Prove that $d$ must be a perfect square

Question

Prove that if $a,b,c,d$ are integers such that $$(a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c)^2 = d,$$ then $d$ is a perfect square.

In order for $(a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c)^2$ to be an integer, either $a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c$ has to be the square root of a positive integer or $a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c$ must be a perfect square. How do we deal with those conditions?

Answer 1

We expand and get $$ d=(a+2^{1/3}b+2^{2/3}c)^2\\ =(a^2+4bc)+(2c^2+2ab)2^{1/3}+(b^2+2ac)2^{2/3} $$ If $d$ is to be an integer, then from the expression above we must have $2c^2+2ab=b^2+2ac=0$. Assuming $a\neq0$, this means that $$ b=-\frac{c^2}{a}\\ c=-\frac{b^2}{2a} $$ Inserting one of these into the other, we get $$ b=-\frac{\frac{b^4}{4a^2}}{a}=-\frac{b^4}{4a^3}\\ b(b^3+4a^3)=0 $$ so since $a\neq0$, and $\sqrt[3]4\notin \Bbb Q$, we must have $b=0$, which gives $c=0$, and $d=a^2$, a perfect square.

On the other hand, if $a=0$, we have $c^2=b^2=0$, so $d=0$. Whether you count this as a perfect square is up to your definitions, but I would say that it is.

Answer 2

Here's an alternative approach to the above answer if you know some field theory:

Let $\alpha = a + 2^{1/3}b + 2^{2/3}c$. We want to show that $\alpha$ is rational, from which it follows that $\alpha$ is integer since $\alpha^2 = d$ is integer. Assume otherwise. The field $\mathbb{Q}(2^{1/3})$ has degree $3$ over $\mathbb{Q}$, Since $\alpha$ is not rational and $\alpha \in \mathbb{Q}(2^{1/3})$, $\mathbb{Q}(\alpha)$ must have degree $3$ over $\mathbb{Q}$. Since $\alpha^2 = d$ is integer, the polynomial $f(x) = x^2 - \alpha^2$ is in $\mathbb{Q}[x]$. Now $f(\alpha) = 0$, so $\mathbb{Q}(\alpha)$ has degree at most $2$. This is a contradiction, so $\alpha$ must be rational.

Answer 3

Theorem: Suppose that $a+2^{1/3}b+2^{2/3}c=0$ where $a,b,c\in\mathbb{Z}$. Then $a=b=c=0$.

Proof: Note that $$ \begin{align} &\left(a+2^{1/3}b+2^{2/3}c\right)\left(\left(a^2-2bc\right)+2^{1/3}\left(2c^2-ab\right)+2^{2/3}\left(b^2-ac\right)\right)\\ &=a^3+2b^3+4c^3-6abc\tag{1} \end{align} $$ Thus, if $a+2^{1/3}b+2^{2/3}c=0$, then $(1)$ says that $$ a^3+2b^3+4c^3-6abc=0\tag{2} $$ If it is not the case that $a=b=c=0$, we can assume that $\gcd(a,b,c)=1$.

Looking at $(2)$ mod $2$, we see that $\phantom{2}a^3\equiv0\pmod2$; therefore $a\equiv0\pmod2$.
Looking at $(2)$ mod $4$, we see that $2b^3\equiv0\pmod4$; therefore $b\equiv0\pmod2$.
Looking at $(2)$ mod $8$, we see that $4c^3\equiv0\pmod8$; therefore $c\equiv0\pmod2$.

Therefore, $2\mid\gcd(a,b,c)$. Contradiction. Thus, $a=b=c=0$.

QED

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Answer to the Question

Suppose that $$ \left(a+2^{1/3}b+2^{2/3}c\right)^2=d\tag{3} $$ Expanding $(3)$ gives $$ \left(a^2+4bc-d\right)+2^{1/3}\left(2ab+2c^2\right)+2^{2/3}\left(b^2+2ac\right)=0\tag{4} $$ The Theorem guarantees that $2abc+2c^3=0$ and $b^3+2abc=0$. That is, $2c^3=b^3$. Since $2^{1/3}\not\in\mathbb{Q}$, we must have that $b=c=0$.

The Theorem also guarantees that $a^2+4bc-d=0$. Since $b=c=0$, this says that $d=a^2$, which is a perfect square.