# Rational number trivial case

Let $a,b,c$ denote rational numbers, such that $(a+b\sqrt[3]2+c\sqrt[3]4)^3$ is also rational. Prove that at least two of the numbers $a,b,c$ must be zero. Actually I confused of the beginning steps for this proof. Should I apply the strategies of contradictory proofs

• First expand the expression. Notice that the product of a non zero rational number with an irrational number is an irrational number. Also notice that $\sqrt[3]{2}$ and $\sqrt[3]{4}$ are irrational numbers. – Scientifica Jul 18 '14 at 16:27

Since $$(a+bx+cx^2)^3\equiv (6bc^2+3a^2c+3ab^2)x^2+(6ac^2+6b^2c+3a^2b)x+(4c^3+2b^3+a^3+12abc)\pmod{(x^3-2)}$$ if $(a+b\sqrt[3]{2}+c\sqrt[3]{2})^3$ belongs to $\mathbb{Q}$ then $$(2bc^2+a^2c+ab^2)=0, \qquad (2ac^2+2b^2c+a^2b)=0,\tag{1}$$ hence: $$a^2 b^2 c^2 = (a^2c+ab^2)(ac^2+b^2 c) = a^3 c^3 + 2a^2 b^2 c^2 + ab^4 c,$$ $$a^3 c^3 + a^2 b^2 c^2 + ab^4 c = 0,$$ $$ac (a^2 c^2 + a c b^2 + b^4) = 0.$$ Since the discriminant of $y^2+y+1$ is negative, the last equation implies $a=0$ or $c=0$. If we plug this identities back into $(1)$, we get that the only possibilities are that at least two variables among $\{a,b,c\}$ are zero.

Oh an answer while I was writing! (I was writing in Notepad not directly in the forum).

Anyway, before answering the question, lets prove these:

Lemma 1: $\sqrt[3]{2}\not\in\mathbb{Q}^+$.

Corollary: $\sqrt[3]{4}\not\in\mathbb{Q}^+$.

Lemma 2: $\forall (a,b)\in\mathbb{Z}^2,\,a\sqrt[3]{2}+b\sqrt[3]{4}\in\mathbb{Q}\Longleftrightarrow (a,b)=(0,0)$

Theorem: $\forall (a,b)\in\mathbb{Q}^2,\,a\sqrt[3]{2}+b\sqrt[3]{4}\in\mathbb{Q}\Longleftrightarrow (a,b)=(0,0)$

Proof of lemma 1:

Assume that $\sqrt[3]{2}\in\mathbb{Q}$. Then $\exists !(a,b)\in\mathbb{N}^2,\,\sqrt[3]{2}=\dfrac{a}{b}$ such as $a,b\neq 0$ and $a∧b=1$.

Then $$a^3=2b^3\,\,\,\,(1)$$and so $a$ is an even number. Then $\exists k\in\mathbb{N},\,a=2k$ and so in $(1)$ we have $b^3=4k^3$ and so $b$ is also an even number which contradicts $a∧b=1$.

Proof of the corollary:

Assume that $\sqrt[3]{4}\in\mathbb{Q}^+$. Then $\sqrt[3]{4^2}\in\mathbb{Q}^+$ and so $2\sqrt[3]{2}\in\mathbb{Q}^+$ which contradicts the lemma 1.

Proof of lemma 2:

If $(a,b)=(0,0)$ then $a\sqrt[3]{2}+b\sqrt[3]{4}=0\in\mathbb{Q}$.

Now lets prove that $\forall (a,b)\in\mathbb{Z}^2,\,a\sqrt[3]{2}+b\sqrt[3]{4}\in\mathbb{Q}\Longrightarrow (a,b)=(0,0)$.

To prove it, lets prove that $\forall (a,b)\in\mathbb{Z}^2,\,(a,b)\neq (0,0)\Longrightarrow a\sqrt[3]{2}+b\sqrt[3]{4}\not\in\mathbb{Q}$

If $a=0$, and so $b\neq 0$, or $b=0$ and so $a\neq 0$, it's trivial that $a\sqrt[3]{2}+b\sqrt[3]{4}\not\in\mathbb{Q}$ using the lemma 1 or the corollary.

Assume that $\exists (a,b)\in\mathbb{Z}^2$ such as $a,b\neq 0$ and: $$a\sqrt[3]{2}+b\sqrt[3]{4}\in\mathbb{Q}\,\,\,\, (2.1)$$

Then $(a\sqrt[3]{2}+b\sqrt[3]{4})^2\in\mathbb{Q}$ and $(a\sqrt[3]{2}+b\sqrt[3]{4})^2=a^2\sqrt[3]{4}+2b^2\sqrt[3]{2}+4ab$.

