3 ms·
Suppose the smallest (positive) number of rotations it takes to return is d. Then the numbers of rotations that have that effect are 0, d, 2d, 3d, etc. But clea
by gjm11 11y ago
Suppose the smallest (positive) number of rotations it takes to return is d. Then the numbers of rotations that have that effect are 0, d, 2d, 3d, etc. But clearly 5 rotations take you back where you started, so 5 is one of those numbers; that is, 5 is a multiple of d. Since 5 is prime this means either d=1 or d=5.
My second sentence above is a bit of a handwave. If you want to get formal about it, here's one way: suppose d isn't a multiple of 5; then note that there's an integer a such that ad = 1 (mod 5). Then ad rotations have the same effect as 1 rotation; but ad rotations do nothing, so d=1.
Now the handwaving is concentrated in the "then note that ..."; if that isn't sufficiently obvious, consider ad for a=0,1,2,3,4 and note that no two can be equal mod 5 because if ad=bd mod 5 then (a-b)d is a multiple of 5, but neither factor is a multiple of 5 and 5 is prime. So: five numbers, all of them different mod 5, so we must have one each of each congruence class mod 5; in particular, one of them is 1 mod 5.
- S4M 11y agoYes, that's a good explanation (I suspect your first paragraph is at the core of Fermat's little theorem's demonstration), it's just a shame it was left out of the video - but maybe because this part would require its own video.