Generate many examples of Ramanujam's Diophantine Equation 1.0

This programme is about finding many examples of Ramanujam's Diophantine Equation. For mathematical formulas used in making this programme, I have referred to the article MATHEMATICAL MINIATURE 9.pdf by John Butcher, butcher@math.auckland.ac.nz.To circumvent the problem of editing the Greek symbols, I have replaced them with suitable alphabets.The basic equation we are trying to solve is to get the numbers :x & y and u & v and N that satisfy the Diophantine equation :x^3 + y^3 = u^3 + v^3 = Nwhere the four integers x, y, u, v have no common factor.For eg, the "lowest" Diophantine Number N of Ramanujam is :1729 = 9^3 + 10^3 = 1^3 + 12^3Another example is :(-107766)^3 + (-634932)^3 = (-2013055)^3 + 1991671^3= -257217167508536664 (Calculator) { -2.572171675085367e+017 (Matlab) }ie, 2013055^3 = 107766^3 + 634932^3 + 1991671^3I wish to thank Prof John Butcher for his article which triggered and enabled me to write this MATLAB code for "generating" Ramanujam's Diophantine Equation Numbers. I hope that this programme will be useful to many the world over. Time permitting, I may be improving on this programme to make it suitable for generating a series of Diophantine Numbers automatically. But the base groundwork is already laid now, and we need only to build further.Q1a : I am curious to know why Prof John Butcher has said that gcd (a, b) and gcd (c, d) should be 1.In the case of his own example : 9^3 + 15^3 = 2^3 + 16^3 = 4104,intermediate calculations give a = 12, b = 3, c = 9, d = 7.See Usage Eg Case 4 below. Obviously, gcd (a, b) = 3 (not 1).But, gcd (x, y, u, v) = 1 (var name used in my programme is G_Gxy_Guv.)Also, in his second example : 33^3 + 15^3 = 2^3 + 34^3 = 39312,calculations give a = 3, b = 9, c = 9, d = 8. See Usage Eg Case 5 below.Obviously, gcd (a, b) = 3 (not 1). But, gcd (x, y, u, v) = 1.Similarly, in his 3rd example too, gcd (a, b) = 3 (not 1) !In many of my own examples with randomly chosen values, I have mixed results:In Usage Eg 2, gcd (c, d) = 9 (not 1), but, gcd (x, y, u, v) = 1.In Usage Eg 3, gcd (a, b) = 3 (not 1), but, gcd (x, y, u, v) = 3 (not 1) !Egs 7 and 8 are similar to Eg 2 ; Egs 9, 10, and 11 are similar to Eg 3.Q1b) Therefore, I would like to know the condition or constraint, which when fulfilled, will ensure that we will certainly obtainx, y, u, v such that gcd (x, y, u, v) = 1Q2 : I would also like to know how the article's Theorem 1 is used in deriving the formulae for Diophantine Numbers.It says : ... "although beneath the surface" ... ************Now added in the suite :find_x_y__p_1_mod_6.m : This function finds x and y pairs of numbers such that x2 + 3y2 = p where mod(p, 6) = 1

Requirements:	No special requirements
Platforms:	Matlab
Keyword:	Calculations,  Curious,  Intermediate,  Randomly,  Similarly,  Usage
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