While $4ab\in\mathbb{N}$ then: $$a^2\sqrt[3]{4}+2b^2\sqrt[3]{2}\in\mathbb{Q}\,\,\,\, (2.2)$$

By $(2.2)-\dfrac{a^2}{b}(2.1)\in\mathbb{Q}\,\,\,\,\left(\dfrac{a^2}{b}\in\mathbb{Q}^+\right)$ we get: $$\left( 2b^2-\dfrac{a^3}{b}\right)\sqrt[3]{2}\in\mathbb{Q}\,\,\,\, (2.3)$$

If $\left( 2b^2-\dfrac{a^3}{b}\right) =0$ then $a=\sqrt[3]{2}b\not\in\mathbb{N}$ which is false.

And so $\left( 2b^2-\dfrac{a^3}{b}\right)\neq 0$, but while $\left( 2b^2-\dfrac{a^3}{b}\right)\in\mathbb{Q}$ it means using $(2.3)$ that $\sqrt[3]{2}\in\mathbb{Q}$ which contradicts the lemma 1.

We conclude that $\forall (a,b)\in\mathbb{Z}^2,\,(a,b)\neq (0,0)\Longrightarrow a\sqrt[3]{2}+b\sqrt[3]{4}\not\in\mathbb{Q}$ which proves lemma 2.

Proof of the theorem:

If $(a,b)=(0,0)$ it's as same as in lemma 2.

Let $(a,b)\in\mathbb{Q}$ such as $a\sqrt[3]{2}+b\sqrt[3]{4}\in\mathbb{Q}$.

$(a,b)\in\mathbb{Q}$ so $\exists ((p,p'),(q,q'))\in\mathbb{Z}^2\times\mathbb{N}^{+2}\, (a,b)=\left(\dfrac{p}{q},\dfrac{p'}{q'}\right)$.

We have then:$$\dfrac{p}{q}\sqrt[3]{2}+\dfrac{p'}{q'}\sqrt[3]{4}\in\mathbb{Q}\,\,\,\, (3)$$

By $qq'\times (3)$ still $qq'\in\mathbb{N}^+$ we get $pq'\sqrt[3]{2}+p'q\sqrt[3]{4}\in\mathbb{Q}$. Using lemma 2 we get $(pq',p'q)=(0,0)$ and so $(p,p')=(0,0)$ and so $(a,b)=(0,0)$ which proves the theorem.

We know that: $$\forall (x,y,z)\in\mathbb{R}^3,\,(x+y+z)^3=x^3+y^3+z^3+3x^2(y+z)+3y^2(x+z)+3z^2(x+y)+6xyz\,(4)$$.

Now let $(a,b,c)\in\mathbb{Q}^3$ such as $(a+b\sqrt[3]{2}+C\sqrt[3]{4})^3\in\mathbb{Q}$.

By using $(4)$: $(a+b\sqrt[3]{2}+c\sqrt[3]{4})^3=a^3+2b^3+4c^3+12abc+3(a^2b+2b^2c+2ac^2)\sqrt[3]{2}+3(a^2c+ab^2+2bc^2)\sqrt[3]{4}$.

While $a^3+2b^3+4c^3+12abc\in\mathbb{Q}$, $(a+b\sqrt[3]{2}+C\sqrt[3]{4})^3\in\mathbb{Q}$ if and only if $3(a^2b+2b^2c+2ac^2)\sqrt[3]{2}+3(a^2c+ab^2+2bc^2)\sqrt[3]{4}\in\mathbb{Q}$.

Using the theorem: $(a+b\sqrt[3]{2}+C\sqrt[3]{4})^3\in\mathbb{Q}$ if and only if $a^2c+ab^2+2bc^2=0$ and $a^2b+2b^2c+2ac^2=0$.

Then see Jack D'Aurizio's post.

Step #$1$:

• $a+b\sqrt[3]{2}\neq0 \implies c\notin\mathbb{Q}$

• $a+c\sqrt[3]{4}\neq0 \implies b\notin\mathbb{Q}$

• $b\sqrt[3]{2}+c\sqrt[3]{4}\neq0 \implies a\notin\mathbb{Q}$

Step #$2$:

• $a+b\sqrt[3]{2}=0 \wedge a\neq0 \implies b\notin\mathbb{Q}$

• $a+c\sqrt[3]{4}=0 \wedge a\neq0 \implies c\notin\mathbb{Q}$

• $b\sqrt[3]{2}+c\sqrt[3]{4}=0 \wedge b\neq0 \implies c\notin\mathbb{Q}$