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I want to find integer solutions of the following inequality by using Mathematica

$$\bigg| \sqrt[3]{2}-\frac{p}{q} \bigg | <\frac{1}{q^{5/2}}$$

Reduce[Abs[Surd[2, 3] - p/q] < 1/q^(5/2), {p, q}, Integers] //ToRadicals // TraditionalForm

but there is a strange result

enter image description here

I also tried FindInstance but there is the only one solution

FindInstance[Abs[Surd[2, 3] - p/q] < 1/q^(5/2), {p, q}, Integers, 2]

enter image description here

There are 3 solutions according the Wolfram Alpha

enter image description here

but...

WolframAlpha["Abs[Surd[2,3]-p/q]<1/q^2.5)", {{"IntegerSolution"}, "Content"}]

{}

It's still not working..

I tried "Open Code" in WolframAlpha but it didn't help me.

what's going on here? :)

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  • 2
    $\begingroup$ If you don't mind specifying a range of integers to search (which, understandably, you probably do) you could use a region: reg = ImplicitRegion[Abs[Surd[2, 3] - p/q] < 1/q^(5/2), {{p, 1, 5}, {q, 1, 5}}]; and then Reduce[Element[{p, q}, reg], {p, q}, Integers] gives you the desired output. $\endgroup$ Sep 30, 2017 at 12:46
  • $\begingroup$ A graphical solution is possible with Show[Graphics[{LightGray, Table[Point[{i, j}], {i, 0, 6}, {j, 0, 6}]}, Axes -> True], RegionPlot[Abs[Surd[2, 3] q - p] < 1/q^(3/2), {p, 0, 6}, {q, 0, 6}, Frame -> True], Graphics[{Red, Point[{1, 1}], Point[{2, 1}], Point[{5, 4}]}, Axes -> True]] to make yourself shure this are all solutions of integer pairs. $\endgroup$ Mar 26, 2020 at 21:56

1 Answer 1

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See KennyColnago answer.

Union[Flatten[
Select[Table[
Quiet[FindInstance[Abs[Surd[2, 3] - p/q] < 1/q^(5/2), {p, q}, 
Integers, 2, RandomSeed -> k]], {k, 1, 200}], 
Head[#] =!= FindInstance &], 1]]

(* {{p -> 1, q -> 1}, {p -> 2, q -> 1}, {p -> 5, q -> 4}} *)
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1
  • $\begingroup$ Too slow....... $\endgroup$
    – vito
    Oct 1, 2017 at 5:23

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