Question:

Step 1: Analyze the first equation, $|x| + |y| = 2$. From this equation, it follows that $0 \le |x| \le 2$ and $0 \le |y| \le 2$. Step 2: Analyze the second equation, $\sin\left(\frac{\pi x^2}{4}\right) = 1$. For $\sin(\theta) = 1$, we must have $\theta = (2k + \frac{1}{2})\pi$ for some integer $k$. Thus, $\frac{\pi x^2}{4} = (2k + \frac{1}{2})\pi$. Dividing by $\pi$ and multiplying by 4, we get $x^2 = 4(2k + \frac{1}{2}) = 8k + 2$. Step 3: Determine possible values for $x$. Since $0 \le |x| \le 2$, we have $0 \le x^2 \le 4$. We need to find integer values of $k$ such that $0 \le 8k + 2 \le 4$. If $k=0$, then $x^2 = 8(0) + 2 = 2$. This value satisfies $0 \le 2 \le 4$. If $k=1$, then $x^2 = 8(1) + 2 = 10$. This value does not satisfy $x^2 \le 4$. If $k=-1$, then $x^2 = 8(-1) + 2 = -6$. This value does not satisfy $x^2 \ge 0$. Therefore, the only possible value for $x^2$ is $2$. Step 4: Find the values of $x$ and $y$. From $x^2 = 2$, we have $x = \sqrt{2}$ or $x = -\sqrt{2}$. Substitute $|x| = \sqrt{2}$ into the first equation: $\sqrt{2} + |y| = 2$ $|y| = 2 - \sqrt{2}$ This implies $y = 2 - \sqrt{2}$ or $y = -(2 - \sqrt{2})$. Step 5: List all ordered pairs $(x, y)$. The possible values for $x$ are $\sqrt{2}$ and $-\sqrt{2}$. The possible values for $y$ are $2 - \sqrt{2}$ and $-(2 - \sqrt{2})$. Combining these, we get the following four ordered pairs: $(\sqrt{2}, 2 - \sqrt{2})$ $(\sqrt{2}, -(2 - \sqrt{2}))$ $(-\sqrt{2}, 2 - \sqrt{2})$ $(-\sqrt{2}, -(2 - \sqrt{2}))$ Thus, there are 4 ordered pairs satisfying both equations.