Question

The number of elements in the set $\left\{\left(a,b\right):2a^2+3b^2=35,a,b\in Z\right\}$, where $Z$ is the set of all integers, is

• A. $2$
• B. $4$
• C. $8$
• D. $12$
• E. $16$

Answer 1

The correct option is C

$8$

Explanation for the correct option:

Step 1: Calculate the possible value of $b$

The given set is,

$\left\{\left(a,b\right):2a^2+3b^2=35,a,b\in Z\right\}$

We get,

$2a^2+3b^2=35$

$\Rightarrow$ $3b^2=35-2a^2$

The maximum value of the right-hand side is $35$.

So, $3b^2\le 35$

$\Rightarrow$ $b^2\le \frac{35}{3}$

$\Rightarrow$ $b^2\le 11.66$

$\Rightarrow$ $b=0,\pm 1,\pm 2,\pm 3$

Step $2$: Find the value of $b$ where $3b^2$ is odd

Any number multiplied by $2$ gives an even number.

We know that the sum of an odd number and an even number is an odd number.

Since $2a^2$ is an even number. Then $3b^2$ must be an odd number.

So, the value of $3b^2$ is odd.

If $b=0$, then $3b^2$ is even.

If $b=\pm 1$, then $3b^2$ is odd.

If $b=\pm 2$, then $3b^2$ is even.

If $b=\pm 3$, then $3b^2$ is odd.

Hence, the possible values are $b=\pm 1$ and $b=\pm 3$.

Step $3$: Calculating the value of $a$

If $b=\pm 1$, then $b^2=1$

$3b^2=35-2a^2$

$\Rightarrow$$3\left(1\right)=35-2a^2$

$\Rightarrow$ $2a^2=32$

$\Rightarrow$ $a^2=16$

$\Rightarrow$ $a=\pm 4$

If $b=\pm 3$, then $b^2=9$

$3b^2=35-2a^2$

$\Rightarrow$$3\left(9\right)=35-2a^2$

$\Rightarrow$ $2a^2=8$

$\Rightarrow$ $a^2=4$

$\Rightarrow$ $a=\pm 2$

Therefore, the elements of the given set are:

$\left\{\left(2,3\right),\left(2,-3\right),\left(-2,3\right),\left(-2,-3\right),\left(4,1\right),\left(4,-1\right),\left(-4,1\right),\left(-4,-1\right)\right\}$

There are total $8$ elements present in the set.

Hence, the correct option is $C)8$