Fermat’s HUGE little theorem, pseudoprimes and Futurama


Welcome to a special Halloween edition
of Mathologer. Recently I did a video on Fermat’s mega famous Last Theorem
Fermat’s theorem says that the pretty Pythagorean integer identities like 3
squared plus 4 squared equals 5 squared or 5 squared plus 12 squared equals 13
squared don’t have nontrivial counterparts if the exponent 2 is
replaced by any integer greater than 2. Today I’d like to tell you about a
second theorem associated with the great Pierre de Fermat. It goes by the name of
Fermat’s little theorem and is also about powers of integers. Very important, at
least for people obsessed with movie maths like me and Marty, both of Fermat’s
theorems have featured in the Simpsons and Futurama. In 1995, around the
time the proof of Fermat’s Last Theorem was announced Homer stumbles across a
pretend counterexample to Fermat’s Last Theorem. A really fun maths troll by the
Simpsons writers that had millions of viewers reach for their calculators. Then
in the 2014 Simpsons Futurama crossover episode the equation at the core of a
Fermt’s little theorem is discovered by my alter-ego
Professor Frink. There in the upper left corner of the hatch in Bender’s head some
of you may not be familiar with the three-stroke equal sign and the mod
thingo in brackets. I’ll explain later but luckily we can
also express Fermat’s little theorem in simpler terms like this. Here the
vertical green line means “divides into” or “is a factor of”. So what this says is
that if you pick any prime number and any positive integer a then the prime
must divide the difference on the right. For a really simple example, let’s choose
our prime to be 3 and integer 4. Fermatis little theorem then tells us that 3
divides 4 cubed minus 4. Of course, that’s easy to check directly
since 3 divides 64 minus 4 which is equal to 60. But without doing the
arithmetic and just by looking at the expression on
the right it’s not at all obvious why this should be true, right? How can we be
sure that Fermat’s little theorem is always true, no matter what monster-sized prime
and integer we’ve chosen. And why should we care?
Of course mathematicians care simply because that’s the way we’re built. And
in a minute I’ll show you a really really really beautiful proof of
Fermat’s little theorem. But just in case that’s not sufficient incentive — though
it really should be — I also promise to show you two really cool application of
Fermat’s little theorem. Our first application is an unexpected and really
weird way to hunt down prime numbers something which is very important for
things like credit card encryption schemes. Along the way, we’ll also
encounter some of my favourite numbers, the mysterious firm our pseudo primes. sA
a second application I’ll turn Fermat’s little theorem into a tool that you can
use to solve really frightening Helloween problems like this one here, Oh
and we’ll also have some more Futurama 🙂 But first the proof. It turns out that to prove Fermat’s little theorem all we
have to do is to count some pretty necklaces. What do necklaces have to do
with any of this? Don’t worry, of course I’ll explain. As regular viewers are
probably now very used to I’ll focus on one example which is sufficiently
generic to illustrate the general proof. So let’s again choose the integer to be
4 and let’s prove that the prime 7 divides 4 to the power of 7 minus 4 in a
way that will also clearly work for any prime and any positive integer. Now what
we’re going to do is to count the number of necklaces of length 7 where each of
the 7 beads can be one of 4 colours. Except not all of the beads can be the
same colour. Here’s an example … and here’s another one. So it’s ok to not use all
the colours. The only thing we avoid is choosing all the beads to be of one
colour so that’s not allowed. Ok seven beads, four colours, not all the same,
okay? Also important: we consider rotated versions of a necklace to count as one
and the same necklace. So, all of these guys here they all are same necklace.
However, we consider two necklaces that are mirror images to be different as
long as they cannot be rotated to coincide. Okay we’ll show that the total
number of such necklaces is 4 to the power of 7 minus 4 divided by 7. Now of course the
number of different necklaces must be an integer, right? But for that to be the
case 7 must divide 4 to the power 7 minus 4 which is what we want to prove!
Sneaky huh? Ok time to begin counting necklaces. To do this we first cut each
necklace and straighten it out like this. Of course there
are other ways to cut the necklace giving rise to different strings. For
example cutting here gives this string here. There are 7 gaps between the beads
and so there are 7 possible strings that we can cut and these seven cuts
correspond to the following 7, and this is really really important,
