How to determine that the solution to the following problem is unique?

Question

The question is:

[BMO2 2000 Q3] Find positive integers a and b such that $$(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = 49 + 20\sqrt[3]{6}$$

It suffices to find one solution to gain full marks, which I did by expanding and assuming that neither $a$ nor $b$ are perfect cubes, so that $2\sqrt[3]{ab} = 48$. The solution I get from this method is $(a, b) = (48, 288)$. However, from this two questions arise in my mind:

a) Is this is the only solution? I.e. are there $c$ and $d$ such that $$(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = (\sqrt[3]{c} + \sqrt[3]{d} - 1)^2$$

$\Rightarrow \sqrt[3]{a} + \sqrt[3]{b} = \sqrt[3]{c} + \sqrt[3]{d}$ or $\sqrt[3]{a} + \sqrt[3]{b} - 1 = -\sqrt[3]{c} - \sqrt[3]{d} + 1$

b) Can one solve the original problem in such a way that it is clear that the solutions found are unique?

EDIT: Thank you to gammatester for his remark that there is another solution: $(a, b) = (288, 48)$, but I am referring to distinct solutions which are not permutations of other solutions.

Answer 1

I apologize, this solution is a bit hand-wavy and (it may in fact be wrong) as my skills/knowledge in this topic is rather weak. Perhaps someone else can later provide a more rigorous version of this argument if it is correct.

We have, $$\sqrt[3]{a^2}+\sqrt[3]{b^2}-2(\sqrt[3]{a}+\sqrt[3]{b}) = 20\sqrt[3]{6}.$$

Consider the highest power of $6$ that divides $a$ and $b$ modulo $3$. I believe (likely using some argument about the linear independence of cuberoots of integers over $\mathbb{Q}$) that it can be shown that the highest power of $6$ (modulo $3$) that divides $a$ is $6^1$ and similarly $6^2$ for $b$. Such an argument would probably use the fact that the righthand side doesn't contain any $\sqrt[3]{6^2}$ terms.

Also, we have, like you said, $$2\sqrt[3]{ab} = 48 = 2^4 \cdot 3$$ $$\implies 2^3 | \sqrt[3]{ab}$$ $$\implies 2^9 | ab.$$

That means that there are $9$ factors of $2$ between $a$ and $b$. Since $a$ contains $1$ factor of $6$ and $b$ contains $2$ factors of $6$, that means that there are a total of $6$ factors of $2$ left between $a$ and $b$ after we subtract out the factors of $2$ belonging to the $6$'s. In order to write $\sqrt[3]{a} = k\sqrt[3]{6}$ and $\sqrt[3]{b} = l\sqrt[3]{6^2}$ for some integers $k,l$; then those $6$ factors of $2$ must be distributed between $a$ and $b$ in multiples of $3$. There are three ways to do this:

1. Give all six factors of $2$ to $a$
2. Give all six factors of $2$ to $b$
3. Give both $a$ and $b$ three factors of $2$ each.

This gives us three candidates for our choices of $a,b$:

1. $a = 6\cdot 2^6$ and $b=6^2$
2. $a = 6$ and $b=6^2\cdot 2^6$
3. $a = 6\cdot 2^3$ and $b=6^2\cdot 2^3$

Of these three, only the last one solves the equation, hence the solution is unique.

As a side note, I've verified that $(48,288)$ is the only solution up to reordering in Mathematica: Solve[(x^(1/3) + y^(1/3) - 1)^2 == 49 + 20*6^(1/3) && x > 0 && y > 0, {x, y}, Integers]

Hope this helps!

Answer 2

We do not need to account for your second case, since it is trivially impossible due to the fact that for positive integers $ a, b, c, d $ the left hand side is positive and the right hand side is negative. Your first case is settled by the following theorem:

Theorem. Let $ a, b \in \mathbf Q^{\times} $ be non-perfect cubes such that $ a/b $ is not a perfect cube in $ \mathbf Q $. Then, $ \sqrt[3]{a} + \sqrt[3]{b} = \sqrt[3]{c} + \sqrt[3]{d} $ for $ c, d \in \mathbf Q $ if and only if $ \{ a, b \} = \{ c, d \} $ (i.e the solution is unique up to permutation).

Proof. Write $ b = qa $ for $ q \in \mathbf Q $ not a perfect cube. Then, we have an equality

$$ \sqrt[3]{a} + \sqrt[3]{qa} = \sqrt[3]{c} + \sqrt[3]{d} $$

These cube roots all lie in some field extension $ L $ of $ \mathbf Q $, and we can talk about the field trace $ T = \textrm{Tr}_{L/\mathbf Q} $. Since $ x^3 - a $ is irreducible in $ \mathbf Q[x] $ if $ a $ is not a perfect cube, and the trace is a multiple of the sum of conjugates, we find that the trace of all irrational cube roots of rational numbers vanish. This is the key idea of the proof. Multiply both sides by $ a^{2/3} $ to get

$$ a + a\sqrt[3]{q} = \sqrt[3]{a^2 c} + \sqrt[3]{a^2 d} $$

If neither cube root on the right hand side was rational, taking traces would yield the absurd $ a = 0 $. Therefore, at least one of $ a^2 c, a^2 d $ is a perfect cube. Assume that it is $ a^2 c = u^3 $ without loss of generality, and now multiply both sides of the original equation by $ (qa)^{2/3} $ to get

$$ a \sqrt[3]{q^2} + qa = \sqrt[3]{q^2 a^2 c} + \sqrt[3]{q^2 a^2 d} $$

By the same argument, one of the cube roots on the right hand side must be rational, since $ q^2 $ is not a perfect cube. If it was $ q^2 a^2 c $, then $ q^2 $ would also be a perfect cube as the ratio of two perfect cubes, which is a contradiction. Therefore $ q^2 a^2 d = v^3 $ is a perfect cube. Substituting into the original equation gives

$$ \sqrt[3]{a} + \sqrt[3]{qa} = \frac{u \sqrt[3]{a}}{a} + \frac{v\sqrt[3]{qa}}{qa} $$

However, $ \sqrt[3]{a} $ and $ \sqrt[3]{qa} $ are linearly independent over $ \mathbf Q $, which means that the only vanishing linear combination is the trivial one, and we have

$$ \frac{u}{a} = \frac{v}{qa} = 1 $$

Now, $ u^3 = a^2 c $ gives $ c = a $, and $ v^3 = q^2 a^2 d $ gives $ d = qa = b $. QED.

Now, since $ 288/48 = 6 $ is not a perfect cube in $ \mathbf Q $ (and neither are $ 288 $ and $ 48 $), the result follows.