Consider the following problem from 1987 IMO Show that there is no function $f:\mathbb N→\mathbb N$ Such that
$f(f(n))=n+1987$
In the book that I was referring the solution is given as follows:
" We show that infact there is no function that satisfies this condition
$f(f(n))=n+k$ where k is odd
Assume such an $f$ exists. Then $f(n+k)=f(n)+k$ for all n in N and $f(n-k)=f(n)-k$ for all $n\geq k$ Consider $g(n)=f(n)-k$
Verify that $g(f(n))=n$ for all n and $f(g(n))=n$ for all $f(n) \geq k$
Consider the sets $A=${$n|0\leq n<k$ and $f(n)<k$} and $B=${$n|0\leq n<k$ and $f(n)\geq k$}
Show that there is a bijection between A and B and conclude
$|A|=|B|$
This implies $2|A|=k$ and this forces a contradiction "
The only part I am having trouble is finding such a bijection
Can someone please help
Here $\mathbb{N}$ is the set of all natural numbers and 0