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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!

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    $\begingroup$ A cubic equation has one or three (edit: distinct) roots $\endgroup$ – basket Oct 15 '16 at 17:24
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    $\begingroup$ @basket a double- or triple-root is also possible. $\endgroup$ – Peter Oct 15 '16 at 17:25
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    $\begingroup$ If the cubic had more than one integer root, all roots would be integers, and since $1995$ is squarefree, they would be distinct. So, we need to rule out this possibility. $\endgroup$ – Peter Oct 15 '16 at 17:40
  • $\begingroup$ @Peter I hope you do understand, that the equation has exactly one root. $\endgroup$ – martina Oct 15 '16 at 17:45
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    $\begingroup$ @martina I understand this. I look for a contradiction. If you mean my first comment : I just corrected basket's general statement. $\endgroup$ – Peter Oct 15 '16 at 17:46
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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.

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  • $\begingroup$ Thank you very much! so, just for sure, if i understand it good: this equation cannot have one integer root. Moreover the coefficient $r$ has to be integer in order to the roots are integers. And when coefficient is real than the roots are also real and they are not the same at all. Is it right? $\endgroup$ – martina Oct 15 '16 at 18:35
  • $\begingroup$ @martina The equation can have exactly one integer root. It can't have two or three. Whether they are distinct doesn't matter here $\endgroup$ – user378953 Oct 15 '16 at 18:51
  • $\begingroup$ but if the equation has exactly one root, the root has to be even as well as odd at the same time. let´s say, that our root is $m$, than according to vieta´s formulas $3m=1996$ and $m^{3}=1995$ so from the first I see, that m has to be even and from the second that m has to be odd. $\endgroup$ – martina Oct 15 '16 at 19:02
  • $\begingroup$ @martina OK. I am considering that "three integer roots" that happen to coincide. Or three integer roots counting multiplicity. $\endgroup$ – user378953 Oct 15 '16 at 19:15

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