Family of functions satisfying a particular criterion

[issacnewton]

Homework Statement

Let $f:\mathbb{Z}\to \mathbb{Z}$ be a family of functions following the criterion. $$f(2a) + 2 f(b) = f(f(a+b)) $$. Find the family of functions

Relevant Equations

definition of function

Since, both the domain and the range is set of integers, we must have just operations of addition and multiplication only in the function. That means, function should be some kind of a polynomial. Plugging $a=0$ and $b=0$, I can deduce that $3 f(0) = f(f(0))$. Also I can deduce that $3 f(2a) = f(f(3a))$. But after this, I am stuck. So, any guidance will be helpful. This is a math Olympiad question.

Thanks

 

[fresh_42]

Since $1$ generates the integers, it is certainly useful to consider $f(1)$. E.g. compare $f^2(0+1)$ and $f^2(1+0)$.

 

[issacnewton]

Ok, we would get the following equations
$$ f(2) + 2 f(0) = f(f(1)) $$
$$ f(0) + 2 f(1) = f(f(1)) $$
And subtracting one from the another, we would get
$$ f(2) + f(0) = 2 f(1) $$

 

[fresh_42]

I would also try multiplication: Is $x \stackrel{f}{\longmapsto} c\cdot x$ a solution? The tricky part is probably to get rid of $f^2$. So what happens if you consider $f^2(0)=:c$ or $f^2(1)=:c\,?$ Or consider $f(-1)$.

I don't see the solution, but I think deriving other rules as the one you just got is helpful. E.g. what happens if you repeat the calculation which led to $f(2)+f(0)=f(1)$, i.e. $f^2(2)=f(2)+2f(1)=f(4)+2f(0)=f(0)+2f(2)$ and so on. Once you have found a recursion, you will be done.

 

[Stephen Tashi]

Since, both the domain and the range is set of integers, we must have just operations of addition and multiplication only in the function.

It may turn out that the answer obeys those conditions, but there are many types of functions from $\mathbb{Z}$ to $\mathbb{Z}$ that are not polynomials.

 

[issacnewton]

I am also able to prove that $f(f(0)) = 3 f(0)$, $f(2f(0)) = 5 f(0)$, $f(3f(0)) = 7 f(0)$. So, we can prove with induction that if $n \in \mathbb{N}$
$$ f(n\cdot f(0)) = (2n+1) f(0) $$

 

[Stephen Tashi]

$$f(2a) + 2 f(b) = f(f(a+b)) $$.

One observation is that the right hand side of the equation remains the same if you swap $a$ and $b$.

 

[Stephen Tashi]

$$f(2a) + 2 f(b) = f(f(a+b)) $$. Find the family of functions

I see how to guess a family of functions informally. I don't yet see how to deduce it.

For a given number $c$, if we want a function that can gives the same value in a calculation that involves applying it to each pair $a,b$ such that $a+b = c$, nothing comes to mind except $f(x) = k x$. Making that guess, we have

$k2a + 2kb = k(k(a+b))$
$2k(a+b) = k^2(a+b)$
$2 = k$ or $0 = k$

 

[Antarres]

Hmm, how about this. Setting $b=0$, we get:
$$f(2a)+2f(0)=f(f(a))$$
Setting $a=0$ and substituting $b$ for aa, we get:
$$f(0)+2f(a)=f(f(a))$$
Right hand sides are equal, so we can identify them:
$$f(2a)=2f(a)−f(0)$$
This looks like a useful relation.
Let's try to put another value in the initial equation, for example $a=1$. We get:
$$f(2)+2f(x)=f(f(x+1))=f(0)+2f(x+1)$$
Where we used the second equation from the top to simplify the right hand side.
From there we get:
$$f(x+1)−f(x)=const.$$
So we have an arithmetic progression, and therefore $f$ is linear. Finally, we substitute $f(x)=ax+b$ for some constants $a$ and $b$. Solving, we get the solution.

EDIT: Oops, I just realized this was in the homework section, so I cut the solution down. This should be enough for a hint.

 

[WWGD]

How about considering the cases a=b and a=-b?

 

[Antarres]

Well sure, you can do that, but those relations won't contribute much, I think. The problem becomes relatively easy once it's concluded that the function is linear. Generally in these types of problems, substituting 0, 1, -1, usually helps acquire some good relations to start off. Either that or switching places of $a$ and $b$ looks interesting since the right side is symmetrical with respect to them.

 

[fresh_42]

$f(x)=2x$ is obviously a solution, and by symmetry we have $f(2x)=2f(x)-f(0)$. If I haven't made a mistake, we also have from $f^2(3x)$
\begin{align*}
f(2x)&=2f(x)-f(0)\\
f(4x)&=4f(x)-3f(0)\\
f(6x)&=6f(x)-f(0)\\
f(8x)&=8f(x)-7f(0)
\end{align*}
which is a bit strange. The questions now are whether $f(0)\neq 0$ is possible, and whether $f(x)=2x$ is the only solution.

 

[Antarres]

@fresh_42 Actually, I cut the solution down, since it's homework, but after plugging in the second condition, that it is linear, you obtain precisely whether $f(x) = 2x$ is the only solution(turns out it isn't, actually, although at first I thought that it is). Solving you get that solutions have the form of $f(x) = 2x + c$, for arbitrary $c$.