I have a problem with this exercise:
Prove, that in an equation of the form
$x^{3} - 1996x^{2} + rx + 1995 = 0$
where $r$ is a real coefficient, there is no more than one integer root.
I tried to document myself on the Internet. I found some methods.
- the first one:
Cubic equation has one (triple) or two (double and simple) roots, when disriminant is 0, with these formulas:
$D=\frac{q^{2}}{4}+\frac{p^{3}}{27}$
$q=c+\frac{2a^{3}-9ab}{27}$
$p=b-\frac{a^{2}}{3}$
I tried to follow these formulas, but the result was, that there are more than two roots.
- the second method
If both $\Delta$ and $\Delta_{0}$ are equal to 0, then the equation has a single root (which is a triple root):
$\displaystyle -{\frac {b}{3a}}{\text{}}$
$\displaystyle \Delta =18abcd-4b^{3}d+b^{2}c^{2}-4ac^{3}-27a^{2}d^{2}$
$\displaystyle \Delta _{0}=b^{2}-3ac{\text{}}$
I tried to follow these formulas, but they were too difficult for me to solve. Because when I substituted the first formula I got another cubic equation moreover the coefficients were too large.
Can you help me to choose the best method? Or do you know any easier method?
Thank you for your time!
If your polynomial has two integer roots, then the third root is also an integer (since the sum of the roots is $1996$). In particular, $r$ is also an integer, since it is the sum of all products of two roots.
The rational root theorem now implies that every root is a divisor of $1995.$ But there is no way to write $1996$ as the sum of three divisors of $1995$, since all divisors of $1995$ are odd.