Complex System of Equations

First consider the cases a=0 or b=0.

If a=0, the first equation is x(xy-1)=0, which has 2 solutions: x=0 or xy=1. With x=0, second equation is y=b. With xy=1, second equation is 2y=b. The only way that the equation has only one solution is that b=0 (the solutions are (0,b) and (0,b/2)).

So, there is a pair (0,0), which has only one solution.

When b=0, second equation is y(xy+1)=0. If y=0, then -x=a. If xy=-1, then -2x=a. Again, the only possibility which has a unique solution is a=0.

So, we can consider that a and b are not 0. In this case, x and y are also different from 0. I can write y=k/x. Then the equations are:

kx-x=a $k^2/x+k/x=b$

From the second equation, $x=\frac{k(k+1)}{b}$ , and putting it on the first equation:

(k-1)k(k+1)=ab.

For each k, we have a x and a y that are solutions to the system of equations. But this last equation has 3 different complex solutions, so the system has 3 solutions.

Then, the only pair of parameters that have only one solution is (0,0).

[This solution is essentially identical to Laura's. It explains how I was approaching it.]

Note that the system can be rewritten as $\begin{cases} x(xy-1)=a\\ y(xy+1)=b \end{cases}$ If $xy=t,$ then multiplying the two equations, we obtain $t(t-1)(t+1)=ab,$ which can be rewritten as $t^3-t-ab=0.$ Conversely, if $t$ is a root of this equation, not equal to $1$ or $-1$ , then (with some work) $x=\frac{a}{t-1}$ and $y=\frac{b}{t+1}$ gives a solution to the original system of equations.
Note: Because we manipulated the system of (non-linear) equations, it is not guaranteed that any final solution will work. For example, for a given $t$ , we likely do not have $x = y = \sqrt{t}$ as a solution.

For any $a$ and $b$ , by the Fundamental Theorem of Algebra, the polynomial $t^3-t-ab$ has three roots, if counted with multiplicity. If $ab\neq 0,$ then these roots cannot be $0$ , $1$ or $-1.$ Also, by Vieta's formula, the sum of the roots is zero, so, because $0$ is not a root, the three roots cannot be all the same. Therefore, there are at least two distinct values of $t$ that satisfy the equation $t^3-t-ab=0,$ and, as explained above, each produces a solution of the original system. Note that these solutions cannot be the same because they have a different value of $xy.$ Hence $ab = 0$ , in order for there to be a unique solution.

If $a=0$ and $b\neq 0,$ then both $(0,b)$ and $(\frac{2}{b},\frac{b}{2})$ are solutions of the system. Similarly, if $a\neq 0$ and $b=0$ , then $(-a,0)$ and $(-\frac{a}{2}, \frac{2}{a})$ are solutions.

Finally, if $a=b=0,$ then note that $xy-1$ and $xy+1$ cannot be both zero. So at least one of the variables $x$ and $y$ is $0$ . Therefore $xy=0,$ so the system implies that both $x$ and $y$ are equal to zero.

This shows the system has exactly one solution if and only if $(a,b)=(0,0)$ , implying the answer is $1.$

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This is an example of a map from the affine complex plane to itself which is quasi-finite, but not finite.

Let us assume that $x$ and $y$ are both non-zero. For the first equation, we will treat $y$ as a constant in a quadratic of $x$ and vice-versa for the second. Note that for any $y$ or $x$ , the first and second equations have at least one solution in complex numbers for $x$ and $y$ , respectively. Since we are looking for pairs $a,b$ that yield exactly one solution, this implies that each equation has exactly 1 solution. Using the quadratic formula on both equations, we get

$x=\frac{1\pm \sqrt{1+4ay}}{2y} , y=\frac{-1\pm \sqrt{1+4bx}}{2x}$

Since there is exactly one solution, the discriminants are zero, so ignoring the discriminants we have $x=\frac{1}{2y}$ and $y=\frac{-1}{2x}$ , which is impossible. Thus at least one of $x, y$ must be zero.

Say $x=0$ . It immediately follows from the first equation that $a=0$ . We wish to find constants $b$ such that the system

$x^2y-x=0$ $y^2x+y=b$

has only one solution for $(x,y)$ . Factoring out an $x$ from the first equation, we get $x(xy-1)=0$ . Therefore either $x=0$ or $xy-1=0$ . If $x=0$ , then plugging that in to the second equation yields $y=b$ . If $xy-1=0$ , then we have that $xy^2+y=y(xy+1)=y \cdot 2=b$ . Since there can only be one solution, we can conclude that the two solutions generated are the same, and therefore $b=2y=y \Rightarrow y=0, b=0$ .

