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 4 of 4: Combine with the involution to pin Δ = 1

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f(f(n))=1+(n-1)\Delta^2=n \implies \Delta=1 \implies f(x)=x

Detailed analysis

Apply $f$ to $f(n)=1+(n-1)\Delta$ using the formula from Step 3 again: $f(f(n))=1+(f(n)-1)\Delta=1+(n-1)\Delta^2$ . By Step 1, $f(f(n))=n$ , so $1+(n-1)\Delta^2=n$ for every $n$ , i.e. $(n-1)(\Delta^2-1)=0$ for every $n$ ; taking $n=2$ gives $\Delta^2=1$ , so $\Delta=1$ (as $\Delta>0$ ). Hence $f(n)=1+(n-1)=n$ for all $n$ . Conversely $f(x)=x$ works: the triple $a,b,a+b-1$ satisfies $a+b-1<a+b$ , $a<b+(a+b-1)$ and $b<a+(a+b-1)$ (both reduce to $a,b\ge1$ ), so it is always a non-degenerate triangle. Hence $f(x)=x$ is the unique solution. $\blacksquare$