Problem 13: IMO 1987, Day 2, Problem 4

by henderson, May 26, 2016, 5:07 PM

$\bf\color{red}Problem \ 13\$ Prove that there is no function $f$ from the set of non-negative integers into itself such that $f(f(n))=n+1987$ for all $n$ .
$\bf\color{red}Solution\$ Define $a(n)=f(n)-n$ . Then, from $f(f(n))=n+1987$ we get $(f(f(n))-f(n))+(f(n)-n)=1987$ , or $a(f(n))+a(n)=1987$ for every $n$ natural.

If we replace $n$ by $f(n)$ in $a(f(n))+a(n)=1987$ , we get $a(f(f(n)))+a(f(n))=1987=a(f(n))+a(n)$ , and then $a(n+1987)=a(n)$ for every natural $n$ . Thus, $a$ is an 1987-periodic funcion.

Now, let $a_1=a(1), a_2=a(2),...,a_{1987}=a(1987)$ be the $1987$ possible values of $a(n)$ (Since $a$ is periodic, this is possible). We must pair the 1987 $a_i$ in the following mannner: $a_i$ is paired with $a_j$ if $a_i +a_j =1987$ (we can always do this, because $a(f(n))+a(n)=1987$ for every $n$ natural). If there's two or more $j$ satisfying the pairment, anyone serve. So, we can divide $1987$ numbers in pairs, and this is a contradiction, because $1987$ is odd. Finally, the problem is solved.

Note. Accually, the above solution shows that there doesn't exist a function for any odd non-negative integer instead of $1987.$

This post has been edited 4 times. Last edited by henderson, Sep 12, 2016, 2:32 PM

functional equation

Comment

0 Comments

Submit

Math Path

Problem 13: IMO 1987, Day 2, Problem 4

by henderson, May 26, 2016, 5:07 PM

0 Comments