## Task B-2.1.

Determine all integer values of $x$ for which $\left(x^{2}-5 x+5\right)^{\left(x^{2}-4 x-5\right)}=1$.
## Solution.

We distinguish three cases in which the value of the power is 1. The first possibility is that the base of the power is 1, the second is that the exponent of the power is 0, and the base is different from 0, while the third possibility is that the base of the power is -1, and the exponent is of the form $2k, k \in \mathbb{Z}$.

1. case: the base of the power is 1.

Then $x^{2}-5 x+5=1$, or, it holds that $x^{2}-5 x+4=0$.

By solving the equation, we get $x_{1}=1$ and $x_{2}=4$.

For $x=1: \quad 1^{-8}=1$

For $x=4: \quad 1^{-5}=1$

2. case: the exponent of the power is 0, and the base is different from 0.

Then $x^{2}-4 x-5=0$.

By solving the equation, we get $x_{1}=5$ and $x_{2}=-1$.

Check by substitution that for $x_{1}=5$ and $x_{2}=-1$ the base of the power is not 0.

For $x=5: \quad 5^{0}=1$

For $x=-1: \quad 11^{0}=1$

3. case: the base of the power is -1, and the exponent is of the form $2k, k \in \mathbb{Z}$.

First, determine the values of the variable $x$ for which $x^{2}-5 x+5=-1$, and then check whether the exponent of the power is an even number for these values.

By solving the equation $x^{2}-5 x+6=0$, we get $x_{1}=3$ and $x_{2}=2$.

For $x=3: \quad(-1)^{-8}=1$

For $x=2: \quad(-1)^{-9}=-1$

Therefore, the value of the power is 1 for $x=3$, but it is not 1 for $x=2$.

1 point

Finally, we conclude that $x \in\{-1,1,3,4,5\}$.