Question

The number of distinct integer solutions (x, y) of the equation |x + y| + |x - y| = 2, is

Answer 1

We are given the equation:
$|x + y| + |x - y| = 2$

Case 1: $x + y \ge 0$ and $x - y \ge 0$. In this case, the equation becomes:
$(x + y) + (x - y) = 2 \implies 2x = 2 \implies x = 1$
Substitute $x = 1$ into $x + y \ge 0$ and $x - y \ge 0$:
$1 + y \ge 0 \quad \text{and} \quad 1 - y \ge 0$
Solving these inequalities gives:
$y \ge -1 \quad \text{and} \quad y \le 1$
Thus, $y$ can be $-1, 0, 1$, giving 3 solutions for $x = 1$.

Case 2: $x + y \ge 0$ and $x - y \le 0$. In this case, the equation becomes:
$(x + y) + (-x + y) = 2 \implies 2y = 2 \implies y = 1$
Substitute $y = 1$ into $x + y \ge 0$ and $x - y \le 0$:
$x + 1 \ge 0 \quad \text{and} \quad x - 1 \le 0$
Solving these inequalities gives:
$x \ge -1 \quad \text{and} \quad x \le 1$
Thus, $x$ can be $-1, 0, 1$, giving 3 solutions for $y = 1$.

Case 3: $x + y \le 0$ and $x - y \ge 0$. In this case, the equation becomes:
$(-x - y) + (x - y) = 2 \implies -2y = 2 \implies y = -1$
Substitute $y = -1$ into $x + y \le 0$ and $x - y \ge 0$:
$x - 1 \le 0 \quad \text{and} \quad x + 1 \ge 0$
Solving these inequalities gives:
$x \le 1 \quad \text{and} \quad x \ge -1$
Thus, $x$ can be $-1, 0, 1$, giving 3 solutions for $y = -1$.

Case 4: $x + y \le 0$ and $x - y \le 0$. In this case, the equation becomes:
$(-x - y) + (-x + y) = 2 \implies -2x = 2 \implies x = -1$
Substitute $x = -1$ into $x + y \le 0$ and $x - y \le 0$:
$-1 + y \le 0 \quad \text{and} \quad -1 - y \le 0$
Solving these inequalities gives:
$y \le 1 \quad \text{and} \quad y \ge -1$
Thus, $y$ can be $-1, 0, 1$, giving 3 solutions for $x = -1$.

Conclusion: From all four cases, we get a total of $3 + 3 + 3 + 3 = 12$ distinct integer solutions. Therefore, the correct answer is 12.

Answer 2

We are given the equation:
$|x + y| + |x - y| = 2$

Case 1: $x + y \ge 0$ and $x - y \ge 0$. In this case, the equation becomes:
$(x + y) + (x - y) = 2 \implies 2x = 2 \implies x = 1$
Substitute $x = 1$ into $x + y \ge 0$ and $x - y \ge 0$:
$1 + y \ge 0 \quad \text{and} \quad 1 - y \ge 0$
Solving these inequalities gives:
$y \ge -1 \quad \text{and} \quad y \le 1$
Thus, $y$ can be $-1, 0, 1$, giving 3 solutions for $x = 1$.

Case 2: $x + y \ge 0$ and $x - y \le 0$. In this case, the equation becomes:
$(x + y) + (-x + y) = 2 \implies 2y = 2 \implies y = 1$
Substitute $y = 1$ into $x + y \ge 0$ and $x - y \le 0$:
$x + 1 \ge 0 \quad \text{and} \quad x - 1 \le 0$
Solving these inequalities gives:
$x \ge -1 \quad \text{and} \quad x \le 1$
Thus, $x$ can be $-1, 0, 1$, giving 3 solutions for $y = 1$.

Case 3: $x + y \le 0$ and $x - y \ge 0$. In this case, the equation becomes:
$(-x - y) + (x - y) = 2 \implies -2y = 2 \implies y = -1$
Substitute $y = -1$ into $x + y \le 0$ and $x - y \ge 0$:
$x - 1 \le 0 \quad \text{and} \quad x + 1 \ge 0$
Solving these inequalities gives:
$x \le 1 \quad \text{and} \quad x \ge -1$
Thus, $x$ can be $-1, 0, 1$, giving 3 solutions for $y = -1$.

Case 4: $x + y \le 0$ and $x - y \le 0$. In this case, the equation becomes:
$(-x - y) + (-x + y) = 2 \implies -2x = 2 \implies x = -1$
Substitute $x = -1$ into $x + y \le 0$ and $x - y \le 0$:
$-1 + y \le 0 \quad \text{and} \quad -1 - y \le 0$
Solving these inequalities gives:
$y \le 1 \quad \text{and} \quad y \ge -1$
Thus, $y$ can be $-1, 0, 1$, giving 3 solutions for $x = -1$.

Conclusion: From all four cases, we get a total of $3 + 3 + 3 + 3 = 12$ distinct integer solutions. Therefore, the correct answer is 12.