The sum of the squares of two consecutive integers is 41. The integers are :

1. 4 and -5 or -4 and 5

2. 4 and 5 or -4 and -5

3. 3 and 4 or -4 and -3

4. 6 and 3 or -6 and -3

Let two consecutive integers be x and x + 1.

Given,

Sum of squares of two consecutive integers = 41.

⇒ x² + (x + 1)² = 41

⇒ x² + x² + 1 + 2x = 41

⇒ 2x² + 2x = 41 - 1

⇒ 2x² + 2x = 40

⇒ 2(x² + x) = 40

⇒ x² + x = $\dfrac{40}{2}$

⇒ x² + x = 20

⇒ x² + x - 20 = 0

⇒ x² + 5x - 4x - 20 = 0

⇒ x(x + 5) - 4(x + 5) = 0

⇒ (x - 4)(x + 5) = 0

⇒ x - 4 = 0 or x + 5 = 0

⇒ x = 4 or x = -5

If x = 4,

⇒ x + 1 = 4 + 1 = 5.

If x = -5,

⇒ x + 1 = -5 + 1 = -4.

Numbers = 4 and 5 or -4 and -5.

Hence, Option 2 is the correct option.