Let (R = \mathbb{Z}[\sqrt{2}]) and (M = {a + b\sqrt{2} \in R \mid 5 \mid a \text{ and } 5 \mid b}). We first check whether (M) is a subring of (R). For that, we need to verify if (M) is closed under addition, subtraction, and multiplication, and contains (0). 1. Closure under addition: Take (x = a_1 + b_1\sqrt{2} \in M) and (y = a_2 + b_2\sqrt{2} \in M). Then (a_1, a_2, b_1, b_2) are divisible by 5. [ x + y = (a_1 + a_2) + (b_1 + b_2)\sqrt{2} ] Since (5 \mid a_1, a_2) and (5 \mid b_1, b_2), we have (5 \mid (a_1 + a_2)) and (5 \mid (b_1 + b_2)). Therefore (x + y \in M). 2. Closure under subtraction: [ x - y = (a_1 - a_2) + (b_1 - b_2)\sqrt{2} ] Again, (5 \mid (a_1 - a_2)) and (5 \mid (b_1 - b_2)), so (x - y \in M). 3. Closure under multiplication: [ x \cdot y = (a_1 + b_1\sqrt{2})(a_2 + b_2\sqrt{2}) = (a_1 a_2 + 2 b_1 b_2) + (a_1 b_2 + a_2 b_1)\sqrt{2} ] Since (5 \mid a_1, a_2, b_1, b_2), write (a_1 = 5k_1), (a_2 = 5k_2), (b_1 =