Let's determine the real roots of the following equation:

$$
x^{10} + 36 x^{6} + 13 x^{2} = 13 x^{8} + x^{4} + 36
$$
Reducing our equation to 0

$$
x^{10}-13 x^{8}+36 x^{6}-x^{4}+13 x^{2}-36=0
$$

which can also be written as

$$
x^{4}\left(x^{6}-1\right)-13 x^{2}\left(x^{6}-1\right)+36\left(x^{6}-1\right)=0
$$

or

$$
\left(x^{6}-1\right)\left(x^{4}-13 x^{2}+36\right)=0
$$

Thus, either

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

or

$$
x^{4}-13 x^{2}+36=0
$$

From (1),

$$
x_{1}=1, \quad x_{2}=-1
$$

The expression $x^{4}+x^{2}+1=0$ is greater than 1 for real $x$, as none of its terms can be negative.

From (2),

$$
x_{3,4}^{2}=9, \quad x_{5,6}^{2}=4
$$

and thus,

$$
x_{3}=3, \quad x_{4}=-3, \quad x_{5}=2, \quad x_{6}=-2
$$

Tolnai Tibor (Szombathely, Nagy Lajos g. IV. o. t.)