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I have been trying to solve a problem where OP seeks Integer roots to cubic equation. There are similar problems ( see 1, 2, 3, 4, 5, 6, 7) in this forum.

To solve the general cubic$$x^3+ax^2+bx+c=0\tag{i}$$, Remove the $ax^2$ term by substituting $x=\dfrac {y-a}3$. Let the transformed equation be$$y^3+py+q=0\tag{ii}$$,we get a root as

$$y=\left(-\frac q2+\sqrt{\frac {q^2}4+\frac {p^3}{27}}\right)^{\frac 13}+\left(-\frac q2-\sqrt{\frac {q^2}4+\frac {p^3}{27}}\right)^{\frac 13}\tag{iii}$$

If the the root is integer then the surds $\sqrt{\frac {q^2}4+\frac {p^3}{27}}$, and $\left(-\frac q2-\sqrt{\frac {q^2}4+\frac {p^3}{27}}\right)^{\frac 13}$ has to be integer. If $\sqrt{\frac {q^2}4+\frac {p^3}{27}}$ is integer we can write, $\sqrt{\frac {q^2}4+\frac {p^3}{27}}=c \implies \frac {q^2}4+\frac {p^3}{27}=c^2$. Now, from this post we know, that if $p/3=n$ and $q=n(n-1)$ then $\sqrt{\frac {q^2}4+\frac {p^3}{27}}$ is an integer, then we get-

$$y=n^{\frac13}-(n^2)^{\frac13}$$

If we substitute $n=n_1^3$, then we get, $y=n_1-n_1^2$, which is a negative number.

Is it a correct solution? Is the integer solution of a cubic always negative? This seems not to be true, but then what am I missing?

If this solution is correct but gives some integer solution, then , how we get all the integer solutions of cubic putting constrain on $p,q$ like $p/3=n$ and $q=n(n-1)$? Note, $p, q$ are nonzero.

Servaes
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Consider Non-Trivial Cases
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  • The integer solution of a cubic is not always negative. For example ${{x}^{3}}-2{{x}^{2}}-2 x+4=0$ has one integer root and it is positive – Henry Mar 24 '20 at 16:37

2 Answers2

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  1. "If the root is integer, then the surds... has to be integer" This is not correct. The irrational parts could cancel out, leaving an integer. Try $p=0$, $q=-1$.

  2. The case $p/3 = n$ and $q = n(n-1)$ considered in the other post is not representative of all cubics, only that particular family of cubics. So I don't see how you can draw the conclusion "integer solution of a cubic [is] always negative."

Ted
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  • $p=0$, $q=-1$ seems to be trivial case, consider nonzero $p, q$, can give us an example then? – Consider Non-Trivial Cases Mar 24 '20 at 16:45
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    If $p=3n,q=n(n-1)$ gives an integer for the square root, then do does $p=3n,q=-n(n-1)$; and the sign change on $1$ reverses the signs of all roots. – Oscar Lanzi Mar 24 '20 at 18:07
  • @Andrew Take any 3 integers a,b,c that sum to 0, let $p = ab+bc+ac$ and $q = -abc$. Then the roots of the cubic will be the integers $a$, $b$, $c$, and with almost any case you try, the $\sqrt{q^2/4 + p^3/27}$ will not be an integer. – Ted Mar 25 '20 at 16:10
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It is clearly untrue as the cubic equation $x^3=0$ has a nonnegative integral root. In general, for any three integers $a$, $b$ and $c$ with $c\geq0$ the cubic equation $$(x^2+ax+b)(x-c)=0,$$ has a nonnegative integral root. As noted in Andrews answer, your argument fails when you claim that the surds must be integers; the two algebraic numbers $$\left\{-\frac q2+\sqrt{\frac {q^2}4+\frac {p^3}{27}}\right\}^{\frac 13} \qquad\text{ and }\qquad\left\{-\frac q2-\sqrt{\frac {q^2}4+\frac {p^3}{27}}\right\}^{\frac 13},$$ can very well both be irrational and still sum to an integer.

To find integral roots of rational cubic equations, the rational root theorem is very effective in determining whether an integral solution exists. As far as I know, there is no algebraic invariant of the cubic that determines whether the cubic has an integral root.

Servaes
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