[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:01.52,0:00:03.06,Default,,0000,0000,0000,,Voiceover: Bob discovered\Nsomething very interesting
Dialogue: 0,0:00:03.06,0:00:04.93,Default,,0000,0000,0000,,while making multicolored earrings
Dialogue: 0,0:00:04.93,0:00:07.20,Default,,0000,0000,0000,,out of beads for his store.
Dialogue: 0,0:00:07.20,0:00:08.97,Default,,0000,0000,0000,,Now, his customers like variety,
Dialogue: 0,0:00:08.97,0:00:13.25,Default,,0000,0000,0000,,so he decides to make every\Npossible style for each size.
Dialogue: 0,0:00:13.25,0:00:16.04,Default,,0000,0000,0000,,Starting with size three, he begins
Dialogue: 0,0:00:16.04,0:00:19.39,Default,,0000,0000,0000,,by figuring out all possible styles.
Dialogue: 0,0:00:21.76,0:00:24.68,Default,,0000,0000,0000,,Each earring begins as a string of beads,
Dialogue: 0,0:00:24.68,0:00:28.21,Default,,0000,0000,0000,,and then the ends are\Nattached to form a ring.
Dialogue: 0,0:00:29.10,0:00:32.95,Default,,0000,0000,0000,,So first, how many\Npossible strings are there?
Dialogue: 0,0:00:32.95,0:00:35.26,Default,,0000,0000,0000,,With two colors and three beads,
Dialogue: 0,0:00:35.26,0:00:38.78,Default,,0000,0000,0000,,there are three choices,\Neach from two colors.
Dialogue: 0,0:00:38.78,0:00:42.62,Default,,0000,0000,0000,,So two times two times two equals eight
Dialogue: 0,0:00:42.62,0:00:45.38,Default,,0000,0000,0000,,possible unique strings.
Dialogue: 0,0:00:45.38,0:00:46.89,Default,,0000,0000,0000,,And then he subtracts the strings
Dialogue: 0,0:00:46.89,0:00:49.94,Default,,0000,0000,0000,,which have only one color,\Nor monocolored strings,
Dialogue: 0,0:00:49.94,0:00:53.92,Default,,0000,0000,0000,,since he's only building\Nmulticolored earrings.
Dialogue: 0,0:00:53.93,0:00:56.72,Default,,0000,0000,0000,,Then he glues them all\Ntogether to form rings.
Dialogue: 0,0:00:56.72,0:00:59.82,Default,,0000,0000,0000,,He was assuming he would end\Nup with six different earrings,
Dialogue: 0,0:00:59.82,0:01:01.54,Default,,0000,0000,0000,,but something happened.
Dialogue: 0,0:01:01.54,0:01:05.11,Default,,0000,0000,0000,,He could no longer tell the\Ndifference between most of them.
Dialogue: 0,0:01:05.11,0:01:08.19,Default,,0000,0000,0000,,It turns out he only has two styles,
Dialogue: 0,0:01:08.19,0:01:10.97,Default,,0000,0000,0000,,because each style is now part of a group
Dialogue: 0,0:01:10.97,0:01:13.79,Default,,0000,0000,0000,,with two identical partners.
Dialogue: 0,0:01:14.64,0:01:19.71,Default,,0000,0000,0000,,Notice you can always match\Nthem up based on rotations.
Dialogue: 0,0:01:19.71,0:01:22.03,Default,,0000,0000,0000,,So the size of these groups must be based
Dialogue: 0,0:01:22.03,0:01:26.72,Default,,0000,0000,0000,,on how many rotations it takes\Nto return to the original.
Dialogue: 0,0:01:26.72,0:01:30.03,Default,,0000,0000,0000,,Or how many rotations to complete a cycle.
Dialogue: 0,0:01:31.41,0:01:33.56,Default,,0000,0000,0000,,So this means that the original set
Dialogue: 0,0:01:33.56,0:01:36.69,Default,,0000,0000,0000,,of all multicolored strings divides evenly
Dialogue: 0,0:01:36.69,0:01:40.50,Default,,0000,0000,0000,,into groups of size three.
Dialogue: 0,0:01:44.24,0:01:48.28,Default,,0000,0000,0000,,Now, would this be true for other sizes?
Dialogue: 0,0:01:48.28,0:01:50.08,Default,,0000,0000,0000,,That would be convenient,\Nsince he always wants
Dialogue: 0,0:01:50.08,0:01:52.14,Default,,0000,0000,0000,,the same amount of each style.
Dialogue: 0,0:01:52.14,0:01:54.62,Default,,0000,0000,0000,,So he tries this with four beads.
Dialogue: 0,0:01:54.62,0:01:57.54,Default,,0000,0000,0000,,First he builds all possible strings.
