2. 

(a) Write the natural number 12 as the difference of squares of non-zero natural numbers.

(b) Show that the number $A=24 \cdot 2^{2 n-1}+12 \cdot 2^{2 n+2}+6 \cdot 2^{2 n-3}, n \geqslant 2, n \in \mathbb{N}$ can be written as the sum of two differences of squares of non-zero natural numbers.
## Solution.

(a) $12=4 \cdot 3=4 \cdot(4-1)=4^{2}-4=4^{2}-2^{2}$. (2p)

(b)

$$
\begin{aligned}
24 \cdot 2^{2 n-1} & =3 \cdot 2^{3} \cdot 2^{2 n-1}=(4-1) \cdot 2^{2 n+2}=2^{2} \cdot 2^{2 n+2}-2^{2 n+2} \\
& =\left(2^{n+2}\right)^{2}-\left(2^{n+1}\right)^{2} \\
12 \cdot 2^{2 n+2} & =3 \cdot 2^{2} \cdot 2^{2 n+2}=(4-1) \cdot 2^{2 n+4}=2^{2} \cdot 2^{2 n+4}-2^{2 n+4} \\
& =\left(2^{n+3}\right)^{2}-\left(2^{n+2}\right)^{2} \\
6 \cdot 2^{2 n-3} & =3 \cdot 2 \cdot 2^{2 n-3}=(4-1) \cdot 2^{2 n-2}=\left(2^{n}\right)^{2}-\left(2^{n-1}\right)^{2}
\end{aligned}
$$

It follows that $A=\left(2^{n+3}\right)^{2}-\left(2^{n+1}\right)^{2}+\left(2^{n}\right)^{2}-\left(2^{n-1}\right)^{2}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_b494d2c46493426c59c1g-2.jpg?height=45&width=1585&top_left_y=1985&top_left_x=247)
$A_{1} A_{3}=2 A_{1} A_{2}, A_{1} A_{4}=2 A_{1} A_{3}, A_{1} A_{5}=2 A_{1} A_{4}, \ldots, A_{1} A_{2015}=2 A_{1} A_{2014}$.

A straight road, passing over the bridge in city $A_{2014}$, connects two cities $B$ and $C$, located on opposite sides of the river, equally distant from the bridge.

(a) If the distance between the towns $A_{1}$ and $A_{2}$ is $2 \mathrm{~km}$, find the distance between $A_{4}$ and $A_{8}$.

(b) Compare the distance between $B$ and $A_{1}$ with the distance between $C$ and $A_{2015}$.

## Solution.

(a)

$$
\begin{aligned}
& A_{1} A_{2}=2 \mathrm{~km} \\
& A_{1} A_{3}=2 A_{1} A_{2}=2 \cdot 2 \mathrm{~km}=4 \mathrm{~km} \\
& A_{1} A_{4}=2 A_{1} A_{3}=2 \cdot 4 \mathrm{~km}=8 \mathrm{~km} \\
& A_{1} A_{5}=2 A_{1} A_{4}=2 \cdot 8 \mathrm{~km}=16 \mathrm{~km} \\
& A_{1} A_{6}=32 \mathrm{~km}, \quad A_{1} A_{7}=64 \mathrm{~km}, \quad A_{1} A_{8}=128 \mathrm{~km} .(\mathbf{2 p})
\end{aligned}
$$

Then, $A_{4} A_{8}=A_{1} A_{8}-A_{1} A_{4}=128 \mathrm{~km}-8 \mathrm{~km}=120 \mathrm{~km}$. (1p)

![](https://cdn.mathpix.com/cropped/2024_06_07_b494d2c46493426c59c1g-3.jpg?height=69&width=1587&top_left_y=365&top_left_x=243)

![](https://cdn.mathpix.com/cropped/2024_06_07_b494d2c46493426c59c1g-3.jpg?height=57&width=1245&top_left_y=417&top_left_x=243)