# $a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$

Q. If $$a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$$, prove that $$a = b = c = 0$$ ($$a$$, $$b$$, $$c$$ and $$p$$ are rational and $$p$$ is not a perfect cube.)

My approach:

$$p^\frac{1}{3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}$$

Case 1: If the $$b^2 - 4ac$$ is a perfect square, I get the LHS as irrational and the RHS as rational, which is a contradiction.

Case 2: If $$b^2 - 4ac$$ is not a perfect square, $$b = \pm \sqrt{b^2 - 4ac} - 2cp^\frac{1}{3}$$

Here, the LHS is rational and the RHS is irrational, contradiction again. (Edit: The answer of @GNUSupporter has the proper proof.)

So the equation is not quadratic and $$c = 0$$.

$$a + bp^\frac{1}{3} = 0$$

$$-\dfrac{a}{b} = p^\frac{1}{3}$$

This is a contradiction and hence $$b = 0$$ and $$a = 0$$

Is there any other way to solve this?

• You mean $p^{1/3} = \cdots.$
– mjw
Dec 21 '20 at 15:46
• Why is $b = ± \sqrt{b^2 - 4ac} - 2cp^⅔$ a contradiction? Dec 21 '20 at 15:48
• There is, a priori, no immediate reason to conclude that the RHS is irrational. It is a difference between irrational numbers, the result could be rational. Dec 21 '20 at 15:50
• $\sqrt 2-\sqrt 2$, for one. Any constructed example is going to look similarly trivial, unless I go out of my way to obfuscate it (like with $\sqrt{3+2\sqrt2}-\sqrt2=1$). But they prove that it can be done. Dec 21 '20 at 15:57
• But can you find $x^{1/2} - y^{1/3} = 0$ (where x and y are not perfect squares and cubes respectively)?
– Shub
Dec 21 '20 at 16:05

I have somewhat weird way to see this. Consider the system \begin{align*} a + b p^{1/3} + c p^{2/3} & = 0\\ cp + a p^{1/3} + b p^{2/3} & = 0\\ bp + cp p^{1/3} + a p^{2/3} & = 0 \end{align*} or $$\begin{pmatrix} a & b & c \\ cp & a & b \\ bp & cp & a \end{pmatrix} \begin{pmatrix} 1 \\ p^{1/3} \\ p^{2/3} \end{pmatrix} =0.$$

So the coefficient matrix has zero determinant, i.e. $$a^3 + b(b^2-3ac)p + c^3 p^2=0.$$ Now we can proceed with infinite descent.

(The essence is until here, and below is just some calculations.)

Note that we can assume that $$p$$ is an integer; If $$a + b (n/d)^{1/3} + c (n/d)^{2/3}=0$$ then $$ad^2 + bd\cdot d^{2/3} n^{1/3} + c d^{4/3}n^{2/3}=0,$$ i.e. $$ad^2 + bd\cdot (d^2n)^{1/3} + c (d^2n)^{2/3}=0$$ so we are reduced to the integer $$p$$ case.

Let $$q$$ be a prime factor or $$p$$. One can assume that $$q^3 \not\mid p$$; in this case $$q$$ factor is absorbed into coefficients $$b$$ and $$c$$. Assume $$(a, b, c)$$ is a nontrivial integer solution. We have two cases;

• Case 1: Let $$p = qN$$ with $$q\not\mid N$$. Then $$a^3 + b(b^2-3ac)qN + c^3 q^2N^2=0,$$ i.e. $$a = qA$$. Then $$q^2A^3 + b(b^2-3qAc)N + c^3 qN^2=0,$$ i.e. $$b = qB$$. Then again $$qA^3 + B(q B^2-3Ac)qN + c^3 N^2=0,$$ i.e. $$q|c$$, i.e. $$c = qC$$, and $$A^3 + B(B^2-3AC)qN + C^3 q^2 N^2 = A^3 + B(B^2-3AC)p + C^3 p^2 =0.$$ Thus, if $$(a, b, c)$$ is an integer solution then $$(a/q, b/q, c/q)$$ also is an integer solution; this descent cannot be done infinitely since $$a, b, c$$ are finite, i.e. a contradiction.

• Case 2 : Let $$p = q^2 N$$ with $$q\not\mid N$$. Then $$a^3 + b(b^2-3ac)q^2 N + c^3 q^4 N^2=0.$$ One can assert $$a = q A$$, then $$q A^3 + b(b^2-3qAc) N + q^2c^3 N^2=0,\quad \mathbf{(**)}$$ i.e. $$b = qB$$. Thus $$A^3 + B(q B^2-3Ac)q N + q c^3 N^2=0,$$ i.e. $$A$$ can be divided by $$q$$ once again. Let $$A = qA'$$ to have $$q^2 A'^3 + B( B^2-3A'c)q N + c^3 N^2=0,$$ i.e. $$c = qC$$, $$q A'^3 + B( B^2-3qA'C) N + q^2 C^3 N^2=0 \quad \mathbf{(**)}$$ Compare two equations marked by (**); If $$(A, b, c)$$ satisfies $$q A^3 + b(b^2-3qAc) N + q^2c^3 N^2=0$$ then we have another integer solution $$(A/q, b/q, c/q)$$. So, again by infinite descent, there is no such $$(a, b, c)$$.

