24
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For an alternative approach, notice that $$f(2(x-1))+2f(y+1)=f(f(x+y))=f(2x)+2f(y)$$ so that $$2(f(y+1)-f(y))=f(2x)-f(2x-2).$$ Setting $x=0$ say gives $f(y+1)-f(y)=C$ is a constant. Therefore $f(x)=Cx+D$ for constants $C$ and $D$.

Now put this into the original equation to find all possible pairs $(C,D)$, etc.

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Another solution:

Note that substituting $a=0$ implies that $$f(f(b))=f(0)+2f(b).$$ So setting $b=0$ implies $f(2a)+2f(0)=f(f(a))=f(0)+2f(a) \implies f(2a)=2f(a)-f(0).$

Then rewriting the original equation with these two properties in mind, $$\begin{align*} f(2a)+2f(b)&=f(f(a+b)) \\\iff 2f(a)-f(0)+2f(b)&=f(0)+2f(a+b) \\ \iff 2f(a)+2f(b) &=2f(a+b)+2f(0) \\ \iff f(a)+f(b)&=f(a+b)+f(0)\end{align*}$$ Now define a new function $g(x)=f(x)-f(0)$ to see that $g(a)+g(b)=g(a+b)$, and hence $g$ satisfies Cauchy's functional equation. Thus $g(x)=mx$ for some constant $m$, and so $f(x)=mx+n$ for a constant $n=f(0)$.

Putting this into the original equation yields that the only solutions are $f(x)=0$ and $f(x)=2x+n$ for an integral constant $n$, and we have already verified that these two work.

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