7 different strings of beads. Now if we had started with all possible necklaces
and imagine cutting each of them in all possible ways then we’d obviously end up
with all possible strings and the total number of strings is very easy to count.
Each string has 7 beads and each bead can have any of 4 colors. So the total
number of different strings is 4 times 4 times 4, 7 times which comes to 4 to
power of 7 strings. Now, remember, we didn’t begin with absolutely all
necklaces, we left out the boring one color ones which also means we miss out
on the boring one color strings. How many of those are there. There’s 4 of those, and so our total number of strings is 4 to power 7 minus 4.
Almost there. Now, every necklace under consideration also gives rise to 7
different strings as in our example and that means that the number of necklaces
we began with is the number down there divided by 7. Tada that’s it. How sneaky
and how absolutely ingenious was that! Yes but something a little bit weird
here, right? You think you’ve got it? Yes what’s peculiar is that at no point did
I use the fact that 7 is a prime number. Hmm so where can our proof go wrong if we don’t use a prime number?
Well the problem is that if we replace our prime number 7 with a composite
number, then not all strings that we get by cutting a necklace are necessarily
different. And obtaining 7 different strings for each necklace was absolutely
critical for our counting argument. For example, if he use 9 beads instead of 7
beads one possible necklace looks like this. In this necklace things repeat
every three beads which means that when we cut this necklace in all possible
ways we only get three different strings rather than nine and this invalidates
the last step in our proof. I’ll leave it as a first challenge for you to show in
the comments that this sort of periodic necklace cannot occur if a necklace has
a prime number of beads and at least two colors. Oh and just in case you’re
wondering 9 doesn’t divide 4 to the power of 9 minus 4. But wait a minute
we’re actually onto something extra interesting here which means it’s time
for another t-shirt… Tada Now as I was saying, 9 not dividing 4 to
the power of 9 minus 4 is also interesting and leads to our first
application of Fermat’s little theorem. Fermat’s little theorem tells us that if 9
was prime then 9 would divide 4 to power of
9 minus 4. But we know that actually 9 does not divide 4 to the
power of 9 minus 4 so nine cannot be a prime. I know
Wow, incredible, stop the presses, right 🙂 However what this suggests is that we
can investigate whether or not some mystery number m is prime by swapping
m for 9 and checking for divisibility. And if the difference on the right is not
divisible by m then m is definitely not prime. On the other hand, what if the
difference on the right is divisible by m? Does this guarantee that m is prime?
Hmm, sadly the answer is `No’. For example, if we test m equals 4, then the
difference is definitely divisible by 4 since both terms on the right are
divisible by 4. But of course 4 is very much not prime. Bummer. Still 4 is a
pretty special number for this setup so maybe this is all worth a little more
exploration. So let’s do an experiment and let Mathematica check divisibility
for the first 1,000 integers. Let’s see how many non-primes and maybe primes
get discovered this way. However before firing up the machine I’ll make
an adjustment. Those powers here will get very big very fast there’s also nothing
special about using 4 in Fermat’s little theorem and we could use any integer
greater than 1. So let’s start at the start and keep those powers as small as
possible by replacing 4 with 2. Now we let Mathematica do its thing that’s
the output of my program and as you can see at the beginning it’s spot-on.
All the non primes 4, 6, 8, etc. fail our divisibility test and so Fermat’s little
theorem proves they are non-primes. In fact, it’s only when we get to 341
that we have a non-prime that our test thinks may be prime and there’s only three such numbers
under 1000. But we don’t have to put up with nonsense from those three
troublemakers. 2 was just one possible choice for our integer. So let’s throw
3 at them. And that works. Using 3 identifies 341 and 645 as non-prime numbers. One annoying number to go. So we can throw 4 at 561
and if that doesn’t work then 5 and then 6 and 7, and so on. Something
is going to work, right? Wrong!! It turns out that 561 is infinitely annoying. None
of our Fermat’s little theorem tests will show that 561 is composite. Numbers like
561 which Fermat’s little theorem cannot distinguish from primes are called Fermat pseudoprimes or Carmichael numbers, named after the American mathematician
Robert Daniel Carmichael. Actually they were missnamed since Carmichael was
neither the first to discover these numbers nor the first to seriously