This method also works for $$p^{1/4}$$ case.

I think I have never seen the matrix of the form $$\begin{pmatrix} a & b & c \\ cp & a & b \\ bp & cp & a \end{pmatrix}$$ or its variants. Are there any reference?

• I have never seen it either, +1 for the nice method! Dec 22 '20 at 3:13
• Consider $p^{1/3} = \alpha$ and $K = \mathbb{Q}(\alpha)$ a field extension, which is a vector space over $\mathbb{Q}$ generated by $1, \alpha, \alpha^2$. Then any $t = a + b \alpha + c \alpha^2$ generates a map $m_t \colon K \to K$; $m_t(x) = tx$. This map is linear as a vector space map, and the matrix above is essentially $m_{a + b \alpha + c \alpha^2}$ with appropriate basis choice for $K$. Aug 29 '21 at 15:13
• What i have done last year is just finding contradiction under assumption of $N_{K/\mathbb{Q}}=0$; This is somewhat clear since any $\sigma \colon K\hookrightarrow \mathbb{C}$ doesn't map $x\ne0$ into $0$; but the procedure assumes that the minimal polynomial of $\alpha$ is cubic polynomial, so concrete calculation above is somewhat needed. Aug 29 '21 at 15:14
• Thanks for your explanation. It is some kind of a miracle that I understood everything you've said now, because I would not have understood it had you responded on Dec 22,2020! Yes, it's a great idea all round, something that I saw and used in other examples as well. Aug 29 '21 at 16:23

There's a typo in the first step. @mjw caught it. Note that you're applying the quadratic formula on $$p^{1/3}$$, so $$p^{1/3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}, \text{ if }c \ne 0.$$ Another missing link in your proof is your lack of argument about the claimed irrationality of $$± \sqrt{b^2 - 4ac} - 2cp^{2/3}$$.

As you handled the degenerate case "$$c = 0$$ and $$b \ne 0$$" well, we'll keep the assumption "$$c \ne 0$$ or $$b = 0$$" for the rest of the proof. Also we assume that $$\sqrt{b^2-4ac}$$ is irrational. Multiply $$2c$$ on both side of the above equality, then cube it.

\begin{align} 8c^3p =& -b^3 \pm 3b^2 \sqrt{b^2-4ac} - 3b(b^2-4ac) \pm (b^2-4ac)^{3/2} \\ =& -4b^3+12abc \pm 4(b^2-ac) \sqrt{b^2-4ac} \\ 2c^3p =& -b^3 + 3abc \pm (b^2-ac) \sqrt{b^2-4ac} \end{align}

Make $$\sqrt{b^2-4ac}$$ the subject of the above equality. If $$b^2-ac \ne 0$$,

$$\sqrt{b^2-4ac} = \pm \frac{2c^3p + b^3 - 3abc}{b^2-ac} \in \mathbb{Q},$$ contradicting our assumption on the irrationality of $$\sqrt{b^2-4ac}$$ if $$b^2 - ac \ne 0$$.

Assume that $$b^2 - ac = 0$$. Then $$b^2 - 4ac = b^2 = -3b^2$$, and $$p^{1/3} = \frac{(-b\pm\sqrt3|b|i)}{2c} = \begin{cases} \frac{b}{c} e^{2\pi i/3} \text{ or } \frac{b}{c} e^{4\pi i/3} \text{ if } b > 0 \\ \frac{b}{c} e^{\pi i/3} \text{ or } \frac{b}{c} e^{5\pi i/3} \text{ if } b < 0 \end{cases}.$$

In this case, if $$b \ne 0$$, we don't have the desired conclusion, since $$p = \left(\dfrac{b}{c}\right)^3$$ might not be an integer, so it's not a perfect cube.

If $$b = 0$$, we use the assumption $$b^2 = ac$$, we've $$a = 0$$ or $$c = 0$$.

• If $$a = 0$$, only the term $$cp^{2/3} = 0$$ is left in the original equation, but $$p \ne 0$$ as $$p$$ can't be a perfect cube, so $$c = 0$$.
• If $$c = 0$$, the original equation becomes $$a = 0$$. Done.
• Thank you for proving the irrationality of $± \sqrt{b² - 4ac} - 2cp^{2/3}$ :)
– Shub
Dec 22 '20 at 3:51
• @Shub Thx for comment. Oops using Wolfram Alpha to verify an instance of the last case $4p^{2/3}-2p^{1/3}+1=0$, I found that $p^{1/3} = \frac12 e^{\pi i/3}$ or its conjugate, which gives $p = \frac18$. Dec 22 '20 at 8:30
• If $b^2 - ac = 0$ then $$b^2 - 4ac<0$$ $$p^\frac{1}{3} \in \mathbb{C}$$ $$a + bp^\frac{1}{3} + cp^\frac{2}{3} \in \mathbb{C}$$, which is not possible.
– Shub
Dec 22 '20 at 8:57
• @Shub My previous numerical example with $a = 1$, $b = -2$, $c = 4$ and $p^{1/3} = \frac12 e^{\pi i/3}$ shows that $a+bp^{1/3}+cp^{2/3} \in \mathbb{R} \subseteq \mathbb{C}$. Note that the set of real numbers is included in that of complex numbers, so that's possible. Dec 22 '20 at 9:05

