If a ∈ R and the equation (x - [x])² + 2(x - [x]) + a² = 0 (where [x] denotes the greatest integer ≤ x) has no integral solution but has real solutions, then all possible values of a lie in the interval:

Let {x} = x - [x], where [x] denotes the greatest integer less than or equal to x. Then {x} represents the fractional part of x, and 0 ≤ {x} < 1. The given equation becomes
{x}² + 2{x} + a² = 0
This is a quadratic equation in {x}. Since it has real solutions, the discriminant must be non-negative:
(2)² - 4(1)(a²) ≥ 0
4 - 4a² ≥ 0
1 - a² ≥ 0
a² ≤ 1
-1 ≤ a ≤ 1
The solutions for {x} are given by the quadratic formula:
{x} = (-2 ± √(4 - 4a²)) / 2 = -1 ± √(1 - a²)
Since 0 ≤ {x} < 1, we must have 0 ≤ -1 ± √(1 - a²) < 1.
If we consider -1 + √(1 - a²), then
0 ≤ -1 + √(1 - a²) < 1
1 ≤ √(1 - a²) < 2
1 ≤ 1 - a² < 4
-3 ≤ -a² < 0
0 < a² ≤ 3
If we consider -1 - √(1 - a²), then this will always be negative, so it is not in the range [0, 1). Hence we only consider the first case.
Since the equation has no integral solutions, {x} cannot be 0 or 1. Therefore, we must exclude the values of a that give {x} = 0 or {x} = 1. That is, we need to find values of a such that
-1 + √(1-a²) ≠ 0 and -1 + √(1-a²) ≠ 1
If -1 + √(1 - a²) = 0, then √(1 - a²) = 1, which means 1 - a² = 1, so a = 0.
If -1 + √(1 - a²) = 1, then √(1 - a²) = 2, which means 1 - a² = 4, so a² = -3 (no real solution).
Thus, a cannot be 0. Combining this with 0 < a² ≤ 3, we get 0 < a² ≤ 1. This means -1 < a < 1 and a ≠ 0. Therefore, the interval for a is (-1, 0) ∪ (0, 1).