The number of elements in the set {(a, b) : 2a^2 + 3b^2 = 35, a, b ϵ Z}, where Z is the set of all integer, is

The number of elements in the set {(a, b) : 2a² + 3b² = 35, a, b ϵ Z}, where Z is the set of all integer, is

1) 2

2) 4

3) 8

4) 12

5) 16

Solution: (3)

2a² + 3b² = 35

3b² = 35 – 2a²

Maximum value on RHS is 35.

3b² ≤ 35

b² ≤ 11.66

b = 0, ± 1, ± 2, ± 3….

3b² = 35 – 2a²

Any number multiplied by 2 gives even number.

3b² is odd.

If b = 0, then 3b² is not odd.

If b = ± 1, then 3b² is odd.

If b = ± 2, then 3b² is even.

If b = ± 3, then 3b² is odd.

3b² = 35 – 2a²

If b = ± 1, then b² = 1,

3 = 35 – 2a²

2a² = 32

a² = 16

a = ± 4

If b = ± 3, then b² = 9

27 = 35 – 2a²

2a² = 8

a² = 4

a = ± 2

Therefore there are 8 elements in total.