investigate them. So maybe we should call them pseudo Carmichael numbers:) It
appears that the real discoverer of these numbers was a Czech mathematician
but Václav Šimerka. Anyway Carmichael numbers are very rare. Among the first
billion positive integers there are over 50 million primes but only 646
Carmichael numbers. Still it was proved in 1994 that overall they’re infinitely
many Carmichael numbers, really tricky stuff. Actually knowing about these
numbers and the fact that they are relatively rare is also of practical
importance. Being able to easily generate very large primes primes that have a
thousand plus binary digits is of key importance for all sorts of encryption
algorithms such as the algorithms that keep your credit card transactions safe.
Well, keeps them safe until some company stuffs things up, right:) And
despite the crazy powers that appear in Fermat’s little theorem and despite the
existence of those infinitely annoying pseudoprimes,
primarily tests based on Fermat’s little theorem turn out to be more useful than
pretty much anything else, making Fermat’s little theorem of huge practical
importance. Now before moving on to our second Fermat’s little theorem
application, let me just mention a characterization of these mysterious
pseudoprimes. This really beautiful characterization was discovered (well
before Carmichael of course 🙂 by the German mathematician Alvin Reinhold
Korselt. At first glance proving that a number like 561 is a pseudoprime looks
insanely difficult because it involves showing that 561 divides
the difference on the right for all infinitely many choices of the integer a.
However Korselt proved that if you know the prime factorization of a number then
you can pretty much tell at a glance whether the number is a Carmichael
number. Just subtract 1 from every one of the numbers in sight okay then our
number is a Carmichael number if and only if all the numbers on the left are
different and if they all divide the number on the right.
So number is a Carmichael number if none of its prime factors repeat and if all
its prime factors minus one divided the number minus one. Pretty amazing
characterization isn’t it? I love it. Okay an arithmetic challenge for you: use
Korselt’s characterization to prove that 1729 is a Carmichael number (Marty: no
calculators!) Marty’s got something with calculators 🙂 And now a quick pop culture
maths puzzle: why do you think the Futurama writers kept sneaking the
number 1729 into the program? Is it because it is a pseudoprime? And while
we’re here why are my initials BP on the Futurama spaceship Nimbus? Hmm. Now for our second Fermat’s little theorem application and time for yet another
t-shirt … Tada Okay, finally, to get into the Halloween
spirit let me show you how he can figure out by
hand what the remainder is when you divide the seriously monstrous number 11
to the power of 666 by the unlucky prime 13. For this we need to recast Fermat’s
little theorem into the form that appears in Bender’s head. What do we do?
Well we first take the common factor a out of the difference, like that. This
shows that if the prime o does not divide the first factor, our integer a,
then it must divide the second factor, right? Pretty obvious.
Okay now to our unlucky prime 13 and the base 11 of our monster integer. Of course
13 does not divide 11 and so Fermat’s little theorem promises us that 13
divides 11 to the power of 12 minus 1. In other words, when you divide 11 to the
power of 12 by 13 you get a remainder of .. what well? 1 of course. In maths lingo
we write that succinctly like this. Okay and what we are interested in is the
value of the pumpkin in this equation. We can sneak up on the pumpkin by starting
with our tiny 11 to the power of 12. Now comes the trick. Since 11 to the power of
12 has remainder 1 when divided by 13, so does any power of 11 to the power of 12.
We leave that for you as an easy challenge. On the left, we can pull the n
into the brackets and we’ve almost revealed our pumpkin. 12 times 55 is 660,
so if we choose n equal to 55, then we get this. Okay almost there. The power is still 6 short of our 666 but you can check by hand
that 11 to the power of 6 has remainer 12 on division by unlucky 13. Finally, we
multiply the left sides together and the right sides together to arrive at our final answer, a remainder of 1 times 12 which is equal to 12. Can you justify the last step of this calculation? Okay, last challenge for you:
what is the remainder of the super unlucky evil 666 to the power of 666
when you divide it by 13. I should warn you that if you don’t succeed in submitting
the correct answer in the comments by midnight on Halloween Jason will pay you
a visit. And don’t try to be smart, only one answer, otherwise I tell Jason. And
don’t use a calculator, otherwise I’ll tell Matty. And that’s it for today.
Happy Halloween!