Let $$K= \mathbb{Q}(p^{\frac{1}{3}}) \cong \mathbb{Q}[x]/(x^3-p).$$ We have the following:

1. $$\text{Tr}_{K/\mathbb{Q}} (p^{\frac{1}{3}}) = 0$$, as the minimal polynomial of $$p^{\frac{1}{3}}$$ is $$x^3-p.$$
2. $$\text{Tr}_{K/\mathbb{Q}} (p^{\frac{2}{3}}) = 0$$, as the minimal polynomial of $$p^{\frac{2}{3}}$$ is $$x^3-p^2.$$
3. $$\text{Tr}_{K/\mathbb{Q}} (a) = 3a$$, for all $$a \in \mathbb{Q}.$$

Now applying the trace to your equation we get \begin{align*} \text{Tr}_{K/\mathbb{Q}}(a + bp^\frac{1}{3} + cp^\frac{2}{3}) &= \text{Tr}_{K/\mathbb{Q}}(a)+ b \text{Tr}_{K/\mathbb{Q}}(p^\frac{1}{3} ) + c \text{Tr}_{K/\mathbb{Q}}(p^\frac{2}{3})\\ &= \text{Tr}_{K/\mathbb{Q}}(a)+0+0 = 3a = 0= \text{Tr}_{K/\mathbb{Q}}(0),\\ \end{align*} thus $$a=0.$$ Next multiply your equation by $$p^{\frac{1}{3}}$$, apply the trace, and conclude that $$c=0.$$ Repeat.

Edit: As Paramanand Singh correctly points out the problem reduces to showing that $$f(x)= x^3-p$$ is the minimal polynomial of $$p^{\frac{1}{3}},$$ which I assumed. However, this follows directly from Eisenstein's Criterion and the fact that $$f(p^{\frac{1}{3}})=0.$$ In light of this information, the polynomial given by the OP is of degree $$2$$, thus must be the zero polynomial.

• Your first point that $x^3-p$ is the minimal polynomial for $p^{1/3}$ is what we need to prove here. It can't be assumed. More generally the problem boils down to showing that if $x^3-p\in\mathbb {Q} [x]$ has no roots in $\mathbb {Q}$ then it is irreducible over $\mathbb{Q}$. Dec 27 '20 at 3:01
• @ParamanandSingh You're right! I was so eager to use traces so I jumped. Thank you for your comment Dec 27 '20 at 8:25

Here's another way. If $$a+bq+cq^2=0$$, then $$a=-(bq+cq^2)$$, hence $$a^2=b^2q^2+2bcq^3+c^2q^4$$. So if $$q^3=p$$, then we have two equations:

\begin{align} a+bq+cq^2&=0\\ (a^2-2bcp)-c^2pq-b^2q^2&=0 \end{align}

Multiplying both sides of the first equation by $$b^2$$ and the both side of the second equation by $$cp$$, and then adding the resulting equations, we have

$$(ab^2+a^2c-2bc^2p)+(b^3-c^3p)q=0$$

Now if $$q$$ is irrational (i.e., if $$p$$ is not a perfect cube) and the other variables are rational, then we must have $$ab^2+a^2c-2bc^2p=b^3-c^3p=0$$. But $$b^3-c^3p=0$$ for a non-cube $$p$$ implies $$b=c=0$$, in which case the original equation, $$a+bq+cq^2=0$$, implies $$a=0$$.

Since $$q=p^{1/3}$$ is irrational it follows that minimal polynomial of $$q$$ over $$\mathbb{Q}$$ is of degree greater than $$1$$. If we have $$a+bq+cq^2=0$$ for some rational $$a, b, c$$ with $$c\neq 0$$ then $$f(x) =(a/c)+(b/c)x+ x^2\in\mathbb{Q}[x]$$ is the minimal polynomial of $$q$$.

And therefore it must divide the polynomial $$g(x) =x^3-p\in\mathbb{Q} [x]$$ (which has $$q$$ as a root) leading to a linear factor of this polynomial $$g(x)$$. But $$g(x)$$ has no rational root so we get a contradiction.

We must have thus $$c=0$$ and then it is easy to conclude $$a=b=0$$ just from irrationality of $$q$$.

The fact that we are dealing with cube roots and polynomials of degree $$3$$ makes the algebra a lot simpler. If a polynomial of degree $$3$$ is reducible it must have a linear factor and hence a rational root. And thus if a third degree polynomial with rational coefficients has no rational roots then it is irreducible.

The same can't be said of a general polynomial of degree greater than $$3$$. But for polynomials of the form $$x^n-p$$ the same result holds.

Let $$n$$ be a positive prime and $$p$$ be a positive rational number which is not an $$n$$-th power then $$x^n-p$$ is irreducible over $$\mathbb{Q}$$.