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

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 $$\sqrt 1=\sqrt[3]1$$.

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.