Using the modulo approach is much simpler, but as I had taken a different approach...
The difference between any perfect square and any other perfect square is given as k(2n - k), such that k is a integer difference between n, the root of the perfect square, and any integer.
Therefore, if there is a rational root for the quadratic for odd a, b and c, (4ac) must be able to be expressed as k(2b - k), where k is an integer.
As any integer multiplied by an integer composed of the prime factor 2 will be even, both (4ac) and 2b must be even. Any odd integer multiplied by another odd integer will be odd and any odd integer subtracted from an even integer will be odd, therefore, in order for (4ac) to be able to be expressed as k(2b - k), k must be even.
As k must be even, let k = 2x. (2x)(2b - (2x)) can be rewritten as (4x)(b - x)), and therefore (4ac) must be able to equate to (4x)(b - x)). Letting (4ac) = (4x)(b - x)), then dividing each side by 4, results in (ac) = (x)(b - x)).
As (ac) must be odd, (x)(b - x)) must be odd. Thus x must be odd, but as an odd integer minus an odd integer must be an even integer, there is a contradiction. Therefore, in conclusion, no rational root exists for odd integers a, b, and c for the quadratic equation.
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Posted by Dej Mar
on 2010-08-23 04:09:24 |
answer
