I'm in the subject 'Matrices', but there still is a use of the Fields topic.
I was asked to find a system of equations with exactly 49 different solutions. (no further information included)
Now, I thought of the following system:
Z7 is the following field:
By looking at the system of equations, there are infinite possibilities. If we are limited only to Z7, then it means that there are 7 options for x, and 7 options for y, and together - 49.
(0,0), (0,1), (0,2), ... (0,6)
(1,0) . . . (1,6)
(6,0), . . . . . (6,6)
and so on...
Now, this is the subject of matrices, and I'm really not sure that it's right to use such ways to solve this problem.
Can you please help me?
In fact, in , 7, 14, and 48 are all congruent to 0, so your equations become just 3x= 0 and 6x= 0 (mod 7). While y can be any number, x must be 0. There are only 7 solutions, not 49.