(Olympiad Question) $f(2x) + 2f(y) = f(f(x+y))$

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?

• Where do you get $f(nx)=nf(x)$? – Lord Shark the Unknown Jul 18 at 2:30
• @LordSharktheUnknown By induction. Thankfully, I found it on my Olympiad training book – Supakorn Srisawat Jul 18 at 2:36
• Induction? There are solutions $f$ to this problem where $f(2x)\ne 2f(x)$. – Lord Shark the Unknown Jul 18 at 2:37
• @SupakornSrisawat Please use the tag [contest-math] when asking about problems relating to Olympiads. – user574848 Jul 18 at 6:24

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.
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.