\section*{Problem 3A - 311043A}

Let \(f\) be a function defined for all positive integers \(n\) and only for these, and whose function values \(f(n)\) are all integers. Furthermore, it is assumed that for all positive integers \(m, n\), the equation \(f(f(m)+f(n))=m+n\) holds.

Determine all the numbers that can occur as the function value \(f(1992)\) for such a function \(f\).
}

All the following variables \(m, n, k\) are positive integers.

For \(m=n\), the functional equation is \(f(2 f(n))=2 n\).

The argument \(2 f(n)\) of the outer function must be a positive integer, so \(f(n)\) is always a positive integer.

\(f\) is also injective: from \(f(n)=f(k)\) it follows that \(2 n=f(2 f(n))=f(2 f(k))=2 k\), hence \(n=k\).

Since in \(f(f(m)+f(n))=m+n\) the right side can take all natural numbers greater than 1, every natural number greater than 1 must have a preimage.

From the functional equation and the injectivity of \(f\) it follows that

\[
2 f(n)=f(n+1)+f(n-1)=f(n+2)+f(n-2)=\ldots=f(1)+f(2 n-1)
\]

because if one applies \(f\) to each side of this system of equations, one always gets \(2 n\).

Thus, \(2 f(n)\) has \(n-1\) different representations as the sum of two different summands (different because of the injectivity of \(f\)). [Two representations are called different if they do not transform into each other by swapping the summands.] Therefore, it must be that \(f(n) \geq n\) for all \(n \geq 2\).

If \(f(1)>1\), then \(2=f(f(1)+f(1)) \geq 4\), a contradiction, so \(f(1)=1\).

If now \(f(2)>2\), the number 2 would have no preimage, a contradiction, so \(f(2)=2\). Inductively, one gets \(f(n)=n\) for all natural \(n\), which satisfies the functional equation.

Note: One already obtains the solution from \(2 f(n)=f(n+1)+f(n-1)\): the characteristic polynomial has the double root 1, so \(f(n)=a+b n\). Since \(f\) has a natural range, \(a\) is an integer and \(b \geq 0\). Due to the injectivity, \(b \neq 0\). Therefore, and since every natural number greater than or equal to 2 has a preimage, \(b=1\) and \(a=1\) or \(a=0\). Thus, and after substituting into the functional equation, one gets \(a=0\).