IMO 2019: Find all functions $f: \mathbb Z \to \mathbb Z$ such that $f(2a) + 2\, f(b) = f \big( f (a+b)\big)$.

Question

Source: International Mathematical Olympiads, 2019.

The question is:

Let $\mathbb{Z}$ be the set of integers. Find all functions $f: \mathbb Z \to \mathbb Z$ such that, for all integers $a$ and $b$, the following is true $$f(2a) + 2\,f(b) = f\big(f (a+b)\big)\,.$$

I only managed to do this:

Suppose some function f has the form : f(x)= n.x ( with n, integer).

We would get :

n(2a)+ 2(nb) = n(n(a+b))

--> 2na + 2nb = n²(a+b)

--> 2n (a+b) = n²(a+b) ( Dividing by a + b, provided a+b is not equal to 0)

--> 2 n = n²

--> n² = 2 n

--> n = 2. ( Although I wonder whether dividing by number n is legal, even if n is not 0)

So, apparently, if a function satisfying the f-conditions has form f(x)= n.x, then

this function must have form f(x) = 2x.

However, apparently, f must have any sum ( a + b) in its domain, in order to be from Z to Z; for the expression f( a+b) featured in the expression defining f. But, while solving the equation, I excluded the case in which (a+b)= O. Notwithsatnding (0 + 0) is a sum of integers.

So, my hypothesis fails, and I am left with a negative conclusion :

there is no function f ( satisfying the f formula) such that f(x) = n.x.

Answer 1

We say that $f:\mathbb{Z}\to\mathbb{Z}$ satisfies the condition $P(a,b)$ for some integers $a$ and $b$ if $$f(2a)+2\,f(b)=f\big(f(a+b)\big)\,.$$ We want to find all functions $f:\mathbb{Z}\to\mathbb{Z}$ such that $f$ has the property $P(a,b)$ for every $a,b\in\mathbb{Z}$. We shall prove that either $f(n)=0$ for every integer $n$, or there exists a constant $c\in\mathbb{Z}$ such that $f(n)=2n+c$ for each integer $n$.

Let $c:=f(0)$. For arbitrary $a,b\in\mathbb{Z}$, the conditions $P(a,b)$ and $P(b,a)$ implies $$f(2a)+2\,f(b)=f\big(f(a+b)\big)=f\big(f(b+a)\big)=f(2b)+2\,f(a)\,,$$ whence $$f(2a)-f(2b)=2\big(f(a)-f(b)\big)$$ for every $a,b\in\mathbb{Z}$. In particular, when $b:=0$, we get $$f(2a)=2f(a)-c\tag{*}$$ for all integers $a$. Therefore, using (*), $P(a,b)$ becomes $$f\big(f(a+b)\big)=2\,f(a)+2\,f(b)-c\tag{$\star$}$$ for all $a,b\in\mathbb{Z}$. In particular, with $b:=0$ in ($\star$), we get $$f\big(f(a)\big)=2\,f(a)+c\tag{#}$$ for every $a\in\mathbb{Z}$. From (#) and ($\star$), we have $$f(a+b)=f(a)+f(b)-c\,.$$ Let $g(n):=f(n)-c$ for every integer $n$. The equation above can be rewritten as $$g(a+b)=g(a)+g(b)$$ for every $a,b\in\mathbb{Z}$. This is Cauchy's functional equation, and the solutions are well known to be $g(n)=kn$ for some constant $k\in\mathbb{Z}$. That is, $$f(n)=g(n)+c=kn+c$$ for some integer constants $k$ and $c$. Now, we must check the property $P(a,b)$. That is, for all integers $a$ and $b$, $$\big(k(2a)+c\big)+2\big(kb+c\big)=f(2a)+2\,f(b)=f\big(f(a+b)\big)=k\big(k(a+b)+c\big)+c\,.$$ Hence, $$2k(a+b)+3c=k^2(a+b)+(k+1)c$$ for all $a,b\in\mathbb{Z}$ Thus, $$(k-2)\big(k(a+b)-c\big)=0$$ for all $a,b\in\mathbb{Z}$. This implies:

• $k=2$, or
• $k(a+b)=c$ for all $a,b\in\mathbb{Z}$.

In the former case, $f(n)=2n+c$ for all integers $n$, which is easily seen to be a solution. In the latter case, $k=c=0$ is the only possibility, so $f(n)=0$ for all integers $n$, which is obviously solution.

Remark. Here is a generalization, a proof of which is obtained largely by the same argument as above. Suppose that $I$ is an integral domain. Let a nonempty set $S\subseteq I$ and a nonzero scalar $\lambda\in I$ be such that

• $S$ is closed under addition (i.e., $S$ is an additive subsemigroup of $I$),
• $S$ is closed under multiplication by $\lambda$,
• if $h:S\to I$ satisfies Cauchy's functional equation $h(x+y)=h(x)+h(y)$ for all $x,y\in S$, then there exists $\mu\in I$ such that $h(x)=\mu x$ for all $x\in S$.

The solutions $f:S\to S$ such that $$f(\lambda x)+\lambda\,f(y)=f\big(f(x+y)\big)$$ for all $x,y\in S$ are

• $f(x)=0$ for all $x\in S$ in the case that $0\in S$, and
• $f(x)=\lambda x+c$ for all $x\in S$, where $c\in I$ is a constant such that $\lambda x+c\in S$ for any $x\in S$.

In the setting where $I=\mathbb{C}$, the set $S$ can be $\mathbb{Z}_{\geq 0}$, $\mathbb{Z}_{>0}$, $\mathbb{Q}$, $\mathbb{Q}_{\geq 0}$, $\mathbb{R}_{\geq 0}$, or $\mathbb{R}_{>0}$. In fact, $S$ can be any additive subsemigroup of $\mathbb{Z}$ or $\mathbb{Q}$, or any divisible additive subsemigroup of $\mathbb{R}_{\geq 0}$ or $\mathbb{R}_{\leq 0}$ will do. An obvious choice of $\lambda\in \mathbb{C}$ that works for all $S$ is a positive integer.

Answer 2

Let's aim for a really short solution. For $n\in\mathbb{Z}$ we have $$f(2)+2f(n)=f(f(n+1))=f(0)+2f(n+1).$$ Because $f(n+1)-f(n)=\frac{f(2)-f(0)}{2}$ is independent of $n$ the function $f$ is linear. It remains to check that $f(n):=A+B\cdot n$ satisfies the original equation for arbitrary $A\in\mathbb{Z}$ and only for $B\in\left\{0,2\right\}$.