Functional Equation Integer Solutions
The function $ f $ satisfies the functional equation
\[f(x) + f(y) = f(x + y) - xy - 1\]for all real numbers $ x $ and $ y $. If $ f(1) = 1, $ then find all integers $ n $ such that $ f(n) = n $. Enter all such integers, separated by commas.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Setting $ x = y = 0, $ we get
\[2f(0) = f(0) - 1,\]so $ f(0) = -1 $.
Setting $ y = 1, $ we get
\[f(x) + 1 = f(x + 1) - x - 1,\]so
\[f(x + 1) - f(x) = x + 2.\]Thus,
\begin{align*}
f(2) - f(1) &= 1 + 2, \\
f(3) - f(2) &= 2 + 2, \\
f(4) - f(3) &= 3 + 2, \\
&\dots, \\
f(n) - f(n - 1) &= (n - 1) + 2.\end{align*}Adding all the equations, we get
\[f(n) - f(1) = 1 + 2 + 3 + \dots + (n - 1) + 2(n - 1) = \frac{(n - 1)n}{2} + 2n - 2 = \frac{n^2 + 3n - 4}{2},\]so
\[f(n) = \frac{n^2 + 3n - 2}{2}\]for all positive integers $ n $.
Setting $ x = -n $ and $ y = n, $ where $ n $ is a positive integer, we get
\[f(-n) + f(n) = f(0) + n^2 - 1.\]Then
\[f(-n) = n^2 - f(n) + f(0) - 1 = n^2 - \frac{n^2 + 3n - 2}{2} - 2 = \frac{n^2 - 3n - 2}{2}.\]Thus, the formula
\[f(n) = \frac{n^2 + 3n - 2}{2}\]holds for all integers $ n $.
We want to solve $ f(n) = n, $ or
\[\frac{n^2 + 3n - 2}{2} = n.\]Then $ n^2 + 3n - 2 = 2n, $ or $ n^2 + n - 2 = 0 $. This factors as $ (n - 1)(n + 2) = 0, $ so the solutions are $ n = \boxed{1,-2} $.
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