## Task 3 - 201243

Given a real number $a \neq 0$ with $|a| \neq 1$.

Determine all real solutions $x$ of the equation

$$
\frac{\left(x^{4}+1\right)\left(x^{4}+6 x^{2}+1\right)}{x^{2}\left(x^{2}-1\right)^{2}}=\frac{\left(a^{4}+1\right)\left(a^{4}+6 a^{2}+1\right)}{a^{2}\left(a^{2}-1\right)^{2}}
$$
A real number $x$ is a solution to the equation

$$
\frac{\left(x^{4}+1\right)\left(x^{4}+6 x^{2}+1\right)}{x^{2}\left(x^{2}-1\right)^{2}}=A \quad \text { with } \quad A=\frac{\left(a^{4}+1\right)\left(a^{4}+6 a^{2}+1\right)}{a^{2}\left(a^{2}-1\right)^{2}}
$$

if and only if $x \neq 0$ and $|x| \neq 1$ and

$$
\begin{gathered}
\left(x^{4}+1\right)\left(x^{4}+6 x^{2}+1\right)=A x^{2}\left(x^{2}-1\right)^{2} \quad \text { thus } \\
x^{8}+(6-A) x^{6}+(2+2 A) x^{4}+(6-A) x^{2}+1=0
\end{gathered}
$$

This equation is equivalent to

$$
x^{4}+\frac{1}{x^{4}}=(6-A)\left(x^{2}+\frac{1}{x^{2}}\right)+2+2 A=0
$$

due to $x \neq 0$. If we set $x^{2}+\frac{1}{x^{2}}=t$, then $x^{4}-t x^{2}+1=0$ (4) and $x^{4}+\frac{1}{x^{4}}=t^{2}-2$. Therefore, $x$ is a solution to (1) if and only if

$$
t^{2}+(6-A) t+2 A=0
$$

From equation (1), it is clear that $x=a$ and $x=-a$ (6) are solutions. If we substitute $x=\frac{1}{a}$ or $x=-\frac{1}{a}$ into the left side of (1), we get a term equal to $A$. Therefore,

$$
x=\frac{1}{a} \quad \text { and } \quad x=-\frac{1}{a}
$$

are also solutions to (1). This implies that the quadratic equation (5) has the solution $t_{1}=a^{2}+\frac{1}{a^{2}}$. By Vieta's formulas, the second solution is

$$
t_{2}=\frac{2 A}{a^{2}+\frac{1}{a^{2}}}=\frac{2\left(a^{4}+6 a^{2}+1\right)}{\left(a^{2}-1\right)^{2}}
$$

Since the quadratic equation (5) has at most two solutions $t_{1}$ and $t_{2}$, and the equation (4) for $t=t_{1}$ has the four solutions $a,-a, \frac{1}{a},-\frac{1}{a}$ and hence no further solutions (due to the conditions, these four solutions are pairwise distinct), all further real solutions $x$ of equation (1) are obtained from the equation

$$
x^{4}-\frac{2\left(a^{4}+6 a^{2}+1\right)}{\left(a^{2}-1\right)^{2}} x^{2}+1=0
$$

This equation has the discriminant

$$
D=\left(\frac{a^{4}+6 a^{2}+1}{a^{4}-2 a^{2}+1}\right)^{2}-1=\frac{16 a^{2}\left(a^{2}+1\right)^{2}}{\left(x^{2}-1\right)^{4}}>0
$$

Therefore, the equation (8) has exactly those $x$ as solutions for which one of the equations

$$
x^{2}=\frac{a^{4}+6 a^{2}+1}{\left(a^{2}-1\right)^{2}} \pm \frac{4 a\left(a^{2}+1\right)}{\left(a^{2}-1\right)^{2}} \quad \text { i.e. } \quad x^{2}=\frac{(a+1)^{4}}{\left(a^{2}-1\right)^{2}} \quad ; \quad x^{2}=\frac{(a-1)^{4}}{\left(a^{2}-1\right)^{2}}
$$

holds. Since the right-hand sides of these equations are positive, the solutions are exactly the numbers

$$
x=\frac{a+1}{a-1} ; \quad x=-\frac{a+1}{a-1} ; \quad x=\frac{a-1}{a+1} ; \quad x=-\frac{a-1}{a+1}
$$

Since $|a| \neq 1$, for each of these four numbers $x \neq 0$. Furthermore, $a+1 \neq a-1$ and since $a \neq 0$, also $a+1 \neq -a+1$, so for each of these four numbers $x \neq 1$ and $x \neq -1$.

Thus, it is proven that the given equation has exactly the solutions listed in (6), (7), and (9), namely the numbers:

$$
a ; \quad -a ; \quad \frac{1}{a} ; \quad -\frac{1}{a} ; \quad \frac{a+1}{a-1} ; \quad -\frac{a+1}{a-1} ; \quad \frac{a-1}{a+1} ; \quad -\frac{a-1}{a+1}
$$

## Adapted from $[3]$