Let $Z$ denote the set of all integers and $A = \{(a,b):a^{2} + 3b^{2} = 28,\ \ \ \ \ \ \ \ \ \ \ a,\ b \in Z\}$ and $B = \{(a,\ b):a b,\ a,\ b \in Z\}$. Then, the number of elements in $A \cap B$ is

Question

Let $Z$ denote the set of all integers and $A = \{(a,b):a^{2} + 3b^{2} = 28,\ \ \ \ \ \ \ \ \ \ \ a,\ b \in Z\}$ and $B = \{(a,\ b):a > b,\ a,\ b \in Z\}$. Then, the number of elements in $A \cap B$ is

Dr. Abubacker · Accepted Answer

(d)
$\because A = \{(a,\ b):a^{2} + 3b^{2} = 28,\ a,\ b \in Z\}$
$=${(5, 1), (-5, -1), (5, -1), (-5, 1), (1, 3), (-1, -3), (-1, 3),
(1, -3), (4, 2), (-4, -2), (4, -2), (-4, 2)}
And $B = \{(a,\ b):a > b,\ a,\ b\  \in Z\}$
$	herefore\ \ \ \ \ \ \ \ \ A \cap B = \left\{ ( - 1,\  - 5),\ (1,\  - 5),\ ( - 1,\  - 3),\ (1,\  - 3),\ (4,\ 2),\ (4,\  - 2) ight\}$
$	herefore$ Number of elements in $A \cap B$ is 6.

Q. Let $Z$ denote the set of all integers and $A = \{(a,b):a^{2} + 3b^{2} = 28,\ \ \ \ \ \ \ \ \ \ \ a,\ b \in Z\}$ and $B = \{(a,\ b):a > b,\ a,\ b \in Z\}$. Then, the number of elements in $A \cap B$ is

Explanation

Reviewed by Dr. Abubacker

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