100 Comments

  • I guess I have to start visiting people, since he didn't specify which Jason.

  • Oh wow, so is 1729 more interesting as a taxicab number or as a Carmichael number? I can't decide.
    "But you can check by hand that 11^6 has a remainder of 12 on division by 13" still using modular arithmetic I hope!

    For the challenge problem: I'll take it as given that mapping a number to itself modulo p (a prime number) is a homomorphism, that is:
    If a = b mod p and c = d mod p then a+c=b+d mod p and a*c = b*d mod p
    This implies that if a = b mod p then a^c = b^c mod p for any natural number c. Also, if 'a' and 'b' are not multiples of p then Fermat's little theorem says the exponent behaves like an element of the integers modulo p-1.
    Our goal is to evaluate 666^666 mod 13. Using arithmetic mod 13 we can replace the base with a smaller number, and with arithmetic mod 13-1 i.e. mod 12 we can replace the exponent with a smaller number.
    In mod 13: 666 = (2*3*3)*(37) = (18)*(-2) = (5)*(-2) = -10 = 3
    In mod 12: 666 = (2*3*3)*(37) = (18)*(1) = 6
    Also note that 3^3 = 27 = 1 mod 13
    Therefore 666^666 = 3^6 = (3^3)^2 = 1^2 = 1 mod 13

  • Whoops, let's not forget the first challenge problem: if a necklace has a prime number of beads and more than two colors then the p different rotations of the necklace are all different.
    For a contradiction, let's assume that for some number m strictly between 0 and p we can shift the beads by m places and get the same arrangement we started with. Let's label the positions 0, 1, 2, 3, …, p-2, p-1 and let's refer to the bead now at position 0 as B. When we rotate the necklace and shift every bead over m places we add m to all the position numbers, so B ends up at position m. And since we are assuming the necklace is symmetric with respect to this rotation the bead that started at position m must be the same color as B (otherwise the rotated necklace would look different). Also note that we should view the position numbers as elements of arithmetic mod p so that they wrap back around after a full rotation. Anyway, if we rotate by 2m then B ends up in position 2m, so the bead that started at 2m must also be the same color. In fact by this argument any bead separated from B by a multiple of m must be the same color, so if not all beads are the same color then there must be some position number n that B never ends up in. This is the same as saying m*x = n mod p has no solution. However, we can prove there is a solution using the fact that the greatest common divisor of two numbers is equal to a linear combination. Since p is prime and 0<m<p we know that GCD(m, p) = 1 so there must be some 'a' and 'b' such that a*m + b*p = 1. This means that a*m = 1 mod p, so going back to our equation m*x = n mod p we can multiply both sides by 'a' to cancel out the m and get x = a*n mod p. Therefore the bead B goes to every possible position under rotations by m, so every bead is the same color. This contradicts our starting assumption, so all rotations of the necklace must be different. QED

  • 666^660=1(mod13), 666^6=1(mod13) => 666^666=1(mod13). The answer is 1.

  • You just went full drive with the T shirts. Well where is the store for viewers to buy them and support Mathologer?

  • At 1.05, a cartoon shows Homer with a false equation of Fermat's last theorem. The numbers are very large, so you can't really calculate them on a hand calculator. However, the first term in the equation is certainly even. Ant the second term is certainly odd. So, if we add them together, the number will be odd. Yet , the term on the RHS is certainly even. So, the equation is false.

