# Generate many examples of Ramanujam's Diophantine Equation (Scripts) 1.0

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## Generate many examples of Ramanujam's Diophantine Equation (Scripts) Publisher's description

### This programme is about finding many examples of Ramanujam's Diophantine Equation

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 = N
where 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 + 12^3

Another example is :
(-107766)^3 + (-634932)^3 = (-2013055)^3 + 1991671^3
= -257217167508536664 (Calculator) { -2.572171675085367e+017 (Matlab) }
ie, 2013055^3 = 107766^3 + 634932^3 + 1991671^3

I 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 obtain
x, y, u, v such that gcd (x, y, u, v) = 1

Q2 : 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

#### System Requirements:

MATLAB 6.5 (R13)
Program Release Status: New Release
Program Install Support: Install and Uninstall

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