Dialogue: 0,0:01:57.54,0:02:00.44,Default,,0000,0000,0000,,With four beads he can\Nchoose from two colors
Dialogue: 0,0:02:00.44,0:02:02.22,Default,,0000,0000,0000,,for each bead, so two times two
Dialogue: 0,0:02:02.22,0:02:06.40,Default,,0000,0000,0000,,times two times two equals sixteen.
Dialogue: 0,0:02:06.40,0:02:10.16,Default,,0000,0000,0000,,Then he removes the two\Nmonocolored necklaces
Dialogue: 0,0:02:10.16,0:02:14.55,Default,,0000,0000,0000,,and attaches all of the\Nothers to form rings.
Dialogue: 0,0:02:14.55,0:02:17.80,Default,,0000,0000,0000,,Now, will they form equal sized groups?
Dialogue: 0,0:02:18.97,0:02:20.15,Default,,0000,0000,0000,,Apparently not.
Dialogue: 0,0:02:20.15,0:02:22.04,Default,,0000,0000,0000,,What happened?
Dialogue: 0,0:02:22.04,0:02:26.94,Default,,0000,0000,0000,,Notice how the initial set of\Nstrings divides into styles.
Dialogue: 0,0:02:26.94,0:02:28.78,Default,,0000,0000,0000,,If strings are of the same style,
Dialogue: 0,0:02:28.78,0:02:30.65,Default,,0000,0000,0000,,it means you can form one into the other
Dialogue: 0,0:02:30.65,0:02:33.00,Default,,0000,0000,0000,,simply by grabbing beads from one end
Dialogue: 0,0:02:33.00,0:02:36.67,Default,,0000,0000,0000,,and sticking them onto the other end.
Dialogue: 0,0:02:36.67,0:02:39.97,Default,,0000,0000,0000,,And there is one style\Nwhich only has two members,
Dialogue: 0,0:02:39.97,0:02:42.24,Default,,0000,0000,0000,,and this is because it's built
Dialogue: 0,0:02:42.24,0:02:45.41,Default,,0000,0000,0000,,out of a repeating unit of length two.
Dialogue: 0,0:02:45.41,0:02:50.25,Default,,0000,0000,0000,,So only two rotations are\Nrequired to complete a cycle.
Dialogue: 0,0:02:50.25,0:02:54.22,Default,,0000,0000,0000,,Therefore this group only contains two.
Dialogue: 0,0:02:54.22,0:02:58.81,Default,,0000,0000,0000,,He cannot split them into\Nan equal number of styles.
Dialogue: 0,0:02:58.81,0:03:00.35,Default,,0000,0000,0000,,What about size five?
Dialogue: 0,0:03:00.35,0:03:04.35,Default,,0000,0000,0000,,Will they break into equal\Nnumber of each style?
Dialogue: 0,0:03:06.88,0:03:10.40,Default,,0000,0000,0000,,Wait, suddenly he realizes he doesn't even
Dialogue: 0,0:03:10.40,0:03:12.90,Default,,0000,0000,0000,,need to build them in order to find out.
Dialogue: 0,0:03:12.90,0:03:15.93,Default,,0000,0000,0000,,It must work, since five cannot be made up
Dialogue: 0,0:03:15.93,0:03:18.51,Default,,0000,0000,0000,,of a repeating pattern, because five
Dialogue: 0,0:03:18.51,0:03:20.91,Default,,0000,0000,0000,,cannot be broken up into equal parts.
Dialogue: 0,0:03:20.91,0:03:23.32,Default,,0000,0000,0000,,It's a prime number.
Dialogue: 0,0:03:23.32,0:03:25.83,Default,,0000,0000,0000,,So no matter what kind\Nof multicolored string
Dialogue: 0,0:03:25.83,0:03:29.40,Default,,0000,0000,0000,,you start with, it will\Nalways take five rotations,
Dialogue: 0,0:03:29.40,0:03:33.44,Default,,0000,0000,0000,,or bead swaps, to return to itself.
Dialogue: 0,0:03:33.44,0:03:37.57,Default,,0000,0000,0000,,The cycle length of every\Nstring must be five.
Dialogue: 0,0:03:37.57,0:03:38.73,Default,,0000,0000,0000,,Well let's check.
Dialogue: 0,0:03:38.73,0:03:41.73,Default,,0000,0000,0000,,First we'll build all possible strings
Dialogue: 0,0:03:42.93,0:03:45.50,Default,,0000,0000,0000,,and remove the two monocolored strings.
Dialogue: 0,0:03:45.50,0:03:47.62,Default,,0000,0000,0000,,Then we separate the strings into groups
Dialogue: 0,0:03:47.62,0:03:50.26,Default,,0000,0000,0000,,which belong to the same style,
Dialogue: 0,0:03:50.26,0:03:53.40,Default,,0000,0000,0000,,and build a single earring for each style.
Dialogue: 0,0:03:53.40,0:03:56.60,Default,,0000,0000,0000,,Notice that each earring\Nrotates exactly five times
Dialogue: 0,0:03:56.60,0:03:59.00,Default,,0000,0000,0000,,to complete a cycle.
