Solving x > 13874  Abs[(x^3) - 3  (y^3)]

by Ed Pegg Jr.

Since the days of Fermat, it has been an ongoing challenge to find simple-appearing equations with large solutions.  The above equation seems simple enough, and can be easily solved with continued fractions.  It can be called a Diophantine Inequality.  I'll start with the answer, and then work back to how it was gotten.

<< NumberTheory`ContinuedFractions`

{x, y}  =  
        With[{k =  Convergents[
                  ContinuedFraction[3^(1/3), 
                    4000]][[3432]]}, {Numerator[k], 
            Denominator[k]}];

We now have the answer! I won't print them out completely, since these are huge numbers.  Mathematica can print the numbers out easily, but it can also express them in a shortened form, with Short[ ].  Each number has 1781 digits.

{Floor[Log[10, x]], Short[x,  .5], Short[y,  .5]}

We can test that they actually work.

x  >  13874  Abs[(x^3) -  3 (y^3)]

We can use the function below to look for other solutions of the form x  >  a Abs[(x^b) -  c (y^b)], where b is the power, and c is the root.

CFRoots[root_, power_,  range_] := 
      Take[Reverse[
          Sort[Transpose[{Map[
                  Floor[Numerator[#]/
                        Abs[root Denominator[#]^power  -  
                            Numerator[#]^power]] &, 
                  Convergents[ContinuedFraction[root^(1/power), range]]], 
                Range[range]}]]], 10]

Can the a term in x  >  a Abs[(x^b) -  c (y^b)]be greater than a million?  Definitely ... but I haven't found an example, yet.  If you find one, send it to me at ed@mathpuzzle.com.  Here's the sort of thing that necessary, in the continued fraction.  First, here's what the cube root of three looks like, as a continued fraction.

ContinuedFractionForm[ContinuedFraction[3^(1/3), 15]]

Since the important part is the set of numbers, we can look at those only.

ContinuedFraction[3^(1/3), 70]

Places where the number is unusually high, in this case 139, indicates a place where the continued fraction is more accurate than might be expected.

Map[
      Floor[Numerator[#]/Abs[3 Denominator[#]^3  -  Numerator[#]^3]] &, 
      Convergents[ContinuedFraction[3^(1/3), 70]]]

{,{3369, 1990},{2209, 2406},{3129, 
          578},{9740, 812},{772, 1020},{418, 
          2240},,{1148, 2741},{422, 1377},{511, 
          532},{1154, 2903},{1317, 1063},{1865, 
          677},{1103, 126}};

I'll close with a few puzzles that are easily solved with Mathematica, using the same techniques.  Find the minimal {x,y} for the below diophantine inequalities.

1.    x  >  57  Abs[(x^3) -  3 (y^3)]
2.    x  >  56  Abs[(x^3) -  4 (y^3)]
3.    x  >  595  Abs[(x^3) -  5 (y^3)]
4.    x  >  93  Abs[(x^3) -  6 (y^3)]
5.    (x^2)  >  214  Abs[(x^4) -  2 (y^4)]
6.    (x^2)  >  3  Abs[(x^5) -  11 (y^5)]