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Welcome to another Mathologer video. Today we'll prove Fermat's Last Theorem. Well,
okay, not quite but we'll do something pretty cool, we'll explain what Fermat
definitely did proof and we'll investigate Euler's conjecture, a vast
and very natural but not very well known generalization of Fermat's super theorem.
Just as a reminder, Fermat's Last Theorem says that the equation over there has no
solutions in positive integers A, B, C as long as the exponent n is an integer
greater than 2, so not possible if n is 3 or 4 or 5, etc. That's plenty to puzzle
over but Euler "outFermats" Fermat. His conjecture goes further and also asserts
the impossibility of positive integer solutions to lots of closely related
equations like for example this one here. As if Fermat's last theorem wasn't hard
enough, right? :) Anyway more details later and back to Fermat. You often hear people
say that the famous French mathematician Pierre de Fermat who gave his name to the
famous theorem actually didn't have a proof. However this is not exactly
correct. True, we don't know of any complete proof by Fermat and probably none
ever existed. However Fermat did leave us with a proof
of a special case where the exponent is 4, which then also happens to take care
of infinitely many other cases of the theorem, namely those corresponding to
the exponents being a multiple of 4 like 8, 12, 16, and so on. And so I thought it
would be nice to tell you how a mathematician would go about figuring
out whether one of Fermat's of Euler's equations has positive integer solutions,
with the video culminating in the simplest Mathologerized proof of the n
equals 4 case, that one here, right. Okay here we go. My last video was about A
squared plus B squared equals C squared, Pythagoras's theorem. One remarkable
property of this equation is that it's actually possible to find positive
integers that satisfy this equation like, for example,
the super easy 3, 4, 5 and the pretty easy 5, 12, 13 and
the much less easy example up there. In fact, there are infinitely many of these
so-called Pythagorean triples. Not too long ago 3blue1brown did
a very nice video about how to construct all of them using a very
simple trick, well worth checking out. Anyway, the example up there together
with many other similarly crazy ones was actually already known to someone who
lived over 3500 years ago in ancient Babylon and who
wrote it down on his clay tablet. There where the arrows pointing. Pretty
mind-blowing isn't it? This and other clay tablets show that
the Babylonians did know about Pythagoras's theorem long before
Pythagoras was born and possibly also knew in principle how to make all
possible Pythagorean triples. Once you know that A squared plus B squared
equals C squared has positive integer solution it's also natural to ask
whether the same is true for closely related equations like this one here or
this one, so all cubed instead of squared, or going even further like this. All
right, well the first equation has infinitely many positive integer
solutions, the simplest one is this beauty here: 1 squared plus 2 squared
plus 2 squared equals 3 squared. The last equation also turns out to have
solutions. Here's one really unforgettable one, 3 cubed plus 4 cubed
plus 5 cubed equals, wait for it,... yes 6 cubed! I dare you to ever forget this
equation now that I've etched it into your mind :) What about the middle equation?
Here we're back with Fermat. So he famously wrote in the margin of one
of his maths books, claiming to have found a marvelous proof too long to fit in
the margin that there are no positive integer solutions to all these equations
here. Now of course everyone knows about Fermat's Last Theorem, right? Well, at least
everyone who watches Mathologer videos. However, not many people seem to know
that the mathematical superstar Leonhard Euler suspected that there was much more
to all this. Euler knew that it is possible to write certain cubes as the sum of
three cubes, like our fantastic 3, 4, 5, 6 example, the big brother of
the 3, 4, 5 Pythagorean triple. On the other hand, he was not able to
find any nice solutions to the corresponding higher order equations.
Well what's in the next red box. Well 3, 4, 5; 3, 4, 5, 6 and
so, if there is a god, then surely 3, 4, 5, 6, 7 is next, right? Sadly
this equation is false, so there's definitely something wrong with the
universe, right? However what we're looking for in this spot does exist and
here's the smallest example. So 4 forth powers adding to another fourth power.
Again Euler could not get any of the higher order equations to work and this
very compelling overall pattern along with this incredible intuition suggested
to Euler that none of the black equations can be solved with positive
integers.This is known as Euler's conjecture. So Euler's conjecture is really
a vast and very cool extension of what Fermat claimed to be true. It's even
cooler if you flip the way you look at it. To flip, we first focus just on the
equations involving cubes. Euler's conjecture then says that at least three
positive integer cubes are required to get a sum that is another integer cube.
