Sophie Germain: mathématicienne extraordinaire - Scripps College

Sophie Germain’s Theorems 53
Cubes modulo 7
n n 3
0 0
1 1
2 8 ≡ 1
3 27 ≡ 6
4 64 ≡ 1
5 125 ≡ 6
6 216 ≡ 6
We can see that no n 3 (mod 7) is congruent to 3, so condition 1 is met.
2. Does x 3 + y 3 + z 3 ≡ 0 (mod 7) imply that 7|xyz Again, we check this
with a table.
Cubes modulo 7
x 3 y 3 z 3 x 3 + y 3 + z 3
0 0 0 0
0 0 1 1
0 0 6 6
0 1 1 2
0 1 6 0
0 6 6 12 ≡ 5
1 1 6 8 ≡ 1
1 6 6 13 ≡ 1
6 6 6 18 ≡ 4
The only instances when x 3 + y 3 + z 3 ≡ 0 (mod 7) occur when all of