Question

Let $p\ge 3$ be an integer and $\alpha ,\beta$ be the roots of $x^2-(p+1)x+1=0$, then the value of ${\alpha }^n+{\beta }^n$, where $nϵN$

• A. is divisible by 'p'
• B. is an integer
• C. is a rational number
• D. both (b) and (c)

Answer 1

The correct option is D both (b) and (c)

Let consider some value of p = 3 (say), then
$\begin{array}{ccc}& x^2-4x+1=0& \\ and& (\alpha ,\beta )=2\pm \sqrt{3}& (\alpha ,\beta aretheroots)\\ Now,& {\alpha }^n+{\beta }^n=(2+\sqrt{3}{)}^n+(2-\sqrt{3}{)}^n& \dots \left(i\right)\left(nϵN\right)\end{array}$
then ${\alpha }^n+{\beta }^n$ will always be an integer, for the validity of statement you can put $n=1,2,3,\dots$ etc. in Eq. (i).
Similarly for $p=4,5,6,\dots$ etc. we can conclude the same results.
Note:
In this question, the discriminant D is always positive. i.e., $b^2-4ac>0$. So the roots will always be real, unequal and irrational.
But the fact is that ${\alpha }^n+{\beta }^n$, for $nϵN$, always yields integral value. This can be easily proved by mathematical induction method. So if the answer is an integer, then it must be a rational number, hence (d) is correct.