# 1979 AHSME Problems/Problem 26

## Problem 26

The function $f$ satisfies the functional equation $f(x) +f(y) = f(x + y ) - xy - 1$ for every pair $x,~ y$ of real numbers. If $f( 1) = 1$, then the number of integers $n \neq 1$ for which $f ( n ) = n$ is $\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }3\qquad \textbf{(E) }\infty$

## Solution

We are given that $f(x) +f(y) = f(x + y ) - xy - 1$ and $f( 1) = 1$, so we can let $y = 1$. Thus we have: $$f(x) + 1 = f(x + 1) - x - 1$$

Rearranging gives a recursive formula for $f$: $$f(x + 1) = f(x) + x + 2$$

We notice that this is the recursive form for a quadratic, so f(x) must be of the form $ax^{2} + bx + c$. To solve for $a, b,$ and $c$, we can first work backwards to solve for the values of f(0) and f(-1): $$f(1) = f(0) + 0 + 2 \Rightarrow 1 = f(0) + 2 \Rightarrow f(0) = -1$$ $$f(0) = f(-1) -1 + 2 \Rightarrow -1 = f(-1) + 1 \Rightarrow f(-1) = -2$$

Since $f(0) = -1$: $$f(0) = a(0)^{2} + b(0) + c = -1 \Rightarrow c = -1$$ Since $f(1) = 1$: $$f(1) = a(1)^{2} + b(1)+ c = a + b - 1 = 1 \Rightarrow a + b = 2$$ Similarly, since $f(-1) = -2$: $$f(-1) = a(-1)^{2} + b(-1)+ c = a - b - 1 = -2 \Rightarrow a - b = -1$$

Thus we have the system of equations: $$a + b = 2$$ $$a - b = -1$$

Which can be solved to yield $a = \frac{1}{2}$, $b = \frac{3}{2}$. Therefore, $f(x) = \frac{1}{2} x^{2} + \frac{3}{2} x - 1$.

Since we are searching for values for which $f(x) = x$, we have the equation $\frac{1}{2} x^{2} + \frac{3}{2} x - 1 = x$. Subtracting $x$ yields $\frac{1}{2} x^{2} + \frac{1}{2} x - 1 = 0$, which we can simplify by dividing both sides by $\frac{1}{2}$: $x^{2} + x - 2 = 0$. This factors into $(x + 2) (x - 1) = 0$, so therefore there are two solutions to $f(x) = x$: $-2$ and $1$. Since the problem asks only for solutions that do not equal $1$, the answer is $\fbox{(B) 1}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 