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Diophantine equations

one man 1330 Maple 17
October 30 2022
5

It seems to me that Draghilev's method can be applied quite successfully to the solution of Diophantine equations. Here is a simple example where we find two solutions at the intersection line of two ellipsoids:
  x1^2-x1*x2+x2^2+x2*x3+x3^2-961=0;
  (x1-3)^2+10*x2^2+x3^2-900=0;
Solutions: (11, -4, -26) and (10, 1, 29).
 


Based on the text of the program, it is possible to solve various examples with Diophantine equations.
3d_1.mw

Explanations.
f3 is an auxiliary equation for finding the starting point, NPar is a procedure that implements the Draghilev method, the red color of the text is the place where the integer values of the points on the integral curve are filtered.

 Can be compared with the solution of the isolve function
 

 restart:
  f1 := x1^2-x1*x2+x2^2+x2*x3+x3^2-961;
  f2 := (x1-3)^2+10*x2^2+x3^2-900;
  isolve({f1, f2})

 
diophantine_equation draghilev

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