Let $$S = \{a + b\sqrt{2} : a, b \in \mathbb{Z}\}$$, $$T_1 = \{(-1 + \sqrt{2})^n : n \in \mathbb{N}\}$$ and $$T_2 = \{(1 + \sqrt{2})^n : n \in \mathbb{N}\}$$. Then which of the following statements is (are) TRUE?

The set $$S$$ consists of all numbers of the form $$a+b\sqrt{2}$$ with $$a,b\in\mathbb{Z}$$, that is $$S=\mathbb{Z}[\sqrt{2}]$$, the smallest ring containing $$\mathbb{Z}$$ and $$\sqrt{2}$$.

Option A: $$\mathbb{Z}\cup T_1\cup T_2\subset S$$

• Every integer $$m\in\mathbb{Z}$$ can be written as $$m+0\sqrt{2}\in S$$.
• For $$T_1=\{(-1+\sqrt{2})^n:n\in\mathbb{N}\}$$ use the binomial theorem: $$(-1+\sqrt{2})^n=\sum_{k=0}^{n}\binom{n}{k}(-1)^{\,n-k}(\sqrt{2})^{\,k}$$ Separate even and odd $$k$$: $$=(-1)^n\!\!\sum_{\substack{k\text{ even}}}\binom{n}{k}2^{k/2}+(-1)^{n-1}\!\!\sum_{\substack{k\text{ odd}}}\binom{n}{k}2^{(k-1)/2}\sqrt{2}$$ The two sums are integers, so the whole expression is of the form $$a+b\sqrt{2}$$ with $$a,b\in\mathbb{Z}$$. Hence every element of $$T_1$$ lies in $$S$$.
• The same reasoning with the sign changed shows that every element of $$T_2=\{(1+\sqrt{2})^n:n\in\mathbb{N}\}$$ also belongs to $$S$$.
Therefore $$\mathbb{Z}\cup T_1\cup T_2\subset S$$ is TRUE.

Option B: $$T_1\cap\left(0,\dfrac1{2024}\right)=\varnothing$$

Put $$\alpha=\sqrt{2}-1\approx0.4142$$. Then $$(-1+\sqrt{2})^n=\alpha^{\,n}$$ because $$\alpha=(-1+\sqrt{2})$$. Since $$0\lt \alpha\lt 1$$, the sequence $$\alpha^{\,n}$$ decreases to $$0$$. Choose $$n$$ so that $$\alpha^{\,n}\lt\dfrac1{2024}$$. Solve: $$n\gt\dfrac{\ln(2024)}{-\ln\alpha}\approx\dfrac{7.61}{0.8814}\approx8.64$$ Taking $$n=9$$ gives $$\alpha^{\,9}\approx0.0032\lt\dfrac1{2024}\approx0.000494$$ false, let’s test $$n=13$$ gives $$\alpha^{\,13}\approx0.00020\lt0.000494$$. Hence an $$n$$ exists with $$\alpha^{\,n}\in\left(0,\dfrac1{2024}\right)$$, so the intersection is NOT empty. Therefore Option B is FALSE.

Option C: $$T_2\cap(2024,\infty)\neq\varnothing$$

Put $$\beta=1+\sqrt{2}\approx2.4142$$. The sequence $$\beta^{\,n}$$ is strictly increasing and unbounded. Estimate $$\beta^{\,9}=\beta^{\,8}\beta\,. $$ Now $$\beta^{\,4}\approx34.0$$, so $$\beta^{\,8}\approx34^{\,2}\approx1156$$. Then $$\beta^{\,9}\approx1156\times2.414\approx2790\gt2024$$. Thus $$\beta^{\,9}\in T_2$$ and $$\beta^{\,9}\gt2024$$, proving the intersection is non-empty. Option C is TRUE.

Option D: For $$a,b\in\mathbb{Z}$$, $$\cos\bigl(\pi(a+b\sqrt{2})\bigr)+i\sin\bigl(\pi(a+b\sqrt{2})\bigr)\in\mathbb{Z}$$ iff $$b=0$$.

Write the complex number in exponential form: $$e^{\,i\pi(a+b\sqrt{2})}=e^{\,i\pi a}\,e^{\,i\pi b\sqrt{2}}=(-1)^{a}\,e^{\,i\pi b\sqrt{2}}.$$ Its magnitude is $$1$$, so the only possible integer values are $$1$$ and $$-1$$. This requires $$e^{\,i\pi b\sqrt{2}}=\pm1\;$$, i.e. $$\pi b\sqrt{2}=k\pi$$ for some $$k\in\mathbb{Z}$$. Hence $$b\sqrt{2}=k$$. Because $$\sqrt{2}$$ is irrational, the only integer multiple of $$\sqrt{2}$$ that equals an integer is $$0$$, forcing $$b=0$$. Conversely, if $$b=0$$, the expression reduces to $$\cos(\pi a)+i\sin(\pi a)=(-1)^{a}\in\mathbb{Z}$$. Therefore the bi-implication holds and Option D is TRUE.

Correct statements: Option A, Option C, Option D.