  • I had a feeling group theory was involved here. 11 is a cyclic generator modulo 13, hence the -1 (here 12) as the remainder of 11^6, and hence 11^666, on division by 13. 3, the reduction of 666 mod 13 (where 663 = 13 * 51), is not a cyclic generator mod 13 (where in fact 3 = 11^4), and hence 3^6 (and thus 666^666) would reduce to 1 mod 13. When we are given an integer T > S, where we are working modulo S, we replace T by (T mod S) the remainder of T upon division by S, call it N. We have effectively moved T beads along our necklace of S beads by instead moving N beads, where N < S.

  • Turkish subtitle, pls

  • So you don't know that 6107 ** 6 + 8919 ** 6 = 9066 ** 6

  • I'd love for my cousin to visit. So, 😛

  • I missed where you explained what the 3-lined equal sign means, or did you actually explain it?

  • All these pretty, intuitive, concrete proofs? Bah! All hail Lagrange's theorem! 🙂

  • 666^666 = ? mod(13)
    P | a (a ^ p-1 – 1) => 13 | 666 ( 666 ^ 12 – 1) => 666^12 = 1 mod (13)
    666^666 => 666^660 * 666^6 => 666^12*55 (1) * 666^6 (2) = ? mod (13)
    (1) 666^12 = 1 mod (13) = > 666^12*55 (1) = 1 mod (13)
    (2) 666^6 = 666^2 * 666^2 * 666^2 = 443556 * 443556 * 443556
    443556 = 9 mod (13) => 9 * 9 * 9 => 729
    729 = 1 mod(13)
    (1) * (2) = ? mod (13)
    1 * 1 = 1 mod (13)
    666^666= 1 mod(13)

  • I'm actually marginally more curious as to the number of Tflops that will be used on this problem between now and midnight October 31st Aussie time.

    Oh, God. What if it's a prime. ?

  • 666^6

  • I had a truly beautiful necklace proving Fermat's little theorem but my neck was too small to wear it.

  • Oh wow I came across Fermat's little theorem and proved a sub-case. For any n coprime to b, n divides b^(n-1)-1, and shows that 1/n is representable in a finite periodicity. It seems b^n -b is more reliable but still doesn't include values like b^2, though b would never be a square root of a prime, I ignored that fix because I wanted all integers

  • I both love ant hate this talent: as you said that we WILL proove it, I got the proof. Because there are 4 types of beads and 7 slots, 4^7, and 4 are excluded, and 7 rotations… But before that point it wouldn't click

  • Ah yes, Parker primes at last!

  • What about the last number at the end? 1010011010
    Looks like a binary number, but isn't byte sized(it's 10 bits, not 8)

  • 1729 is the smallest number that can be expressed as the sum of two different cubes: 10^3 + 9^3 and 1^3 + 12^3 … it is a reflection attributed to Ramanujan in an informal talk with another mathematician, I do not remembert the full details.

  • o_O What did he just say?

  • about the 666^666 mod 13, the answer is… 1

  • This is another mathologer video

  • Carmichael numbers are NOT a problem. I dont understand the hype surrounding them. If youre testing for primeness, using a base of 2, Carmichael numbers pass as possible primes. No big deal. But I imagine the first thing you would have noticed was notice 561 was odd, and not divisible by 2. If it was you wouldnt have wasted your computation time with the Little Theorem. Then comes the base 3. Do you perform Fermats Little Theorem? No. You would notice since youre using 3 anyways that 561 is a multiple of 3, and therefore not prime. How do we test for primeness then? First you choose a base. Test for divisibility by that base. If not divisible THEN use Fermats Little Theorem. If it tests as possible prime still then change the base. Voila. Carmichael numbers disappear and become irrelevant to the primeness testing algorithm.

  • 666^666 has remainder of 1 mod 13. Here is why:
    Since 13 is a prime, 13 is a factor of 666^13-666=666(666^12-1) according to Fermat's Little Theorem.
    Since 666/13=51 and 3/13, 13 cannot be a factor of 666.
    Therefore it should be a factor of 666^12-1.
    Or in other words, 666^12-1 has remainder 0 when divided by 13.
    This means 666^12 has remainder 1 divided by 13.
    Therefore 666^660 which is the same as (666^12)^55 has remainder 1^55 or 1 when divided by 13.
    Thus 666^666 has the same remainder as 666^6 which has the same remainder as 3^6 when divided by 13 since 666=13×51+3.
    Now 3^6=(3^3)^2=27^2
    This has the same remainder as 1^2=1 when divided by 13.
    Therefore 666^666 mod 13 is . . . 1.

