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

The correct answer is 7.

We need to find all integer pairs $(x,a)$ satisfying:

$x^2-2\left|x\right|+|a-2|=0$

Step 1 — Substitute to simplify.
Since $x^2={\left|x\right|}^2$, let $t=\left|x\right|$ where $t\ge 0$. The equation becomes:

$t^2-2t+|a-2|=0$

Rearranging:

$|a-2|=2t-t^2=t(2-t)$

Step 2 — Find valid values of t.
Since the left side $|a-2|\ge 0$, the right side must also be non-negative:

$t(2-t)\ge 0⟹0\le t\le 2$

Since $t=\left|x\right|$ and $x$ must be an integer, the only allowed non-negative integer values are $t\in \{0,1,2\}$.

Step 3 — Enumerate each case.

Case 1: $t=0$ (i.e., $x=0$)
$|a-2|=0(2-0)=0⟹a=2$
Valid pairs: $(0,2)$ → 1 pair

Case 2: $t=1$ (i.e., $x=1$ or $x=-1$)
$|a-2|=1(2-1)=1⟹a-2=\pm 1⟹a=1\text{or}a=3$
Valid pairs:
   For $x=1$: $(1,1)$ and $(1,3)$
   For $x=-1$: $(-1,1)$ and $(-1,3)$
→ 4 pairs

Case 3: $t=2$ (i.e., $x=2$ or $x=-2$)
$|a-2|=2(2-2)=0⟹a=2$
Valid pairs:
   For $x=2$: $(2,2)$
   For $x=-2$: $(-2,2)$
→ 2 pairs

Step 4 — Total count.
All valid integer pairs:

$(0,2)$,  $(1,1)$,  $(1,3)$,  $(-1,1)$,  $(-1,3)$,  $(2,2)$,  $(-2,2)$

Total = $1+4+2=7$

Therefore, the number of ways to choose such a pair of integers is 7.