In how many ways can a pair of integers $(x, a)$ be chosen such that $x^2 - 2 | x | + | a - 2 | = 0$?

We need to find all integer pairs $(x, a)$ that satisfy: $x^2 - 2|x| + |a-2| = 0$

Since we have absolute value expressions, we need to consider different cases based on the signs of $x$ and $(a-2)$.

---

When dealing with absolute values:

• $|x| = x$ when $x \geq 0$

• $|x| = -x$ when $x < 0$

• $|a-2| = a-2$ when $a \geq 2$

• $|a-2| = -(a-2) = 2-a$ when $a < 2$

This gives us 4 cases to analyze.

---

For $x \geq 0$ and $a \geq 2$:

$|x| = x$ and $|a-2| = a-2$

Our equation becomes: $x^2 - 2x + (a-2) = 0$

For a quadratic $ax^2 + bx + c = 0$ to have real solutions, its discriminant $D = b^2 - 4ac$ must be non-negative.

Here: $D = (-2)^2 - 4(1)(a-2) = 4 - 4(a-2) = 4 - 4a + 8 = 12 - 4a$

For real solutions: $12 - 4a \geq 0$

$a \leq 3$

Since $a \geq 2$ and $a \leq 3$, we have $a \in \{2, 3\}$.

When $a = 2$: $x^2 - 2x + 0 = 0$

$x(x-2) = 0$

$x = 0, 2$

Valid pairs: $(0, 2)$ and $(2, 2)$

When $a = 3$: $x^2 - 2x + 1 = 0$

$(x-1)^2 = 0$

$x = 1$

Valid pair: $(1, 3)$

---

For $x \geq 0$ and $a < 2$:

$|x| = x$ and $|a-2| = 2-a$

Our equation becomes: $x^2 - 2x + (2-a) = 0$

Discriminant: $D = 4 - 4(2-a) = 4 - 8 + 4a = 4a - 4$

For real solutions: $4a - 4 \geq 0$

$a \geq 1$

Since $a < 2$ and $a \geq 1$, we have $a = 1$.

When $a = 1$: $x^2 - 2x + 1 = 0$

$(x-1)^2 = 0$

$x = 1$

Valid pair: $(1, 1)$

---

For $x < 0$ and $a \geq 2$:

$|x| = -x$ and $|a-2| = a-2$

Our equation becomes: $x^2 - 2(-x) + (a-2) = 0$

$x^2 + 2x + (a-2) = 0$

Discriminant: $D = 4 - 4(a-2) = 12 - 4a$

For real solutions: $12 - 4a \geq 0$

$a \leq 3$

Since $a \geq 2$ and $a \leq 3$, we have $a \in \{2, 3\}$.

When $a = 2$: $x^2 + 2x + 0 = 0$

$x(x+2) = 0$

$x = 0, -2$

Since we need $x < 0$, only $x = -2$ is valid

Valid pair: $(-2, 2)$

When $a = 3$: $x^2 + 2x + 1 = 0$

$(x+1)^2 = 0$

$x = -1$

Valid pair: $(-1, 3)$

---

For $x < 0$ and $a < 2$:

$|x| = -x$ and $|a-2| = 2-a$

Our equation becomes: $x^2 + 2x + (2-a) = 0$

Discriminant: $D = 4 - 4(2-a) = 4a - 4$

For real solutions: $4a - 4 \geq 0$

$a \geq 1$

Since $a < 2$ and $a \geq 1$, we have $a = 1$.

When $a = 1$: $x^2 + 2x + 1 = 0$

$(x+1)^2 = 0$

$x = -1$

Valid pair: $(-1, 1)$

---

All valid integer pairs $(x, a)$:

$(0, 2)$

$(2, 2)$

$(1, 3)$

$(1, 1)$

$(-2, 2)$

$(-1, 3)$

$(-1, 1)$

Total number of ways = 7

---

When solving equations with absolute values, we always:

Identify the critical points where expressions inside absolute values change sign

Create cases based on these critical points

Solve each case separately

Verify that our solutions satisfy the original conditions for each case