Mathematical induction :

Posted June 11th, 2009 by alpha

Analysis questions

Let $p\ge3$ be an integer. $\alpha$ and $\beta$ are the roots of the equation $x^2-(p+1)x+1=0$ .Using mathematical induction prove that $\alpha^n +\beta^n$ is - (1)an integer
2)not divisible by p

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I've solved the first part

On June 12th, 2009 alpha says:

I've solved the first part of the question.Please solve the second part.

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Second part -- please take a look at my procedure

On June 13th, 2009 alpha says:

Let P(n) be the statement which is to be proved.

$\alpha+\beta=p+1$ ........(1) , $\alpha.\beta=1$ ........(2)
From eqn (1)
$\alpha+\beta$ is not divisible by p.

STEP1-
P(1) is true.
$\alpha+\beta$ is not divisible by p.

To prove that P(2) is true.
$\alpha^2+\beta^2 =(\alpha+\beta)^2 -2\alpha\beta$
= $(\alpha+\beta)^2 -2$ ----------This is not divisible by p.

STEP2 --- { INDUCTION ASSUMPTION }
Let P(k) and P(k-1) be true.
$\alpha^k+\beta^k$ is not divisible by p. .............. (3)

$\alpha^(k-1)+\beta^(k-1)$ is not divisible by p............(4)

STEP3-
To prove that P(k+1) is true.
To prove that $\alpha^(k+1)+\beta^(k+1)$ is not divisible by p.
$\alpha^(k+1)+\beta^(k+1)$ = $(\alpha^k+\beta^k )(\alpha+\beta)-\alpha^k \beta-\beta\alpha^k$
= $(\alpha^k+\beta^k )(\alpha+\beta)-\alpha\beta{\alpha^(k-1)+\beta^(k-1)}$
$\alpha.\beta=1$ , so
= $(\alpha^k+\beta^k )(\alpha+\beta)-{\alpha^(k-1)+\beta^(k-1)}$ is not divisible by p [from (3)]
---> $(\alpha^k+\beta^k )(\alpha+\beta)$ is not divisible by p [from (3)]
and ${\alpha^(k-1)+\beta^(k-1)}$ is not divisible by p [from (4)]

But it isn't necessary that the difference between two integers which aren't divisible by p , should not br divisible by p. HERE LIES THE PROBLEM.
Please help in in proving $(\alpha^k+\beta^k )(\alpha+\beta)-{\alpha^(k-1)+\beta^(k-1)}$ isn't divisible by p.

````Maths````
000
(((())))
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