Question

If p and q are prime numbers satisfying the condition \[{{\text{p}}^2} - 2{{\text{q}}^2} = 1\], then the value of \[{{\text{p}}^2} + 2{{\text{q}}^2}\] is
A. 5
B.15
C.16
D.17

Answer 1

Hint: Here we will use the hit and trial method in the given equation to get the values of p and q and the substitute those values in the desired equation.
Hit and trial method means to substitute random values of the variable which satisfy the given equation.

Complete step-by-step answer:
Here we are given the following equation:
\[{{\text{p}}^2} - 2{{\text{q}}^2} = 1\]
Also, p and q are prime numbers.
We will use the hit and trial method in the above equation to get the values of p and q.
So, Let \[{\text{p}} = 3\]and \[{\text{q}} = 2\]then,
Considering the left hand side of the equation :
\[{\text{LHS = }}{{\text{p}}^2} - 2{{\text{q}}^2}\]
Now, putting the values of p and q in their respective places we get:
\[
  {\text{LHS = }}{3^2} - 2\left( {{2^2}} \right) \\
  {\text{LHS = }}9 - 2\left( 4 \right) \\
  {\text{LHS = }}9 - 8 \\
  {\text{LHS = }}1 \\
 \]
Now considering the right hand side of the equation we get:
\[{\text{RHS = }}1\]
Therefore,
 \[
  {\text{LHS = RHS = }}1 \\
  \therefore {\text{LHS = RHS}} \\
 \]
This implies that the values \[{\text{p}} = 3\]and \[{\text{q}} = 2\] satisfy the given equation.
Now putting these values in \[{{\text{p}}^2} + 2{{\text{q}}^2}\] we get:
\[
  {{\text{p}}^2} + 2{{\text{q}}^2} = {3^2} + 2\left( {{2^2}} \right) \\
  {{\text{p}}^2} + 2{{\text{q}}^2} = 9 + 2\left( 4 \right) \\
  {{\text{p}}^2} + 2{{\text{q}}^2} = 9 + 8 \\
  {{\text{p}}^2} + 2{{\text{q}}^2} = 17 \\
 \]
Hence the desired value of \[{{\text{p}}^2} + 2{{\text{q}}^2} = 17\]
Therefore, Option(D) is the required answer.

Note: If a quadratic equation in two variables is given with no other prior information, always use hit and trial method to get the values.