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

$$
(x-5)^{4}+(x-4)^{4}=97
$$
I. solution: Let us introduce the unknown $z=x-4.5$, thus from our equation we obtain the following equation

$$
\left(z-\frac{1}{2}\right)^{4}+\left(z+\frac{1}{2}\right)^{4}=97
$$

The advantage of this transformation is that if we perform the exponentiation in the fourth power, the coefficients of the even powers of $z$ will be the same in both, while the coefficients of the odd powers will only differ in sign:

$$
z^{4}-2 z^{3}+\frac{3}{2} z^{2}-\frac{1}{2} z+\frac{1}{16}+z^{4}+2 z^{3}+\frac{3}{2} z^{2}+\frac{1}{2} z+\frac{1}{16}=97
$$

Rearranging and removing the fractions:

$$
16 z^{4}+24 z^{2}-775=0
$$

Thus we get a quadratic equation in $z^{2}$. When solving the equation, only the positive root provides real solutions, thus

$$
z^{2}=\frac{25}{4}
$$

from which

$$
\begin{aligned}
& z_{1}=\frac{5}{2}=2.5 \\
& z_{2}=-\frac{5}{2}=-2.5
\end{aligned}
$$

Returning to the original unknown $x=z+4.5$

$$
x_{1}=7, \quad x_{2}=2
$$

Substituting into the equation shows that the obtained roots indeed satisfy the equation.

Gereben Ildikó (Bp. XXI., Jedlik Ányos g. IV. o. t.)

Remark: If we need to find the roots of an equation of the form

$$
(a x-b)^{4}+(a x-c)^{4}=k
$$

we can transform it similarly with $z=a x-\frac{b+c}{2}$, making the equation quadratic in $z^{2}$. This is ensured by the remark we made about the coefficients of the odd powers in the solution.

II. solution: If we reduce the equation to zero, bring it to polynomial form, and simplify by 2, we get the following equation:

$$
x^{4}-18 x^{3}+123 x^{2}-378 x+392=0
$$

If the equation has a rational root, it can only be an integer (since the coefficient of $x^{4}$ is 1), and it can only be one of the divisors of $392=2^{3} \cdot 7^{2}$, i.e., $\pm 1, \pm 2, \pm 4, \pm 7, \ldots$.

By substitution, we find that the numbers $x_{1}=2$ and $x_{2}=7$ are indeed roots.

Dividing the equation by $(x-2)(x-7)=x^{2}-9 x+14$, we get that the further roots can only be the roots of the equation

$$
x^{2}-9 x+28=0
$$

Since the discriminant of this equation is negative, our equation has no more real solutions.

Biborka Tamás (Makó, József A. g. I. o. t.)

III. solution: The number on the right side of the equation can easily be broken down into the sum of two fourth powers:

$$
97=( \pm 2)^{4}+( \pm 3)^{4}
$$

We need to find such an $x$ (if there is one at all) for which one of the numbers $x-5$ and $x-4$ is +2 or -2, and the other is +3 or -3. By trial and error, we can quickly find that the numbers $x_{1}=2$ and $x_{2}=7$ meet the requirement.

Dividing by the factors corresponding to the found roots, the resulting quadratic equation does not provide further real roots, as we have already seen in the previous solution.

Kisvölcsey Jenó (Bp. VIII., Piarista g. III. o. t.)

IV. solution: As we have already seen, our equation can be brought to the following form:

$$
x^{4}-18 x^{3}+123 x^{2}-378 x+392=0
$$

Adding 49 to both sides, we can convince ourselves by calculation that the left side of the resulting equation is a perfect square:

$$
\left(x^{2}-9 x+21\right)^{2}=49
$$

This is true if

$$
x^{2}-9 x+21=7
$$

or if

$$
x^{2}-9 x+21=-7
$$

From the first one

$$
x_{1}=2, \quad x_{2}=7
$$

However, we have already seen that the quadratic equation $x^{2}-9 x+28=0$ has no real roots.

Gémesi Gabriella (Bp. XVI. (Cinkota), Ált. lg. III. o. t.)