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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:

Solving the quadratic, I get:

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

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?

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    $\begingroup$ You mean $p^{1/3} = \cdots.$ $\endgroup$
    – mjw
    Dec 21 '20 at 15:46
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    $\begingroup$ Why is $b = ± \sqrt{b^2 - 4ac} - 2cp^⅔$ a contradiction? $\endgroup$
    – Arthur
    Dec 21 '20 at 15:48
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    $\begingroup$ 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. $\endgroup$
    – Arthur
    Dec 21 '20 at 15:50
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    $\begingroup$ $\sqrt 2-\sqrt 2$, 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. $\endgroup$
    – Arthur
    Dec 21 '20 at 15:57
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    $\begingroup$ But can you find $x^{1/2} - y^{1/3} = 0$ (where x and y are not perfect squares and cubes respectively)? $\endgroup$
    – Shub
    Dec 21 '20 at 16:05
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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?

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  • $\begingroup$ I have never seen it either, +1 for the nice method! $\endgroup$ Dec 22 '20 at 3:13
  • $\begingroup$ 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 $K$. $\endgroup$
    – dust05
    Aug 29 '21 at 15:13
  • $\begingroup$ 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. $\endgroup$
    – dust05
    Aug 29 '21 at 15:14
  • $\begingroup$ 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. $\endgroup$ Aug 29 '21 at 16:23
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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.
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  • $\begingroup$ Thank you for proving the irrationality of $± \sqrt{b² - 4ac} - 2cp^{2/3}$ :) $\endgroup$
    – Shub
    Dec 22 '20 at 3:51
  • $\begingroup$ @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 $p = \frac18$. $\endgroup$ Dec 22 '20 at 8:30
  • $\begingroup$ 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. $\endgroup$
    – Shub
    Dec 22 '20 at 8:57
  • $\begingroup$ @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. $\endgroup$ Dec 22 '20 at 9:05
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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.

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    $\begingroup$ 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} $. $\endgroup$
    – Paramanand Singh
    Dec 27 '20 at 3:01
  • $\begingroup$ @ParamanandSingh You're right! I was so eager to use traces so I jumped. Thank you for your comment $\endgroup$ Dec 27 '20 at 8:25
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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 $a=0$.

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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} $.

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