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 positive integers n for which f(n) = n is:
- 0
- 1
- 2
- 3
- infinite
We are given f(1) = 1. Let’s find the functional values of some other integers.
f(x) + f(y) = f(x + y) – xy – 1
f(x) + f(0) = f(x + 0) – x · 0 – 1 = f(x) – 1
So f(0) = -1
f(x) + f(0) = f(x + 0) – x · 0 – 1 = f(x) – 1
So f(0) = -1
f(1) + f(1) = f(2) – 1 · 1 – 1
1 + 1 = f(2) – 1 – 1
And f(2) = 4
f(2) + f(1) = f(3) – 2 · 1 – 1
4 + 1 = f(3) – 2 – 1
Then f(3) = 8
Let’s make a chart and look for a pattern.
| x | f(x) |
| 0 | -1 |
| 1 | 1 = f(0) + 2 |
| 2 | 4 = f(1) + 3 |
| 3 | 8 = f(2) + 4 |
| 4 | 13 = f(3) + 5 |
| … | |
| n | = f(n – 1) + n + 1 |
So, 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, n > 1, will work. So the answer is (b).