We need to find all integer pairs $(x, y)$ that satisfy this equation. The key insight is that absolute value expressions create different cases based on whether the expressions inside are positive or negative. $$\newline$$

$$\newline$$ Since we have two absolute value expressions, we need to consider when each expression inside is positive or negative: $$\newline$$ $|x + y|$ can be opened as positive ($x+y$), or negative $-(x+y)$ $$\newline$$ $|x - y|$ can be opened as positive ($x - y$), or negative $-(x-y)$ $$\newline$$ This creates 4 main cases to check. Let's work through each systematically. $$\newline$$

$$\newline$$ Case 1: Both Positive $\Rightarrow$ $x + y \geq 0$ and $x - y \geq 0$ $$\newline$$ Our equation becomes: $$\newline$$ $(x + y) + (x - y) = 2$ $$\newline$$ $2x = 2$ $$\newline$$ $x = 1$ $$\newline$$ Now we need to check which values of $y$ satisfy our conditions when $x = 1$: $$\newline$$ $x + y \geq 0$ means $1 + y \geq 0$, so $y \geq -1$ $$\newline$$ $x - y \geq 0$ means $1 - y \geq 0$, so $y \leq 1$ $$\newline$$ Therefore: $-1 \leq y \leq 1$ $$\newline$$ Valid integer solutions when $x = 1$: $(1, -1), (1, 0), (1, 1)$ $$\newline$$

$$\newline$$ Case 2: One Positive, One Negative $\Rightarrow$ $x + y \geq 0$ and $x - y < 0$: $$\newline$$ $|x + y| = x + y$ and $|x - y| = -(x - y) = y - x$ $$\newline$$ Our equation becomes: $$\newline$$ $(x + y) + (y - x) = 2$ $$\newline$$ $2y = 2$ $$\newline$$ $y = 1$ $$\newline$$ Checking conditions when $y = 1$: $$\newline$$ $x + y \geq 0$ means $x + 1 \geq 0$, so $x \geq -1$ $$\newline$$ $x - y < 0$ means $x - 1 < 0$, so $x < 1$ $$\newline$$ Therefore: $-1 \leq x < 1$ $$\newline$$ Integer solutions: $(-1, 1), (0, 1)$ $$\newline$$

$$\newline$$ Case 3: One Positive, One Negative $\Rightarrow x + y < 0$ and $x - y \geq 0$: $$\newline$$ $|x + y| = -(x + y)$ and $|x - y| = x - y$ $$\newline$$ Our equation becomes: $$\newline$$ $-(x + y) + (x - y) = 2$ $$\newline$$ $-2y = 2$ $$\newline$$ $y = -1$ $$\newline$$ Checking conditions when $y = -1$: $$\newline$$ $x + y < 0$ means $x - 1 < 0$, so $x < 1$ $$\newline$$ $x - y \geq 0$ means $x + 1 \geq 0$, so $x \geq -1$ $$\newline$$ Therefore: $-1 \leq x < 1$ $$\newline$$ Integer solutions: $(-1, -1), (0, -1)$ $$\newline$$

$$\newline$$ Case 4: Both Negative $\Rightarrow x + y < 0$ and $x - y < 0$: $$\newline$$ Our equation becomes: $$\newline$$ $-(x + y) - (x - y) = 2$ $$\newline$$ $-2x = 2$ $$\newline$$ $x = -1$ $$\newline$$ Checking conditions when $x = -1$: $$\newline$$ $x + y < 0$ means $-1 + y < 0$, so $y < 1$ $$\newline$$ $x - y < 0$ means $-1 - y < 0$, so $y > -1$ $$\newline$$ Therefore: $-1 < y < 1$ $$\newline$$ Integer solutions: $(-1, 0)$ $$\newline$$

$$\newline$$ Combining all cases, the complete set of integer solutions is: $$\newline$$ $(1, -1), (1, 0), (1, 1), (-1, 1), (0, 1), (-1, -1), (0, -1), (-1, 0)$ $$\newline$$

$$\newline$$ Graphical Approach (risky): $$\newline$$ Modulus equations of the format $|x+y|+|x-y|=n$ represents a square with the origin being the center (this function would make a square for any $n$! $$\newline$$ $|x+y|+|x-y|=n$ represents 4 cases. We have them from above: $$\newline$$ Case 1: Both positive $\newline$ $x=1$ $$\newline$$ Case 2: One positive, one negative $\newline$ $y=-1$ $$\newline$$ Case 3: One positive, one negative $\newline$ $y=1$ $$\newline$$ Case 4: Both negative $\newline$ $x=-1$ $$\newline$$ These are four lines create a square, any point on this square, is the solution. Since we are only looking at integral solutions, we see that there are $8$. $$\newline$$

$$\newline$$ Understanding figures would have gotten you the answer in just 30 seconds! $$\newline$$ The key insight is that the equation $|x + y| + |x - y| = 2$ geometrically represents points whose sum of distances to the lines $x + y = 0$ and $x - y = 0$ equals $2$.

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