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If I have a cubic equation $x^3 + ax^2 + bx + c = 0$, what constraints exist on $a,b,c$ when we have three integer solutions?

How do I choose $a,b,c$ to force integer solutions?

  • I think you should be able to recognize them using Vieta's formula for cubic equations, which states that if a cubic equation $x^3 + ax^2 + bx + c = 0$ has three different roots $x_1,x_2,x_3$, then:

    $$\left{\begin{eqnarray} -a &=& x_1 + x_2 + x_3 \ b &=& x_1x_2 + x_1x_3 + x_2x_3 \ -c &=& x_1x_2x_3 \end{eqnarray}\right..$$

     – Hakim Aug 01 '14 at 14:45
  • Let $x_i$ be the roots, so $x^3+ax^2+bx+c = (x-x_1)(x-x_2)(x-x_3) = x^3-(x_1+x_2+x_3)x^2+(x_1x_2+x_2x_3+x_3x_3)x-x_1x_2x_3$ now you can compare the coefficients and see what you get. The most important property for finding roots 'by hand' is using that $x_i|c$. – flawr Aug 01 '14 at 14:45
  • Are you asking, for example, given $b, c$ how to choose $a$ so that the roots are integers? – Mark Bennet Aug 01 '14 at 14:54

2 Answers2

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John Joy
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Eutherpy
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Hint:

Use the Rational Root Theorem.

user167432
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