25. How many distinct pairs $(x, y)$ of real numbers satisfy the equation $(x+y)^{2}=(x+4)(y-4)$ ?
A 0
B 1
C 2
D 3
E 4
Solution: B
Method 1. The most straightforward method is to rewrite the given equation as a quadratic in $x$, whose coefficients involve $y$, and then use the " $b^{2}-4 a c \geq 0$ " condition for a quadratic, $a x^{2}+b x+c=0$, to have real number solutions.

Now , $(x+y)^{2}=(x+4)(y-4) \Leftrightarrow x^{2}+2 x y+y^{2}=x y-4 x+4 y-16$
$$
\Leftrightarrow x^{2}+x y+4 x+y^{2}-4 y+16=0 \Leftrightarrow x^{2}+(y+4) x+\left(y^{2}-4 y+16\right)=0 .
$$

So here $a=1, b=y+4$ and $c=y^{2}-4 y+16$. We see that $b^{2}-4 a c=(y+4)^{2}-4\left(y^{2}-4 y+16\right)$ $=-3 y^{2}+24 y-48=-3\left(y^{2}-8 y+16\right)=-3(y-4)^{2}$. So there is a real number solution for $x$ if and only if $-3(y-4)^{2} \geq 0$. Now, as for all real numbers $y,(y-4)^{2} \geq 0$, it follows that $-3(y-4)^{2} \geq 0 \Leftrightarrow y-4=0 \Leftrightarrow y=4$. When $y=4$ the quadratic equation becomes $x^{2}+8 x+16=0$, that is $(x+4)^{2}=0$, which has just the one solution $x=-4$. So $(-4,4)$ is the only pair of real numbers, $(x, y)$, for which $(x+y)^{2}=(x+4)(y-4)$.

Method 2. We can simplify the algebra by making the substitution $w=x+4$ and $z=y-4$. Then $x+y=(w-4)+(z+4)=w+z$, and the equation becomes $(w+z)^{2}=w z$.
Now, $(w+z)^{2}=w z \Leftrightarrow w^{2}+2 w z+z^{2}=w z \Leftrightarrow w^{2}+w z+z^{2}=0 \Leftrightarrow\left(w+\frac{1}{2} z\right)^{2}+\left(\frac{\sqrt{3}}{2} z\right)^{2}=0$. The sum of the squares of two real numbers is zero if and only if each real number is 0 . So the only real number solution of $\left(w+\frac{1}{2} z\right)^{2}+\left(\frac{\sqrt{3}}{2} z\right)^{2}=0$ is $w+\frac{1}{2} z=\frac{\sqrt{3}}{2} z=0$. This is equivalent to $w=z=0$ and hence to $x=-4$ and $y=4$. So again we deduce that there is just this one solution.