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 in addition f(1)=1, then the number of positive integers n for which f(n)= n is:
- 0
- 1
- 2
- 3
- infinite
Solution
We are given f(1)=1. Let's find some more of this function's values.
f(x)+ f(y)f(x)+ f(0)== f(x + y)− xy −1 f(x +0)− x ⋅0−1= f(x)−1
So f(0)=−1.
f(1)+ f(1)1+1== f(2)−1⋅1−1 f(2)−1−1
So f(2)=4.
f(2)+ f(1)4+1== f(3)−2⋅1−1 f(3)−2−1
So f(3)=8. We've made a chart (see below) in order to look for a pattern. We see that for f(n)= n, we have to have:
f(n)= f(n −1)+ n +1= n,
so f(n −1)+1=0, and f(n −1)=−1. We see that n =1 works and no other positive integers will work.
The answer is (b).