Task B-1.7. For which integers $p$ does the equation

$$
\frac{1}{(x-4)^{2}}-\frac{p-1}{16-x^{2}}=\frac{p}{(x+4)^{2}}
$$

have a unique integer solution?
## Solution.

$$
\frac{1}{(x-4)^{2}}-\frac{p-1}{16-x^{2}}=\frac{p}{(x+4)^{2}}
$$

Given that $x \neq -4$ and $x \neq 4$, after multiplying the given equation by $(x+4)^{2}(x-4)^{2}$, we get

$$
\begin{aligned}
(x+4)^{2}+(p-1)(x+4)(x-4) & =p(x-4)^{2} \\
x^{2}+8 x+16+(p-1)\left(x^{2}-16\right) & =p\left(x^{2}-8 x+16\right) \\
x^{2}+8 x+16+p x^{2}-16 p-x^{2}+16 & =p x^{2}-8 p x+16 p \\
8 x+32-16 p & =-8 p x+16 p \\
8 x(p+1) & =32 p-32 \\
x(p+1) & =4 p-4
\end{aligned}
$$

(1 point)

For $p \neq -1$, we have $x=\frac{4 p-4}{p+1}$.

$$
x=\frac{4 p-4}{p+1}=\frac{4 p+4-8}{p+1}=4-\frac{8}{p+1}
$$

For $x$ to be an integer, $p+1$ must divide the number 8. Therefore,

$p+1 \in \{\pm 1, \pm 2, \pm 4, \pm 8\}$, which means

(1 point)

$p \in \{-9, -5, -3, -2, 0, 1, 3, 7\}$.

(1 point)

Due to the initial condition $x \neq -4, x \neq 4$, $p$ cannot be equal to 0.

(1 point)

Finally, we get $p \in \{-9, -5, -3, -2, 1, 3, 7\}$.

(1 point)

## ZUPANIA COMPETITION IN MATHEMATICS

## 2nd grade - high school - B variant

## February 15, 2013.

IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.