1996 IMO Problems/Problem 3

Problem

Let $S$ denote the set of nonnegative integers. Find all functions $f$ from $S$ to itself such that

$f(m+f(n))=f(f(m))+f(n)$ $\forall m,n \in S$

Solution

Plugging in m = 0, we get f(f(n)) = f(n) $\forall n \in S$. With m = n = 0, we get f(0) = 0.

Let $n_{0}$ be the smallest fixed point of $f(x)$ such that $n_{0} > 0$. $\implies f(n_{0}) = n_{0}$. Plugging $\m = n = n_{0}$ (Error compiling LaTeX. Unknown error_msg), we get $f(2n_{0}) = 2f(n_{0})$.

By an easy induction, we get $f(kn_{0}) = kf(n_{0}) \forall k \in \mathbb{N}$.

Let $n_{1}$ be another fixed point greater than $n_{0}$. Let $n_{1} = qn_{0} + r$, where $r < n_{0}$.

So, $f(n_{1}) = f(qn_{0} + r) = f(r + f(qn_{0})) = f(r) + qn_{0} = r + qn_{0}$.

$\implies f(r) = r$. But, $r < n_{0} \implies r = 0$.


See Also

1996 IMO (Problems) • Resources
Preceded by
Problem 2
1 2 3 4 5 6 Followed by
Problem 4
All IMO Problems and Solutions