  • I learned to prove Fermat's little theorem in a group theory course. It's a corollary of Lagrange theorem when applied to the multiplicative modular group. I don't know if your proof is equivalent or not.

    I also learned about its application to primality testing in another course I took in year 1, introduction to computer science

  • 36

  • First, notice that, since 663=51×13, then 666≡3 (mod 13), so we only need to calculate the remainder of 3^666 modulo 13.
    Now, since 3 and 13 are coprime, by little Fermat, 3^666≡3^6 (mod 13)≡729 (mod 13). But 728=13×56, so the value of the pumpkin emoji is 1.

  • 13|666(666^(13-1)-1)
    666/13=51+3/13
    666^12-1=0(mod13)
    666^12=1(mod13)
    (666^12)^55=(1(mod13))^55
    666^660=1(mod13)
    7|666(666^(6)-1)
    666/7=95+1/7
    666^6=1(mod13)
    (666^660)(666^6)=1(mod13)1(mod13)
    666^666=1(mod13)

  • if you take this as an integer solution then it's pretty easy :
    666/13=51
    666-(13*51)=663 =666-663 =3
    3 to the power of 3 = 27
    27 / 13 = 2 remainder 1

  • 0:48 Yeah, I saw what you did there 😉

  • Where can I get those t-shirts

  • 13 divides into 666^12-1 evenly, therefore 666^12%13=1
    666^660%13=1
    666%13=3
    666^3%13=1
    666^6%13=1
    666^660×666^6%13=666^666%13=1
    The answer is 1.

  • 666^666 = (666^12)^50 ≡ 666^6 ≡ 3^6 = (27)^2 ≡ 1^2 ≡ 1 mod(13)

  • 666ⁿ666 = 9 (mod 13). In other words 3 + 3 + 3 is the remainder. But you if you see them as digits and double them you get 666 !!!!!! Help!!

  • 17m 40s: Having reduced 11⁶⁶⁶ (mod 13) to 11⁶ (mod 13), we can, because 13 is prime, and (11⁶)² = 11¹² ≡ 1 (mod 13), conclude that:
    11⁶ ≡ ±1 (mod 13).
    This reduces the problem to a binary choice; I feel like there must be some shortcut that eliminates –1, but I don't see it (yet).

    OK, to your challenge:

    666 = 650 + 16 = 650 + 13 + 3 = 13·51 + 3, so 666 ≡ 3 (mod 13)
    So 666⁶⁶⁶ ≡ 3⁶⁶⁶ (mod 13)
    And because, by FLT, 3¹² ≡ 1 (mod 13), we can "cast out" all factors of 3¹² and get
    666⁶⁶⁶ ≡ 3⁶⁶⁶ ≡ 3⁶ = 27² ≡ 1² ≡ 1 (mod 13) – – – [using 27 = 2·13 + 1 ≡ 1 (mod 13)]

    That's my story, and I'm sticking to it!

    [Whew! I've just narrowly avoided a visit from Jason!! Or was it Marty? Or both??]

    Fred

  • mind blown

  • I survived. Cuz its November 1st and Halloween was yesterday.

  • 7,13,19

  • This guy is pretty awesome – he has a relaxed style, he is good-looking, and he has a cool accent.

  • where can i get that shirt

  • So now what if "a" is a matrix?

  • Hey, the Simpsons were only off by a mere 700,212,234,530,608,691,501,223,040,959 … or thereabouts. Hardly even worth mentioning! 😉

  • Hmm sounds like only genius have creative power to discover new math ideas I am not capable of

  • Look at a comment section like this and then from one from a Fortnite video.

