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
$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }3\qquad \textbf{(E) }\infty$
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 $\fbox{(B) 1}$.