4. It is known that there exists a natural number $N$ such that $(\sqrt{3}-1)^{N}=4817152-2781184 \cdot \sqrt{3}$. Find $N$.
Solution. Suppose that raising the number $a+b \sqrt{3}$ to the power $N$ results in the number $A+B \sqrt{3}$ (where $a, b, A, B$ are integers). Expanding the expression $(a+b \sqrt{3})^{N}$, we get a sum of monomials (with non-essential (for us now) integer coefficients) of the form $a^{N-n}(b \sqrt{3})^{n}$. The terms that contribute to the coefficient $B$ are those with an odd exponent $n$. 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 dividing both sides successively by 2 (for example, we can immediately divide each term on the right by 256).

Answer: $N=16$.