  • Is infinity factorial is 0

  • Hi. Please discuss the mystery of Yin and Yang Pappus Chain where the radius of the main circle, divided by the radius of any (not just the main chain) small circle created inside it (as long as its tangent to 3 adjacent circles) will always end up to a whole number. See this PDF –> https://www.dropbox.com/s/0pi68drfmo6xpkl/Yin%20and%20Yang%20Pappus%20Chain.pdf?dl=0

  • 9:34 is this a voice-over?

  • Thank you for your videos. I enjoy them so much

  • Thanks so much for this terrific video – it will make a great addition to our school enrichment resources. Our Year 7 students learn about this theorem and it is challenging at first, so extremely helpful to have your teaching. I love the "why do we care?" discussion ?.

  • If an orchestra is playing a 7 note scale, and half the band is playing x notes for each scale degree and the other is playing y notes for each scale degree, and both groups are playing with the same tempo, how can someone calculate when the two groups will be playing the same note.

  • 8:24 It's a bit obvious, since the number is prime, it has no factoring numbers other than 1 and himself, and there is no expresion than could give back this number other than N·(string of 1s). In the case you took, the 9 being a square number it has factoring number 3, and it can be expresed as 3·3, that its: they are 3 equal strings.

  • spooky

  • I didn't know Herr Starr wears glasses.

  • 1

  • 3:53 he kinda sounds like Stewie when he says "seven"

  • I like you. You put an effort in pronouncing names of people of different languages correctly. My respects

  • That picture had me thinking wtf

  • I know this is off topic but I am hoping you will read this and make a video for one of your subscribers (Umm by that I mean me lol) I am hoping/requesting to pick your brain for the best way to play the Forex market from a mathematical point of view in hopes to figure out the safest risk to reward ratio without blowing ones account to the maximum safest way for a the highest positive expectancy ratio balancing max returns of course 😉 An example of what I am hoping for is idk 3% of account balance on any one trade for risk to reward of 1:3 or 1:10 obviously the higher would be most beneficial but I got no Idea as to approach it from strictly a math mathematical point a view ENTER THE Mathologer.

  • Please can you explain fractional derivatives and fractional integral step by step

  • Amazing Channel. I enjoy it a lot. Sadly, I finished my Computer Science course without the Discrete Mathematics elective. I chickened out and postponed it for the Masters. I watch your videos in hope I can switch my mind into gear for it. Thank you for sharing your ideas and knowledge!

  • I am a researcher and teacher of mathematics. I'm researching on fractals. Maybe you can help me. What is your email?

    Diego Teixeira

    [email protected]

    thank you

  • Pretty cool proof. So obvious as soon as he began. But cool to view it in that way

  • what is the biggest prime number?? can you make a video on prime number….!!!

  • best math teacher I ever came across

  • I didn't know 510 was divisible by 9…

  • 666^666

    = (2 * 3 * 3 * 37) ^ 666

    = (18 * 37) ^ 666

    1. 18^666

    18^12 # 1(mod 13)

    18^6 # 12(mod 13)

    => 18^666 # 12(mode 13)

    2. 37^666

    37^12 # 1(mod 13)

    37^6 # 12(mod 13)

    => 37^666 # 12(mode 13)

    (18 * 37) ^ 666 # 144(mod 13)

    = (18 * 37) ^ 666 # 1(mod 13)

  • I'd like to share to you my thoughts about infinity and division by zero, which I've been contemplating since my secondary school. I have come to some interesting conclusions and workouts, but it's a little too long to place it here just in comments, so I'd be grateful if you told me how can I send you my article. Then, I think, that could be quite an inspiration on releasing another Mathologer video covering that subject. Would be quite interested on you view of that subject and also its development. Please let me know how can I send you that stuff. Best greetings.

  • Sir can u pls pls find 3^3^3 this
    I know the exect answer but i don't how it comes like this

  • hi, is this breakthrough proves that set theory of cardinals is false? https://math.uchicago.edu/~mem/Malliaris-Shelah-CST-new.pdf. also article https://www.scientificamerican.com/article/mathematicians-measure-infinities-and-find-theyre-equal/

  • Hlo sir I have a little problem how can u prove that 0^0 =1
    Pls can u solve it

  • This is beautiful. And it helped me to understand how does the quantum factorization algorithm works (now I get why it's using phase estimation)!
    Thank you very much!!

