Seppo Mustonen                                           9 Nov 2018

About Diophantine equations of form mod(X^n+Y^n,P)=0 or X^n+Y^n=P*Z

In these equations all numbers X,Y,P,Z,n are integers.
The task is to find all integers X,Y so that the sum of n'th powers
of X and Y denoted by X^n+Y^n is divisible by P.
For example, in the case X^4+Y^4=17*Z the pair of integers X=9, Y=67
is a root since then

 X^4=6561  (thus 9*9*9*9)
 Y^4=20151121
 and their sum is X^4+Y^4=20157682.
 This is an integer divisible by P=17 since
 20157682/P=1185746.

To find all the roots it is enough to find them for X,Y values
0,1,2,...,17 since all remaining roots are trivially multiples
of these basic roots.

In the YouTube demo
https://youtu.be/vwsipYJmzeE

all the roots X,Y of this equation have been computed when
X and Y are less or equal to 200 and the solutions (X,Y) have been
plotted in an XY coordinate system.
In this picture the points are scattered according to more or less
regular patterns.
For example the entire set of points can be covered by two
regular square grids as shown in that YouTube demo.

In the most interesting cases more than two regular square grids
with varying square size and orientation are needed for covering
the entire set of solutions.

When n and P are greater and certain rules related to divisibilities
of n and P are met (typically P is a prime number) very interesting
symmetric (kaleidoscopic) structures arise.
In these cases more than two regular square grids with varying square
size and orientation are needed for covering the entire set of
solutions. When points of solution belonging to the same grid are
indicated by the same color and each grid has its own color the
structure of the graph becomes clearer and visually more attractive.

The symmetries of the graph become more evident when it is plotted
for 0 <= X,Y <= 2*P .

In these demos the plotting speed is slowed down so that the user can
see a cumulative process where each layer (points belonging to the same
square grid) is strongly symmetric and covers the preceding status of
the graph.

Remark:
Since the powers X^n and Y^n for values like n=160, X,Y=1000 are really
huge, these values are not actually calculated. It is sufficient to use
modular exponentiation and calculate with residuals of X and Y (mod P).