Dialogue: 0,0:03:59.00,0:04:01.94,Default,,0000,0000,0000,,Therefore, if we glued all\Nthe strings into rings,
Dialogue: 0,0:04:01.94,0:04:06.43,Default,,0000,0000,0000,,they must split into equal\Nsized groups of five.
Dialogue: 0,0:04:06.43,0:04:08.59,Default,,0000,0000,0000,,But then he goes one step further.
Dialogue: 0,0:04:08.59,0:04:12.57,Default,,0000,0000,0000,,Currently he is only using\Ntwo colors, but he realizes
Dialogue: 0,0:04:12.57,0:04:16.18,Default,,0000,0000,0000,,this must hold with any number of colors.
Dialogue: 0,0:04:16.18,0:04:18.84,Default,,0000,0000,0000,,Because any multicolored\Nearring with a prime number
Dialogue: 0,0:04:18.84,0:04:23.78,Default,,0000,0000,0000,,of beads, P, must have\Na cycle length of P,
Dialogue: 0,0:04:23.78,0:04:27.90,Default,,0000,0000,0000,,since primes cannot be broken\Ninto equal sized units.
Dialogue: 0,0:04:30.66,0:04:32.58,Default,,0000,0000,0000,,But if a composite\Nnumber of beads are used,
Dialogue: 0,0:04:32.58,0:04:35.90,Default,,0000,0000,0000,,such as six, we will\Nalways have certain strings
Dialogue: 0,0:04:35.90,0:04:39.12,Default,,0000,0000,0000,,with shorter cycle lengths,\Nsince it's actually
Dialogue: 0,0:04:39.12,0:04:42.14,Default,,0000,0000,0000,,built out of a repeating\Nunit, and therefore
Dialogue: 0,0:04:42.14,0:04:44.61,Default,,0000,0000,0000,,will form smaller groups.
Dialogue: 0,0:04:44.61,0:04:48.88,Default,,0000,0000,0000,,And amazingly he just stumbled\Nonto Fermat's Little Theorem.
Dialogue: 0,0:04:48.88,0:04:55.41,Default,,0000,0000,0000,,Given A colors and strings\Nof length P, which are prime,
Dialogue: 0,0:04:55.41,0:04:58.73,Default,,0000,0000,0000,,the number of possible\Nstrings is A times A
Dialogue: 0,0:04:58.73,0:05:03.28,Default,,0000,0000,0000,,times A, P times, or A to the power of P.
Dialogue: 0,0:05:03.28,0:05:05.72,Default,,0000,0000,0000,,And when he removed the\Nmonocolored strings,
Dialogue: 0,0:05:05.72,0:05:08.86,Default,,0000,0000,0000,,he subtracts exactly A strings,
Dialogue: 0,0:05:08.86,0:05:11.60,Default,,0000,0000,0000,,since there are one for each color.
Dialogue: 0,0:05:11.60,0:05:16.79,Default,,0000,0000,0000,,This leaves him with A to the\Npower of P minus A strings.
Dialogue: 0,0:05:16.79,0:05:19.05,Default,,0000,0000,0000,,And when he glues these strings together,
Dialogue: 0,0:05:19.05,0:05:22.23,Default,,0000,0000,0000,,they will fall into groups of size P,
Dialogue: 0,0:05:22.23,0:05:26.04,Default,,0000,0000,0000,,since each earring must\Nhave a cycle length of P.
Dialogue: 0,0:05:26.04,0:05:31.98,Default,,0000,0000,0000,,Therefore, P divides A to\Nthe power of P minus A.
Dialogue: 0,0:05:31.98,0:05:33.93,Default,,0000,0000,0000,,And that's it.
Dialogue: 0,0:05:33.93,0:05:37.52,Default,,0000,0000,0000,,We can express this statement\Nin modular arithmetic too.
Dialogue: 0,0:05:37.52,0:05:41.06,Default,,0000,0000,0000,,Think of it, if you\Ndivide A to the power of P
Dialogue: 0,0:05:41.06,0:05:45.32,Default,,0000,0000,0000,,by P, you will be left\Nwith a remainder of A.
Dialogue: 0,0:05:45.32,0:05:48.81,Default,,0000,0000,0000,,So we can write this\Nas A to the power of P
Dialogue: 0,0:05:48.81,0:05:53.04,Default,,0000,0000,0000,,is congruent to A mod P.
Dialogue: 0,0:05:53.04,0:05:54.78,Default,,0000,0000,0000,,And here we have stumbled onto one
Dialogue: 0,0:05:54.78,0:05:56.93,Default,,0000,0000,0000,,of the fundamentals\Nresults in number theory
Dialogue: 0,0:05:56.93,0:06:00.00,Default,,0000,0000,0000,,merely by playing with beads.