  • I love when I do something and the exact same thing is done at the video, before I watched 4:30 to 8:43 I did the whole example with 9 and wondered the same thing

  • Can you do a video on the proof of the prime number theorem?

  • 1

  • I didn't know this proof! Very good!!!

  • Just Hello from Russia)

  • My major will be maths if have a chance to to have teachers like you. Best explainer. Thank you. I am just subscriber not student.

  • I came here to solve a programming problem..ended up wondering about the existence of humanity..great video. Subscribed.

  • Who came here after Socratica said search on pseudoprimes?

  • 14:29 That sound redundant. They will ofcourse be different always. Unless you mean the number cannot have any powers of primes greater than one. My intuition is that if you have different numbers and you subtract each of them by one or with any given constant, you will still have different numbers.

  • 666 does not divide 13.
    So 666 ^ 12 / 13 as a reminder of 1.
    and then 666 ^ 660 as a reminder of 1.

    666 ^6 = 87266061345623616
    87266061345623616 / 13 = 6712773949663355,0769230769230769
    Reminder is 0,0769230769230769 * 13 = 0,9999999999999997 = 1

    1 * 1 = 1 is the result !

  • 12:06–12:30 That is normal. Czech intellectuals are kinda neglected by the world…

  • 17:57 Do I get 2 guesses?

  • I am curious to know whether you are acquainted with David S. (X.) Cohen who left his pursuit of a PhD in Math at U.C. Berkeley to become a writer for the Simpsons and went on to executive produce Futurama. I assume he is a big reason why so much math appears in these shows (and perhaps he put your initials on the Nimbus as an easter egg just for you).

  • At 10:43 your table shows 9 as potentially prime. But with 2 as the constant this should be "not divisible -> not prime"

  • Mistake at 16:12: Just because a|bc and a doesn't divide b, a doesn't necessarily divide c. For example: 4|2*6

  • 11:44 that literally destroyed me XDDD

  • Why is there a 6 on the Prime Suspects shirt?

  • 1729=9^3+10^3=12^3+1 is de facto proof of Fermat Last Theorem by S. Ramanujan by showing cubed equality only missed by 1.

  • I’m sure you did it on purpose, but the credit card should end with 5, not 3 ?

  • 666≡3(mod 13)
    666²≡9(mod 13)
    666⁴≡81(mod 13)≡3(mod 13)
    666⁸≡9(mod 13)
    666¹⁶≡3(mod 13)
    666³²≡9(mod 13)
    666⁶⁴≡3(mod 13)
    666¹²⁸≡9(mod 13)
    666²⁵⁶≡3(mod 13)
    666⁵¹²≡9(mod 13)
    666=512+128+16+8+2
    666⁶⁶⁶=666⁵¹²666¹²⁸666¹⁶666⁸666²≡9*9*3*9*9(mod 13)≡(9*9)*3*(9*9)(mod 13)≡3*3*3(mod 13)≡27(mod 13)
    666⁶⁶⁶≡1 (mod 13) QED

  • You explain combinatorics better than anyone I’ve tried to learn from.

  • If 11^660 = 1 (mod 13) then why isn't 11^6 = 1 (mod 13)? Because 660=(6*110) and you said 11^12*n = 1(mod 13)?

  • Here're a few more candidates for you to evaluate.
    Best Confusing Science Toys That Will Definitely Confuse You! https://www.youtube.com/watch?v=wVslP9mvqjM

  • @9:34 Overdub?

  • This video spooked my neurons away.

  • Why was it called "Fermat's last theorem" before it was proven? A theorem is a proven statement of mathematics. You cannot call something a theorem if it isn't proven. In that situation it's a hypothesis or conjecture, not a theorem.

  • 561? Fer-merde!

Leave a Comment

Your email address will not be published. Required fields are marked *