Now say $y=0$ , and it follows that $b=0$ . We can proceed similarly to above and get that $a=-2x=-x \Rightarrow x=0, a=0$ . However, we have already counted $a=0, b=0$ , so this is redundant.

We thus find that the only solution is $a=0, b=0$ and therefore the answer is $\boxed{1}$ .

First we multiply the first equation for y and second equation for y: [; \mathbf{ x^{2}y^{2} - xy = ay } ;] and [; \mathbf { x^{2}y^{2} + xy = bx } ;]

Now we add the first equation to second equation and then we subtract the first from the second equation: [; \mathbf {2xy = bx - ay} ;] and [; \mathbf {x^{2}y^{2} = ay + bx} ;]

Get the square of the first equation and set it equal the second equation multiplied by a factor 4: [; \mathbf {b^{2}x^{2} + a^{2}y^{2} - 2abxy = 4ay + 4bx} ;] [; \mathbf {x^{2}b^{2} - 2x \times (aby - 2b) + a^{2}y^{2} - 4ay = 0} ;]

For [; \mathbf {a \neq 0} ;], we find 2 solutions for y. For [; \mathbf {a = 0} ;] and [; \mathbf {b \neq 0} ;], we find 2 solutions for x.

For [; \mathbf {a = b = 0} ;], we find one solution.

Therefore, the complex system have one solution satisfying the predicted conditions.

There are really 3 cases that we must deal with in this problem: (1) $x \neq 0$ and $y \neq 0$ , (2) $x = 0$ and $y \neq 0$ or $x \neq 0$ and $y = 0$ , and (3) $x = 0$ and $y = 0$ .

Case 1: $x \neq 0$ and $y \neq 0$

We can re-write the two original equations as $yx^2-x-a=0$ and $xy^2+y-b=0$ .

Since $x \neq 0$ and $y \neq 0$ and the coefficients of the quadratic equations are complex, the solutions to the equations can be given as $x = \frac{1 \pm \sqrt{1+4ay}}{2y}$ and $y = \frac{-1 \pm \sqrt{1+4bx}}{2x}$ .

For x and y to only have one solution, the discriminants must be 0, so $x=\frac{1}{2y}$ and $y=\frac{-1}{2x}$ , implying that $2xy=1$ and $2xy=-1$ .

But $1 \neq -1$ , so the discriminants cannot be 0. Therefore, when $x \neq 0$ and $y \neq 0$ , there is not only one complex solution to the system.

Case 2: $x = 0$ and $y \neq 0$ or $x \neq 0$ and $y = 0$

When $x = 0$ and $y \neq 0$ , we can substitute x into the original equation $x^2y-x=a$ to get that $a=0$ .

By rearranging $x^2y-x=0$ , we get that $y=\frac{1}{x}$ .

Plugging this into the original equation $xy^2+y=b$ , we get that $b=y$ . This appears to imply that b can be any complex number. Let us call such an arbitrary complex number c.

We can solve the system of equations $x^2y-x=0$ and $xy^2+y=c$ to check if there is only one complex solution. We factor the first equation to become $x(xy-1)=0$ , so $x=0$ or $x=1/y$ .

Substituting into the second equation, we solve for y and we get that either $y=n$ and $x=0$ or $y=\frac{c}{2}$ and $x=\frac{2}{c}$ .

Since there is more than one solution to the system, there is not merely a single complex solution when $x = 0$ and $y \neq 0$ .

Similar steps can be used to show that there is more than one complex solution when $x \neq 0$ and $y = 0$ . It is shown from substituting y into the second original equation that $b=0$ and that $a=c$ , where c is an arbitrary complex number. When solving the system, $x^2y-x=c$ and $xy^2+y=0$ to check for the number of solutions, it is found that the solution can be either $x=n$ and $y=0$ or $x=\frac{c}{2}$ and $y=\frac{2}{c}$ .

Case 3: $x = 0$ and $y = 0$

We can plug in the values for x and y into the two original equations to get that $a=0$ and $b=0$ .

We then solve the system of equations $x^2y-x=0$ and $xy^2+y=0$ , plugging in the values for a and b, to see how many solutions there are.

We can factor each equation to become $x(xy-1)=0$ and $y(xy+1)=0$ .

Now, we see that (0, 0) is the only solution, since $xy=1$ and $y=0$ , $x=$ and $xy=-1$ , and $xy=1$ and $xy=-1$ are impossible.

Therefore, the only possibility where the original system of equations has one complex solution is when $x=0$ and $y=0$ , or when $a=0$ and $b=0$ .

There is one ordered pair (a,b), so the answer to the problem is $\boxed1$