Re: determine the irreducible polynomial
- From: "Per Vognsen" <per.vognsen@xxxxxxxxx>
- Date: 8 Jun 2005 15:55:20 -0700
Per Vognsen wrote:
[snip]
> latter cases are symmetrical.) We have the coefficient equations a+c =
> 0 and b+d+ac = -16 and together they imply -a^2 = -16-b-d. In the first
> case b+d = 4 and so the equation is -a^2 = -12. In the second case b+d
> = 5, so the equation is -a^2 = -11. Neither 11 nor 12 have integer
> square roots so this is not possible.
>
> I hope I didn't make any algebra errors.
Doh, I most certainly did. The two equations at the end should have
been a^2 = 20 and a^2 = 21. In any case the same reasoning still
applies since 20 and 21 don't have integer square roots.
Per
.
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