Official solution
x2
- x - 1 = 0
\[ \therefore \] \[\alpha \]2
- \[\alpha \] - 1 = 0
\[ \Rightarrow \] \[\alpha \]2 = \[\alpha \] + 1
\[ \therefore \] \[\alpha \]3 = \[\alpha \]2 + \[\alpha \]
= \[\alpha \] + 1 + \[\alpha \]
= 2\[\alpha \] + 1
Now \[\alpha \]4 = 2\[\alpha \]2 + \[\alpha \]
= 2(\[\alpha \] + 1) + \[\alpha \]
= 3\[\alpha \] + 2
Now \[\alpha \]5 = 3\[\alpha \]2 + 2\[\alpha \]
= 3(\[\alpha \] + 1) + 2\[\alpha \]
= 5\[\alpha \] + 3
Given pk
= \[{\left( \alpha \right)^k} + {\left( \beta \right)^k}\]
\[ \therefore \] p5 = \[{\left( \alpha \right)^5} + {\left( \beta \right)^5}\]
= 5\[\alpha \] + 3 + 5\[\beta \] + 3
= 5(\[\alpha \] + \[\beta \]) + 6
= 5(1) + 6 [As \[\alpha \] + \[\beta \] = 1]
= 11
Now p2 \cdot p3
= (\[\alpha \]2 + \[\beta \]2).(\[\alpha \]3 + \[\beta \]3)
= ( \[\alpha \] + 1 + \[\beta \] + 1)(2\[\alpha \] + 1 + 2\[\beta \] + 1)
= ( \[\alpha \] + \[\beta \] + 2)(2(\[\alpha \] + \[\beta \]) + 2)
= (1 + 2)(2 + 2) = 12
\[ \therefore \] p5 \[ \ne \] p2 \cdot p3
So option (D) is wrong.