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matsci0000 · #8 July 5th, 2009, 11:35 AM

Solution given by prove it(MHF contributor):
We know the following:

$\alpha + \beta = p + 1$

$\alpha\beta = 1$

$P(k + 1) = \alpha^{k + 1} + b^{k + 1} = (\alpha^k + \beta^k)(\alpha + \beta) - \alpha\beta(\alpha^{k - 1} + \beta^{k - 1})$ .

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

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

Try dividing $P(k + 1)$ by $p$ .

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

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

Clearly $p + 1$ can not be divided by $p$ exactly, and you have already established that $\alpha^k + \beta^k$ and $\alpha^{k - 1} + \beta^{k - 1}$ are not divisible by $p$ , so $P(k + 1)$ can not be divided by $p$ .
The problem of the question lies HERE.
It is true that $\alpha^k + \beta^k$ and $\alpha^{k - 1} + \beta^{k - 1}$ are not divisible by $p$
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But you can not say that the difference between $\alpha^k + \beta^k$ and $\alpha^{k - 1} + \beta^{k - 1}$
is not divisible by p.

The above argument given by me can be more clear from the following example-
7 is not divisible by 3 and 4 is also not divisible by 3
But their difference which is equal to 3 is divisible by 3.