Question

The number of elements in the set \(\{(x,y): 2x^2 + 3y^2 = 35, x,y \in \mathbb{Z}\}\), where \(\mathbb{Z}\) is the set of all integers, is

• A. 4
• B. 6
• C. 16
• D. 12
• E. 8

Hint:

In equations like $ax^2 + by^2 = c$, try small integer values systematically.

Answer 1

Answer: (E)

Concept: We need integer solutions of a Diophantine equation.

Step 1: Try possible values of $y$
\[ 2x^2 + 3y^2 = 35 \]

Step 2: Check small integer values
For $y = 0$: $2x^2 = 35$ (not possible) For $y = \pm1$: \[ 2x^2 + 3 = 35 \Rightarrow 2x^2 = 32 \Rightarrow x^2 = 16 \Rightarrow x = \pm4 \] For $y = \pm2$: \[ 2x^2 + 12 = 35 \Rightarrow 2x^2 = 23 \quad (\text{not possible}) \] For $y = \pm3$: \[ 2x^2 + 27 = 35 \Rightarrow 2x^2 = 8 \Rightarrow x^2 = 4 \Rightarrow x = \pm2 \]

Step 3: List all solutions
\[ (x,y) = (\pm4, \pm1), (\pm2, \pm3) \] Total = $4 + 4 = 8$ Final Conclusion:
Option (E)