Let $\alpha$ be given positive real number, find all the functions $f: N^{+} \rightarrow R$ such that $f(k + m) = f(k) + f(m)$ holds for any positive integers $k$, $m$ satisfying $\alpha m \leq k \leq (\alpha + 1)m$.
Problem
Source: China TST 2005-6
Tags: function, algebra unsolved, algebra
Bluesea
02.05.2006 10:12
it seems very hard and nice;anybody have solution
toetoe
02.05.2006 11:46
hi, here is my solution : 1-we will find all fonction g such that $g(k+m) = g(k)+g(m) (1)$. lets assume that : $g(1) = \beta$such that $\beta$ is a real number . $g(1)=x \Rightarrow g(2)=2x \dots$ by recurence ,we can prove that $g(x) = ax$ 2- $g$ verifies also the second relation . so $f(x) = g(x)= \beta x$ .
Sardor
27.05.2014 17:43
Any full solution? Please...
alibez
28.05.2014 11:30
toetoe wrote: $g(1)=x \Rightarrow g(2)=2x \dots$ by recurence ,we can prove that $g(x) = ax$ why ? are you sure it is true? toetoe wrote: hi, here is my solution : 1-we will find all fonction g such that $g(k+m) = g(k)+g(m) (1)$. lets assume that : $g(1) = \beta$such that $\beta$ is a real number . $g(1)=x \Rightarrow g(2)=2x \dots$ by recurence ,we can prove that $g(x) = ax$ 2- $g$ verifies also the second relation . so $f(x) = g(x)= \beta x$ . your solution is wrong !
toto1234567890
31.01.2017 05:52
http://www.artofproblemsolving.com/community/q1h448277p2524052 Almost same
Pathological
06.04.2019 18:51
We claim that all solutions are $f(x) \equiv cx,$ for $c \in \mathbb{R}.$ Firstly, it's immediate to verify that these functions $f$ work.
Now, we will show that they are the only such functions which work. Let's select some arbitrary large positive integer $m$, and consider the interval $((m+1)\alpha, m(\alpha + 1))$. Select some integer $k$ such that $k , k+1$ are both within this interval. Then, by the condition of problem, we know that $f(m+1+k+1) = f(m+1) + f(k+1), f(m+1+k) = f(m+1)+f(k), f(m+k+1) = f(m) + f(k+1)$, and $f(m+k) = f(m) + f(k).$
Therefore, this gives that $f(m+k+2) - f(m+k+1) = f(k+1) - f(k) = f(m+k+1) - f(m+k).$
Let's call the natural number $t$ $\mathbf{good}$ if $f(t)$ is the average of $f(t-1), f(t+1)$, so that we know that any number in $(m + \alpha (m+1) + 1, m(\alpha +1) + m)$ is good, for all $m$ sufficiently large. However, it's clear that all sufficiently large numbers are contained in an interval of that form, and so hence we conclude that all sufficiently large numbers are good, say $t \ge C$ are all good.
Then, we know that $f(C), f(C+1), f(C+2), \cdots$ is an arithmetic progression, and so if we apply the condition with $m, k \ge C$, we know that $f \equiv cx$ for some $c \in \mathbb{R}$ when $x \ge C.$
Now, it simply remains to use downwards induction to show that $f \equiv cx, \forall x \in \mathbb{R}.$ (Just set $m =k$ in the condition of the problem)
$\square$
niksic
19.05.2020 20:50
Quite late to post a solution, but the problem is amazing so I'll do it anyways Call an integer $n$ separable if there exist two integers $a,b$ such that $a+b=n$ and $f(a+b)=f(a)+f(b)$. We will prove all sufficiently large integers are separable: it is enough to have $\alpha \leq \frac{a}{b} \leq \alpha +1 \iff \alpha+1 \leq \frac{a+b}{b} \leq \alpha +2 \iff \frac{1}{\alpha +1} \leq \frac{b}{n} \leq \frac{1}{\alpha +2}$. Start with $\frac{1}{n}$ and add $\frac{1}{n}$ to it in every step, and as long as $\frac{1}{n}$ is smaller that the difference between $\frac{1}{\alpha +2}$ and $\frac{1}{\alpha +2}$ we can't jump over the interval $(\frac{1}{\alpha +2}, \frac{1}{\alpha +1})$ so such integer is separable by the problem condition. We use the above claim to conclude that $f(n)$ can be represented as a linear combination of $f(1), f(2), ..., f(k)$ for some fixed $k$ (the $f$-s of $1$ through $k$ being just themselves). Now consider the smallest subset of these such that the linear representation using those is unique (sorry for probable abuse of terminology), and write $f(n)=g_1(n)f(a_1)+...+g_m(n)f(a_m)$ where $\{a_1, a_2, ..., a_m\} \subset \{1,2,...,k\}$. By the uniqueness of the linear representation, we have that the problem condition carries over to the $g_i$ functions (you get the reasoning by thinking of $f$ as a vector field). But we have the functions $g_i$ are from the integers to rationals (I am skipping a bit of reasoning but it should make sense), and every rational can be represented as a linear combination of other rationals, and combine that with the first paragraph we get that $g_i$s have to be linear combinations of just one $g_i(something)$, so it has to take form $g_i(n)=nc_i$ where $c_i$ is constant. So we conclude we can represent $f$ in a comfortable way: $f(n)=n(c_1f(a_1)+...+c_mf(a_m)) = nD$ where $D=const.$, this being true for all $n>k$(although we can be done here it is really hard to explain why). Now for some final touches, divide the problem into two cases: $\alpha \leq1 $ and $\alpha >1$. In the first case we have $f(2n)=2f(n)$ and from here just choose any $n$ and really big $k$ so that $2^kf(n)=f(2^kn)=2^knD \implies f(n)=nD$. For the second case, choose an integer $\alpha \leq a \leq \alpha+1 \implies a>1$. We have $f((a+1)n)=f(n)+f(an)$ and choose $n\in (k/2, k]$ to expand the $f(n)=nD$ to number less that $k$ in an obvious manner, and so we are done $\blacksquare$ I again apologize to anyone who might read this for the terrible explanation and terminology, my aim was to make it fell natural