Let $ f(n)$ be a function defined on the set of all positive integers and having its values in the same set. Suppose that $ f(f(n) + f(m)) = m + n$ for all positive integers $ n,m.$ Find the possible value for $ f(1988).$
Problem
Source: IMO ShortList 1988, Problem 19, Mexico 1, Problem 49 of ILL
Tags: function, algebra, functional equation, IMO Shortlist
ahaanomegas
04.11.2013 18:22
Eight years and one day later . . . the 500th view and
\[ f \left( f(m) + f(n) \right) = m + n \]Substitute $ m = f(a) + f(b) $ and $ n = f(c) + f(d) $ to get \[ f \left( f \left( f(a) + f(b) \right) + f \left( f(c) + f(d) \right) \right) = f(a) + f(b) + f(c) + f(d). \]We know that $ f \left( f(a) + f(b) \right) = a + b $ and $ f \left( f(c) + f(d) \right) = c + d $. Thus: \[ f(a+b+c+d) = f(a) + f(b) + f(c) + f(d), \qquad \forall a, b, c, d \in \mathbb{N} \]If we substitue $a+1$ for $a$ and $b-1$ for $b$, we obtain \[ \begin {eqnarray*} f(a+b+c+d) &=& f(a+1) + f(b-1), \\ f(a) + f(b) &=& f(a+1) + f(b-1) \\ f(a+1) - f(a) &=& f(b) - f(b-1) \]Since this holds for all positive integers $a,b$, we conclude that $f$ is an arithmetic sequence. Hence, it is of the form $f(n) = \alpha \cdot n + \beta$.
We substitute this into our original function equation, and we get \[ \begin {eqnarray*} \alpha \cdot \left( \alpha \cdot m + \beta + \alpha \cdot n + \beta \right) &=& m + n \\ \implies \ \alpha^2 \cdot m + \alpha^2 \cdot n + 2 \alpha \beta + \beta &=& m + n. \]\[ \implies \left\{\begin{array}{lr} \alpha^2 = 1, \\ 2 \alpha \beta + \beta = 0.\end{array} \right. \]The second equation gives us $ \beta \cdot (2 \alpha + 1) = 0 $. If $ 2 \alpha + 1 = 0 $, then $ \alpha = -\dfrac{1}{2} $. But, $\alpha$ and $\beta$ are positive integers, so $ \beta = 0 $. From the first equation, we have $ \alpha = \pm 1 $, but $ \alpha $ must be a positive integer so it is $1$.
Therefore, $ f (n) = n $ and $ f(1988) = \boxed {1988} $.
SMOJ
31.12.2013 16:23
The second solution from the IMO compendium gave that $f(f(1)+n+m)=f(f(1)+f(f(m)+f(n)))=1+f(m)+f(n)$, and hence concluded that $f(m)+f(n)$ is a function of $m+n$. How does this hold true?
chaotic_iak
31.12.2013 19:33
If $f(x)$ is a function of $x$, then $f(x+a)+b$ is also a function of $x$ where $a,b$ are reals. Since $f(m+n)$ is obviously a function of $m+n$, then $f(m+n + f(1)) - 1 = f(m) + f(n)$ is also a function of $m+n$.
grupyorum
02.10.2018 23:14
We claim that, $f(1988)=1988$. We proceed as follows. Let $P(m,n)$ be the assertion given in the problem. Then, if there exists any $n_1,n_2$ with $f(n_1)=f(n_2)$, by inspecting $P(m,n_1)$ and $P(m,n_2)$, we would have deduced that $n_1=n_2$. Thus, $f(\cdot)$ is injective. Next, notice that, for every $k\geq 2$, there exists a $k_0$ such that $f(k_0)=k$. We will now show that the values of $f(1)$ and $f(2)$ uniquely determines the rest of the sequence onwards. To show this, let $f(1)=a$ and $f(2)=b$. Then by considering $P(2,2)=P(1,3)$, together with injectivity, we have $f(1)+f(3)=2f(2)\implies f(3)=2b-a$. Now, $P(4,1)=P(3,2)\implies f(4)+a=3b-a\implies f(4)=3b-2a$. In particular, we claim that $f(k)=(k-1)b-(k-2)a$, for every $k$. For this, we'll rely on induction. Base cases are clear, and assume $f(k-1)=(k-2)b-(k-3)a$. Now, $P(k,1)=P(k-1,2)$ implies, $f(k)+a=(k-2)b-(k-3)a+b\implies f(k)=(k-1)b-(k-2)a$. In particular, we deduce that, $f(k)=uk+v$ for some $u,v\in\mathbb{Z}$, to be determined next. Now, inserting $f(k)=uk+v$ into $P(m,n)$, we arrive at, $u=1$ and $v=0$; hence, $f(k)=k$. In particular, $f(1988)=1988$.
ZETA_in_olympiad
22.05.2022 10:02
$f$ is the identity. So $f(1988)=1988.$ Proof for PEN.