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simplependulum · #4 July 5th, 2009, 12:23 AM

Quote:

Originally Posted by matsci0000

Let p greater than or equal to 3 be an integer.Alpha and beta are the roots of x^2-(p+1)x+1=0.Using mathematical induction ,prove that alpha^n+beta^n
1)is an integer
2)is not divisible by p

We have

$\alpha+ \beta = p+1$
and
$\alpha\beta = 1$

$\alpha^{k+2} + \beta^{k+2} = (\alpha + \beta)(\alpha^{k+1} + \beta^{k+1}) - \alpha \beta(\alpha^{k} + \beta^{k})$

$\alpha^{k+2} + \beta^{k+2} = (p+1)(\alpha^{k+1} + \beta^{k+1}) - (\alpha^{k} + \beta^{k})$

$S_1$ and $S_2$ are true , now , assume $S_{k}$ and $S_{k+1} ~$ are also true .

Obviously , refering to the identity , $S_{k+2}~~$ are true

For part 2 , assume $S_{k-1} , S_{k} , S_{k+1}$ are true.

$\alpha^{k+2} + \beta^{k+2} = (p+1)(\alpha^{k+1} + \beta^{k+1}) - (\alpha^{k} + \beta^{k})$

To prove $S_{k+2}~$ is also true , we have to prove $A_{k+1} - A_{k}=/= 0mod(p)$ Where $A_k = \alpha^{k} + \beta^{k}$

$A_{k+1} - A_{k}= \alpha^{k+1} + \beta^{k+1} - \alpha^{k} - \beta^{k}$

$= \alpha^{k}(\alpha-1) + \beta^{k}(\beta - 1)$
$= \alpha^{k}(p-\beta) + \beta^{k}(p-\alpha)$
$= p(\alpha^{k} + \beta^{k}) - (\alpha \beta)( \alpha^{k-1} + \beta^{k-1})$
$= p(integer) + (\alpha^{k-1} + \beta^{k-1}) =/= 0mod(p)$