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

[(4, 1), (4, -1), (-4, 1), (-4, -1), (2, -3), (2, 3), (-2, 3), (-2, -3)]

Therefore there are 8 elements in total.

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