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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 $a{x}^2+bx+c$. 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 $\boxed{1}$.