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 $\fbox{(B) 1}$ .

1979 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 25
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1979 AHSME Problems/Problem 26

Problem 26

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