Find $a,b\in\mathbb{Z}^{+}$ such that $\large (\sqrt[3]{a}+\sqrt[3]{b}-1)^2=49+20\sqrt[3]{6}$

Question

find positive intergers $a,b$ such that

$\large (\sqrt[3]{a}+\sqrt[3]{b}-1)^2=49+20\sqrt[3]{6}$

Here i tried plugging

$x^3=a,y^3=b$

$(x+y-1)^2=x^2+y^2+1+2(xy-x-y)=49+20\sqrt[3]{6} $

the right hand part is a square hence can be written as $(p+q)^2$

Answer 1

This one is interesting.

Note that not both $a$ and $b$ are perfect cubes. Why? because otherwise the lhs would be a perfect cube!

Now, let's tackle the case where one of them is a perfect cube, say $a$. In that case, $(\sqrt[3]{a} - 1)$ is an integer. By elementary property of surds, square of this integer must match $49$, integer part of rhs, giving us

$$(\sqrt[3]{a} - 1)^2 = 49$$

And thus, $a = 8^3 = 512$.

The remaining part then reduces to $(7 + \sqrt[3]{b})^2 = 49+20\sqrt[3]{6}$, i.e. $b^{2/3} + 14b^{1/3} = 20(6)^{1/3}$

Edit: This quadratic has no integer solution. I was wrong in assuming $6$ as a solution.

This means neither $a$ nor $b$ is a perfect cube

Edit 2: I first thought integer solutions must somewhere involve integer cubes. Although neither $a$ nor $b$ are perfect cubes, it turns out that $ab$ is a perfect cube and that is a key to the solution. See Calvin's solution to prove that $ab$ is a perfect cube and find $a$, $b$

Answer 2

Firstly, if either $a$ or $b$ is a perfect cube, then we have $$ (\sqrt[3]{a}-N)^2 = 49 + 20 \sqrt[3]{6} $$

Convince yourself that the LHS must involve 2 different surd terms, hence this is not possible. Thus, neither $a$ nor $b$ are a perfect cube.

Like you did, expand the terms and consider what they are. Clearly $a, b, a^2, b^2$ are not perfect cubes. If $ab$ is not a perfect cube, then we must have $1 = 49$, which is a contradiction. Hence, we have $ 2\sqrt[3]{ab} + 1 = 49$, or that $ab = 24^3$. We use the substitution $b = \frac{24^3}{a}$ and $x = \sqrt[3]{a}$, and the equation becomes

$$x^2-2x - 48/x + 24^2/x^2 = 20 \sqrt[3]{6}.$$

Multiplying throughout by $x^2$, this equation has real roots of $x = \sqrt[3]{48}$ and $x= \sqrt[3]{288}$, hence $ (a,b) = (48 288) $ and $(288,48)$ are solutions.

Answer 3

I did an Excel search and found $(288,48)=(6^2*2^3,6*2^3)$ and the reverse as solutions. This is confirmed by Alpha

Answer 4

If you don't mind doing a lot of arithmetic, here's an "easy" way to solve things:

Note that $(49+20\sqrt[3]{6})^{3/2} = 788.401\ldots$. Since you're looking for positive integer solutions $a$ and $b$, one need at worst examine numbers from $1$ to $788$: That is, if $b\ge1$, then $\sqrt[3]{a}\le\sqrt[3]{a}+\sqrt[3]{b}-1 =(49+20\sqrt[3]{6})^{1/2}$, hence $1\le a\le788$. In doing so, you will presumably stumble across the solution $(48,288)$ and its reverse, and no others.

You can streamline this a bit by looking for solutions with $a\le b$ so that

$$2\sqrt[3]{a}\le\sqrt[3]{a}+\sqrt[3]{b} = 1+(49+20\sqrt[3]{6})^{1/2}=10.238\ldots$$

leads to $a\le134$. Whether you can easily streamline it further is less clear.