$0<|\sqrt a-\sqrt[3]b|<\epsilon$ for $a,b\in\Bbb Z_+$

Question

I'm trying to solve the following problem:

Given $\epsilon>0$, are there positive integers $a,b$ such that $0<|\sqrt a-\sqrt[3]b|<\epsilon$ ?

My solution: given $n\in\Bbb N$, $$|\sqrt{n^2}-\sqrt[3]{n^3+1}|=\sqrt[3]{n^3+1}-n=\frac1{\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{n^3+1}+n^2}<\frac1{3n}\to 0$$ Thus, the answer is yes.

But I was trying to find an "optimal" solution. That is, now the problem becomes

Given $\epsilon>0$, find the least $b\in \Bbb Z_+$ such that there exists $a\in\Bbb Z_+$ such that $0<|\sqrt a-\sqrt[3]b|<\epsilon$

and now I'm totally lost. Is there some theory about this? Perhaps has it to do with the diophantine equation $a^3-b^2=\pm1$, and hence, to Catalan's conjecture?

Remark: Please note the '$0<$' in the inequality. I'm aware that $\sqrt 1=\sqrt[3]1$.

Answer 1

Listing some easy observations to get the ball rolling.

My guess is that the variable $b$ and the error $\epsilon$ are, asymptotically, related by estimates of the form $$\epsilon\approx \frac C{b^\alpha},$$ where $C$ and $\alpha>0$ are positive constants.

The true relation may be very complicated, but at least we can derive upper and lower bounds of the prescribed form. Your example with $b=n^3+1$, $\root3\of b-\sqrt a\le 1/(3n^2)$, shows that $$ \epsilon\le \frac{1/3}{b^{2/3}} $$ is possible for infinitely many values of $b$.

On the other hand, let $\zeta=(1+i\sqrt3)/2$ be a primitive sixth root of unity. We have the polynomial factorization $$ x^6-y^6=(x-y)\prod_{j=1}^5(x-\zeta^j y).\qquad(*) $$ Assume that integers $b, a$ are chosen in such a way that $\root3\of b-\sqrt a$ is very small (but non-zero). Plug $x=\root3\of b, y=\sqrt a$ into $(*)$. The left hand side $b^2-a^3$ has absolute value $\ge1$ because it is an integer. Predalescu/Catalan says that actually it is $\ge2$ when $b>3$ but that's insignificant, at least for now. Because $x\approx y$, the other factors on the right hand side of $(*)$ have absolute values $\approx x, \sqrt{3}x,2x,\sqrt3 x,x$ for $j=1,2,3,4,5$ respectively. Their product is thus $\approx 6x^5$. The factorization $(*)$ thus gives the estimate $$ |\root3\of b-\sqrt a|\ge\frac{K}{b^{5/3}} $$ with a constant $K\approx 1/6$.

I would summarize this by stating that

$$2/3\le \alpha\le 5/3.$$

Waiting for the experts to show up with something more precise.