Solve the following equation:

$$
\sqrt[n]{(x+1)^{2}}+\sqrt[n]{(x-1)^{2}}=4 \sqrt[n]{x^{2}-1}
$$

Examine the cases for $n=2,3,4$ separately.
Let

$$
\sqrt[n]{x+1}=u, \quad \sqrt[n]{x-1}=v ; \quad \text { then } \quad \sqrt[n]{x^{2}-1}=u v
$$

and divide the thus transformed equation by $v^{2}$ (which would only be possible for $v=0$ if $x=1$, but this is not a root of the equation):

$$
u^{2}+v^{2}=4 u v ; \quad\left(\frac{u}{v}\right)^{2}-4 \frac{u}{v}+1=0
$$

From here, immediately substituting the expressions for $x$

$$
\frac{u}{v}=\sqrt[n]{\frac{x+1}{x-1}}=2 \pm \sqrt{3}, \quad \text { finally } x_{1,2}^{(n)}=\frac{(2 \pm \sqrt{3})^{n}+1}{(2 \pm \sqrt{3})^{n}-1}
$$

Our formula for $n=2,3$ and 4 gives:

$$
\begin{aligned}
& x^{(2)}=\frac{(2 \pm \sqrt{3})^{2}+1}{(2 \pm \sqrt{3})^{2}-1}=\frac{8 \pm 4 \sqrt{3}}{6 \pm 4 \sqrt{3}}=\frac{(4 \pm 2 \sqrt{3})(3 \mp 2 \sqrt{3})}{(3 \pm 2 \sqrt{3})(3 \mp 2 \sqrt{3})}=\frac{\mp 2 \sqrt{3}}{9-12}= \pm \frac{2}{\sqrt{3}} \\
& x^{(3)}=\frac{(2 \pm \sqrt{3})^{3}+1}{(2 \pm \sqrt{3})^{3}-1}=\frac{27 \pm 15 \sqrt{3}}{25 \pm 15 \sqrt{3}} \cdot \frac{5 \mp 3 \sqrt{3}}{5 \mp 3 \sqrt{3}}=\frac{\mp 6 \sqrt{3}}{-10}= \pm \frac{3 \sqrt{3}}{5} \\
& x^{(4)}=\frac{(7 \pm 4 \sqrt{3})^{2}+1}{(7 \pm 4 \sqrt{3})^{2}-1}=\frac{98 \pm 56 \sqrt{3}}{96 \pm 56 \sqrt{3}} \cdot \frac{12 \mp 7 \sqrt{3}}{12 \mp 7 \sqrt{3}}= \pm \frac{7}{4 \sqrt{3}}
\end{aligned}
$$

and in decimal approximate fractions:

$$
x^{(2)} \approx \pm 1.155, \quad x^{(3)} \approx \pm 1.039, \quad x^{(4)} \approx \pm 1.010
$$

Leporisz Gyögy (Budapest, Piarista g. II. o. t.)

Remarks. 1. The two roots are always each other's negatives, indeed

$$
\begin{aligned}
x_{1}^{(n)} & =\frac{(2+\sqrt{3})^{n}+1}{(2+\sqrt{3})^{n}-1}=\frac{\left[(2+\sqrt{3})^{n}+1\right]\left[(2-\sqrt{3})^{n}-1\right]}{\left[(2+\sqrt{3})^{n}-1\right]\left[(2-\sqrt{3})^{n}-1\right]}= \\
& =\frac{(2-\sqrt{3})^{n}-(2+\sqrt{3})^{n}}{2-\left[(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}\right]}=\frac{(2+\sqrt{3})^{n}-(2-\sqrt{3})^{n}}{\left[(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}\right]-2} \\
x_{2}^{(n)} & =\frac{\left[(2-\sqrt{3})^{n}+1\right]\left[(2+\sqrt{3})^{n}-1\right]}{\left[(2-\sqrt{3})^{n}-1\right]\left[(2+\sqrt{3})^{n}-1\right]}=\frac{(2+\sqrt{3})^{n}-(2-\sqrt{3})^{n}}{2-\left[(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}\right]}=-x_{1}^{(n)}
\end{aligned}
$$

This can also be seen from the fact that the two roots of the equation obtained for $u / v$ are reciprocals of each other, and substituting $-x$ for $x$ in $u / v$ results in its reciprocal. (The numerator and denominator of $x_{1}^{(n)}$ are positive.) It is also easily seen that when the powers are expanded, only integer terms remain in the denominator, and the terms remaining in the numerator are integer multiples of $\sqrt{3}$.

2. The nth roots are always real. Indeed, under the square root on the left side of (1), a non-negative number stands for every (real) $x$, and with the found roots, $|x_{1,2}|>1$ always holds, so the number under the right-hand side square root is positive. Indeed,

$$
x_{1}^{(n)}=\frac{(2+\sqrt{3})^{n}+1}{(2+\sqrt{3})^{n}-1}=1+\frac{2}{(2+\sqrt{3})^{n}-1}>1
$$

and thus $x_{2}^{(n)}=-x_{1}^{(n)}<-1$.[^0]


[^0]:    ${ }^{1}$ The superscript indicates that we are talking about the roots of different equations depending on the $n$ root exponent.