Problem 5

Determine all functions

f

from the set of positive integers to the set of positive integers such that, for all positive integers

a

and

b

, there exists a non-degenerate triangle with sides of lengths

a

f(b)

and

f(b+f(a)-1)

. (A triangle is non-degenerate if its vertices are not collinear.)

Step 2 of 4: A second application pins the increment to a constant

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\Delta:=f(2)-1>0,\qquad f(n+1)=f(n)+\Delta \ \text{ or } \ f(n-1)=f(n)+\Delta

Detailed analysis

Since $f(f(n))=n$ , $f$ is injective, so $f(2)\ne f(1)=1$ , giving $\Delta:=f(2)-1\ge1$ . Take $a=2,b=f(n)$ : the triangle with sides $2,f(f(n)),f(f(n)+f(2)-1)=2,n,f(f(n)+\Delta)$ forces $|n-f(f(n)+\Delta)|<2$ , so $y:=f(f(n)+\Delta)\in\{n-1,n,n+1\}$ . Applying $f$ (using $f\circ f=\mathrm{id}$ ) to $y=f(f(n)+\Delta)$ gives $f(y)=f(n)+\Delta$ . If $y=n$ this reads $f(n)=f(n)+\Delta$ , impossible since $\Delta>0$ . So $y=n\pm1$ , giving $f(n+1)=f(n)+\Delta$ or $f(n-1)=f(n)+\Delta$ .