x2−2|x|+|a−2|=0 |x|=‌ |x|=1±√1−|a−2| ‌ If ‌a>2;|a−2|=a−2 |x|=1±√1−(a−2) =1±√3−a since
x is integer
3−a≥0 a≤3 The possible values of a is
=3 Then
x=±1 If
a=2,|x|=|1±1|,⇒x=±2,0
If
a<2,|a−2|=2−a |x|=1±√1−(2−a) |x|=1±√a−1 Since
x is integer
a−1≥0⇒a≥1 ∴ The possible values of
a is 1
If
a=1,|x|=1⇒x=±1 ∴ The possible pairs
=(−1,3),(1,3),(1,1),(−1,1),(2,2),(−2,2),(0,2)
i.e., 7