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).