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Difference between revisions of "1996 IMO Problems/Problem 3"
(→‎Solution)
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Let <math>n_{0} </math> be the smallest fixed point of <math>f(x) </math> such that <math>n_{0} > 0 </math>.  
 
Let <math>n_{0} </math> be the smallest fixed point of <math>f(x) </math> such that <math>n_{0} > 0 </math>.  
 
<math>\implies f(n_{0}) = n_{0} </math>.  
 
<math>\implies f(n_{0}) = n_{0} </math>.  
−
Plugging <math>\m = n = n_{0} </math>, we get <math>f(2n_{0}) = 2f(n_{0}) </math>.  
+
Plugging <math>m = n = n_{0} </math>, we get <math>f(2n_{0}) = 2f(n_{0}) </math>.  
  
 
By an easy induction, we get <math>f(kn_{0}) = kf(n_{0}) \forall k \in \mathbb{N} </math>.
 
By an easy induction, we get <math>f(kn_{0}) = kf(n_{0}) \forall k \in \mathbb{N} </math>.
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<math>\implies f(r) = r </math>. But, <math>r < n_{0} \implies r = 0 </math>.
 
<math>\implies f(r) = r </math>. But, <math>r < n_{0} \implies r = 0 </math>.
  
 +
This means that the set of all fixed points of <math>f(x) </math> is
  
 
==See Also==
 
==See Also==

Revision as of 10:21, 3 June 2024

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 $m = n = n_{0}$, 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, $r < n_{0} \implies r = 0$.

This means that the set of all fixed points of $f(x)$ is

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
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