7. It is known that there exists a natural number $N$ such that

$$
(\sqrt{3}-1)^{N}=4817152-2781184 \cdot \sqrt{3}
$$

Find $N$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c0af7857c77436db1b60g-3.jpg?height=534&width=419&top_left_y=755&top_left_x=1481)
Solution. Suppose that raising the number $a+b \sqrt{3}$ to the power $N$ results in the number $A+B \sqrt{3}$ (here $a, b, A, B$ are integers). Expanding the expression $(a+b \sqrt{3})^{N}$, we get a sum of monomials (with integer coefficients that are not important to us now) of the form $a^{N-n}(b \sqrt{3})^{n}$. The coefficient $B$ will be contributed by those monomials where the exponent $n$ is odd. Therefore, if $(a+b \sqrt{3})^{N}=A+B \sqrt{3}$, then $(a-b \sqrt{3})^{N}=A-B \sqrt{3}$. Multiplying the equations $(\sqrt{3}-1)^{N}=4817152-2781184 \cdot \sqrt{3}$ and $(-\sqrt{3}-1)^{N}=4817152+2781184 \cdot \sqrt{3}$, we get $(-2)^{N}=4817152^{2}-3 \cdot 2781184^{2}$. The exponent $N$ can be found by successively dividing both sides by 2 (for example, each term on the right can be immediately divided by 256).

Answer: $N=16$.