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

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.

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Since we have two absolute value expressions, we need to consider when each expression inside is positive or negative:

$|x + y|$ can be opened as positive ($x+y$), or negative $-(x+y)$

$|x - y|$ can be opened as positive ($x - y$), or negative $-(x-y)$

This creates 4 main cases to check. Let's work through each systematically.

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Case 1: Both Positive $\Rightarrow$ $x + y \geq 0$ and $x - y \geq 0$

Our equation becomes:

$(x + y) + (x - y) = 2$

$2x = 2$

$x = 1$

Now we need to check which values of $y$ satisfy our conditions when $x = 1$:

$x + y \geq 0$ means $1 + y \geq 0$, so $y \geq -1$

$x - y \geq 0$ means $1 - y \geq 0$, so $y \leq 1$

Therefore: $-1 \leq y \leq 1$

Valid integer solutions when $x = 1$: $(1, -1), (1, 0), (1, 1)$

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Case 2: One Positive, One Negative $\Rightarrow$ $x + y \geq 0$ and $x - y < 0$:

$|x + y| = x + y$ and $|x - y| = -(x - y) = y - x$

Our equation becomes:

$(x + y) + (y - x) = 2$

$2y = 2$

$y = 1$

Checking conditions when $y = 1$:

$x + y \geq 0$ means $x + 1 \geq 0$, so $x \geq -1$

$x - y < 0$ means $x - 1 < 0$, so $x < 1$

Therefore: $-1 \leq x < 1$

Integer solutions: $(-1, 1), (0, 1)$

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Case 3: One Positive, One Negative $\Rightarrow x + y < 0$ and $x - y \geq 0$:

$|x + y| = -(x + y)$ and $|x - y| = x - y$

Our equation becomes:

$-(x + y) + (x - y) = 2$

$-2y = 2$

$y = -1$

Checking conditions when $y = -1$:

$x + y < 0$ means $x - 1 < 0$, so $x < 1$

$x - y \geq 0$ means $x + 1 \geq 0$, so $x \geq -1$

Therefore: $-1 \leq x < 1$

Integer solutions: $(-1, -1), (0, -1)$

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Case 4: Both Negative $\Rightarrow x + y < 0$ and $x - y < 0$:

Our equation becomes:

$-(x + y) - (x - y) = 2$

$-2x = 2$

$x = -1$

Checking conditions when $x = -1$:

$x + y < 0$ means $-1 + y < 0$, so $y < 1$

$x - y < 0$ means $-1 - y < 0$, so $y > -1$

Therefore: $-1 < y < 1$

Integer solutions: $(-1, 0)$

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Combining all cases, the complete set of integer solutions is:

$(1, -1), (1, 0), (1, 1), (-1, 1), (0, 1), (-1, -1), (0, -1), (-1, 0)$

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Graphical Approach (risky):

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 $n$!

$|x+y|+|x-y|=n$ represents 4 cases. We have them from above:

Case 1: Both positive

$x=1$

Case 2: One positive, one negative

$y=-1$

Case 3: One positive, one negative

$y=1$

Case 4: Both negative

$x=-1$

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 $8$.

Understanding figures would have gotten you the answer in just 30 seconds!

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 $2$.