The number of integers a in the interval [1, | vedclass

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(C) Given the system of equations:$x+y=a$ $(1)$$\frac{x^2}{x-1}+\frac{y^2}{y-1}=4$ $(2)$From $(2)$,we have $\frac{x^2(y-1)+y^2(x-1)}{(x-1)(y-1)}=4$.Su

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$2013$

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(C) Given the system of equations:$x+y=a$ $(1)$$\frac{x^2}{x-1}+\frac{y^2}{y-1}=4$ $(2)$From $(2)$,we have $\frac{x^2(y-1)+y^2(x-1)}{(x-1)(y-1)}=4$.Substituting $y=a-x$,the denominator becomes $(x-1)(a-x-1) = -(x-1)^2 + (a-2)(x-1)$.The numerator simplifies as $x^2(a-x-1) + (a-x)^2(x-1) = x^2a - x^3 - x^2 + (a^2 - 2ax + x^2)(x-1) = x^2a - x^3 - x^2 + a^2x - a^2 - 2ax^2 + 2ax + x^3 - x^2 = a^2x - a^2 + 2ax - 2x^2$.Setting the equation equal to $4(x-1)(y-1)$,we simplify to $(a-2)(xy - (a-2)) = 0$.If $a=2$,the equation becomes $\frac{x^2}{x-1} + \frac{(2-x)^2}{1-x} = 4$,which simplifies to $\frac{x^2 - (4-4x+x^2)}{x-1} = 4$,so $\frac{4x-4}{x-1} = 4$,which is $4=4$. This holds for all $x 
eq 1$,meaning infinitely many solutions.If $a 
eq 2$,then $xy = a-2$. Since $x+y=a$,$x$ and $y$ are roots of $t^2 - at + (a-2) = 0$.The discriminant $D = a^2 - 4(a-2) = a^2 - 4a + 8 = (a-2)^2 + 4 > 0$,so there are always two distinct real roots for any $a$.However,we must ensure $x 
eq 1$ and $y 
eq 1$. If $x=1$,then $1+y=a \Rightarrow y=a-1$. Substituting into $xy=a-2$ gives $1(a-1) = a-2$,which is $a-1=a-2$,impossible.Thus,for any $a 
eq 2$,the system has exactly two solutions. Since $a \in [1, 2014]$,there are $2014$ integers total,and excluding $a=2$,there are $2013$ values.

The number of integers $a$ in the interval $[1, 2014]$ for which the system of equations $x+y=a$ and $\frac{x^2}{x-1}+\frac{y^2}{y-1}=4$ has finitely many solutions is

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