2580.51 – A Hard Functional Equation

Tags:Problem Set 8Pre-Calculus MathSymbol-Pushing

The function ff satisfies the functional equation

f(x)+ f(y)= f(x + y)− xy −1,f(x) + f(y) = f(x + y) - xy - 1,

for every pair x, yx, y of real numbers. If in addition f(1)=1f(1) = 1, then the number of positive integers nn for which f(n)= nf(n) = n is:

  1. 0
  2. 1
  3. 2
  4. 3
  5. infinite
Solution

We are given f(1)=1f(1) = 1. Let's find some more of this function's values.

f(x)+ f(y)= f(x + y)− xy −1f(x)+ f(0)= f(x +0)− x ⋅0−1= f(x)−1\begin{aligned} f(x) + f(y) &=& f(x + y) - xy - 1 \\ f(x) + f(0) &=& f(x + 0) - x · 0 - 1 = f(x) - 1 \end{aligned}

So f(0)=−1.f(0) = -1.

f(1)+ f(1)= f(2)−1⋅1−11+1= f(2)−1−1\begin{aligned} f(1) + f(1) &=& f(2) - 1 · 1 - 1 \\ 1 + 1 &=& f(2) - 1 - 1 \end{aligned}

So f(2)=4.f(2) = 4.

f(2)+ f(1)= f(3)−2⋅1−14+1= f(3)−2−1\begin{aligned} f(2) + f(1) &=& f(3) - 2 · 1 - 1\\ 4 + 1 &=& f(3) - 2 - 1 \end{aligned}

So f(3)=8f(3) = 8. We've made a chart (see below) in order to look for a pattern. We see that for f(n)= nf(n) = n, we have to have:

f(n)= f(n −1)+ n +1= n,f(n) = f(n - 1) + n + 1 = n,

so f(n −1)+1=0f(n - 1) + 1 = 0, and f(n −1)=−1f(n - 1) = -1. We see that n =1n = 1 works and no other positive integers will work.

The answer is (b).