Question

Let p$\ge$ 3 be an integer and $\alpha$ , $\beta$ be the roots of $x^2$ - (p + 1)x + 1 = 0.Using mathematical induction show that ${\alpha }^n+{\beta }^n$
is an integer

Answer 1

Given $p\ge 3$ is an integer and $\alpha ,\beta$ are the roots of
$x^2$ + (p + 1) x + 1) = 0
so that $\alpha +\beta =p+1,\alpha \beta =1$ .....(1)
${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$
= (p + 1)${}^2$ - 2 = p${}^2$ + 2p - 1 .........(2)
Now p is an integer greater than or equal to 3, therefore from (1) and (2) we conclude that P(1), P(2) hold good and assume that P(n) also holds good i.e., they are all integers .
Now multiplying both sides of
x${}^2$ - (p + 1) x + 1 = 0 by x${}^{n-1}$ , we get
x${}^{a+1}$ = (p + 1) x${}^n$ + x${}^{n-1}$
Its roots are $\alpha$ and $\beta$ also
${\alpha }^{n+1}+{\beta }^{n+1}=(p+1)({\alpha }^n+{\beta }^n)+({\alpha }^{n-1}+{\beta }^{n-1})$
Or P(n + 1) = (p + 1) P(n) + P(n - 1) .......(3)
Since P(n) and P(n - 1) hold good, therefore P(n + 1) also holds good
i.e.,${\alpha }^{n+1}+{\beta }^{n+1}$ is an integer
Again $\frac{\alpha +\beta }{p}=\frac{p+1}{p}=1+\frac{1}{p}$
$\frac{{\alpha }^2+{\beta }^2}{p}=\frac{p^2+2p-1}{p}=(p+2)-\frac{1}{p}$
From above we conclude that $\alpha +\beta$ and ${\alpha }^2+{\beta }^2$ are not divisible by p since 1/p is a proper fraction , therefore P (1), P (2) hold good and assume P (n) also holds good i.e.,P (n) is not divisible by p.
$\therefore$ P(n + 1) = (p + 1) P (n) + P (n - 1) from(3)
since each of P (n) and P (n - 1) are not divisible by 3
$\therefore$ P(n + 1) = (p + 1) P (n) + P (n - 1) is also not divisible (3) .
This proves the 2nd part