Question

Find no. of elements in $S=\left\{\right(a,b\left)\right\}=2a^2+3b^2=35,a,bϵZ$

Answer 1

Basically we have to find no. of solution of the eqn.
$2a^2+3b^2=35$, where a and b are Z
Let us rewrite the equation as
2x + 3y = 35, where x and y are perfect squares
Now, let us list perfect squares less than 35,
$\begin{array}{ccc}Case& x& 2x\\ a& 1& 2\\ b& 4& 8\\ c& 9& 18\\ d& 16& 32\\ e& 25& 50\end{array}$

50 $\to$ This case isn't possible because, overall sum is 35.

Case a: If 2x = 2, 3y = 33
$\Rightarrow$ y = 11, not a perfect square
So not possible

Case b: If 2x = 6, 3y = 27
$\Rightarrow$ y = 5, which is a perfect square
So possible

Case c: If 2x = 18, 3y = 17 or $y=\frac{17}{3}$ which is not an integer
So not possible

Case d: If 2x = 32, 3y = 3
$\Rightarrow$ y = 1, which is a perfect square
So possible

Possible cases: x = 4, y = 9
x = 16, y = 1
$\Rightarrow a=\pm 2$ and $b=\pm 3$
$a=\pm 4$ and $b=\pm 1$
So, $S=\left\{\right(2,3),(2,-3),(-2,3),(-2,-3),(4,1),(4,-1),(-4,1),(-4,-1\left)\right\}$
Total of 8 elements