Find integers $a$ in $[1,2014]$ to get solutions in integers of $x+y=a, \frac{x^2}{x-1}+\frac{y^2}{y-1}=4$

Question

We are to find the number of integers $a$ in the interval $[1,2014]$ for which the system of equations $x+y=a, \frac{x^2}{x-1}+\frac{y^2}{y-1}=4$ will have finite number of solutions in integers.

This is what I have tried so far: since $(x,x-1)=1$ so $(x^2, x-1)=1$. As we see $\frac{x^2}{x-1}+\frac{y^2}{y-1}$ is an integer 4, so it follows that $$\left(\frac{x^2}{x-1}, \frac{y^2}{y-1}\right)\in \{(1,3),(3,1),(2,2),(0,4),(4,0)\}.$$

After this, I am lost. Not sure, if my approach is in correct path. Please help.

Answer 1

Substitute $x=a-y$ to get $$ \frac{(a+2)y^2-(a^2+2a)y+a^2}{y^2-ay+a-1}=4,$$ or after rearrangement, $$\tag1 (a-2)y^2+(2a-a^2)y+(a^2-4a+4)=0.$$ If $a=2$, this beconmes "$0=0$", i.e., every $y$ (integer and $\ne 1$, of course) leads to a solution. Hence for $a=2$, we get infinitely many solutions.

If $a\ne 2$, $(1)$ becomes $$\tag2y^2-ay+(a-2)=0 $$ and for this to have one (and then necessarily two) integer solutions, the discriminant $a^2-4(a-2)=a^2-4a+8$ must be a perfect square, $a^2-4a+8=d^2$. But then $$ (d+a-2)(d-a+2)=d^2-(a-2)^2=4$$ so that we write $4$ as product of two integers with non-zero even difference $2a-4$, which is impossible.

Summary. If $a=2$, there are infinitely many integer solutions $x\in\Bbb Z\setminus\{1\}$, $y=2-x$. And if $a\ne 2$, there are no integer solutions.

Remark. If we plug $a=2$ into $2$, we get the solutions $y=2$ (and $x=0$) and $y=0$ (and $x=2$); these are of course already among the previously found solutions for $a=2$.

Answer 2

An easy trick is to make change of variable $x_1 = x -1$ and $y_1 = y -1$. Then the equations become $x_1 + y_1 = a-2$ and $\frac{(x_1+1)^2}{x_1} + \frac{(y_1+1)^2}{y_1} = 4$. On simplifying we get $(x_1 + y_1)(1 + \frac{1}{x_1y_1}) = 0$. Hence either $x_1 + y_1 = a -2 = 0$ or $x_1 y_1 = -1$. For value $a=2$ we get infinitely many solutions by taking $x1 $ to be any nonzero integer and putting $y_1 = -x_1 $. if $x_1 y_1 = 1$ then $x_1 = 1$ and $y_1 = -1$ or vise a versa, in both cases $a=2$. Hence there is no integer $a$ for which there are finitely many integral solutions.