Question

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

• A. $2$
• B. $6$
• C. $4$
• D. $5$

Answer 1

Answer: (B)

$A = \{(a, b) : a^2 + 3b^2 = 28, a, b \in Z\}$
$(a, b)$ can be $(1, 3)$, $(-1, 3)$, $(1, -3)$, $(-1, -3)$, $(5, 1)$
$(-5, 1)$, $(5, -1)$, $(-5, -1)$, $(4, 2)$, $(-4, 2)$, $(4, -2)$, $(-4, -2)$
$n(A)= 12$, $n(B) =\infty$
$A \cap B = \{(a, b) : a^2 + 3b^2 = 28$ and $a > b, a, b \in Z\} = 6$