Two cubes is definitely not enough. Similarly, focusing on fourth powers
order, Euler says that at least four fourth powers of positive integers are required
to get a sum that is another fourth power. In general our conjecture says
that at least n positive nth powers are required to get a sum that is
another nth power. Does this sound like it must be
true? Well, no, perhaps not to us mere mortals but the demigod Euler was
pretty convinced. For hundreds of years after Fermat and Euler many of the
greatest mathematicians tried and failed to come up with proofs of these
conjectures. In fact whole branches of mathematics were created in the process.
In this way, over time, Fermat's Last Theorem, but strangly not so much Euler's
conjecture evolved from an obscure who-really-gives-a-damn footnote into one of the great unsolved problems of mathematics. Now,
finally, in 1993 the mathematician Andrew Wiles announced a proof of Fermat's
Last Theorem. His proof turned out to be anything but short and when the final
version was published his proof amounted to over a hundred pages of really
high-powered mathematics. Have a look. Alright, anything look familiar?
Don't worry if not. You're not the only one who finds this all scary and totally
incomprehensible. Actually, I doubt that there's more than about 30
mathematicians worldwide, all incredibly smart people who would have studied and
really understood every detail of Wiles's proof, and I'm definitely not one
of them. His proof was really, really big news worldwide at the time it was
announced. But then, just as the world was moving on from learning that Fermat's Last
Theorem had finally been cracked Simpsons fans were treated to a great episode in
which Homer Simpson stumbled across what appeared to be a counterexample to
Fermat's Last Theorem, the equation on the left hovering menacingly over Homer's
head. Could it be true? Well this was 1995 and so many viewers who checked with
their 1995 pocket calculators concluded that the left and right sides of this
equation were equal. Fermat was false and Wiles's proof was clearly rubbish,
right? Well, not so fast, using our 2018 calculators we find that
the left and right sides of the equation pan out to be these monster numbers here.
The two monsters have the same number of digits but on close inspection only
coincide in the first eight digits, which of course was enough to fool many 1995
calculators. Nicely played Simpsons :) Actually, as many cluey Simpsons fans
noted, it's easy to see at a glance that this equation cannot possibly be true.
Why? Because the first term is even, the second one is odd, even plus odd is odd
which means that the left side is odd. However, the right side is clearly even
and so the equation cannot be true. However the Simpsons writers weren't
quite ready to give up. In a later episode we see Homer scribbling another
would be Fermat buster on a blackboard. There it is. In this case the odd-even trick
doesn't help since Homer's equation reduces to odd plus odd is equal to even,
no contradiction. Again nicely played Simpsons :) However the game isn't over yet.
Remember all this is a gentle introduction to my main goal today to
show the simplest possible proof for the simplest case of Fermat. In a couple of
spots this proof makes use of a divisibility by 4 trick. So let's now
have a bit of a warm-up for the hard core part of this video, by using
divisibility by 4 to show that this Simpsons equation cannot be true. First,
when you divide an integer by 4 what are the possible remainders? Well you all
know this. If the number is even then the remainder is either 0 or 2 and if
the number is odd the remainder's either 1 or 3, easy right? Now what about
squares of fourth powers or other even powers such as the 12th powers that just
appeared in the Simpsons clip. What are the possible remainders of such an even
power? When you take an even power of an even number the result is obviously
divisible by 4 and so 0 is the only possible remainder. What about an
even power of an odd number. This will give us an odd number and so the only
possible remainders are still 1 and 3. However, it turns
out that for an even power of an odd number
the only possible remainder is 1. I leave it as an easy puzzle for you to show
that this is the case in the comments. Okay with this divisibility weapon in
hand let's have another look at the second Simpsons equation. So we have two
even powers of odd numbers on the left which both have remainder 1. This means
that the sum on the left has remainder 1 plus 1 is 2. However we also know that
the even power on the right side has remainder of 0. This leads to 2 equals
0 and so Homer's equation must be false. Another puzzle for you: Show how
the second Simpsons equation can also be torpedoed using divisibility by 3. What I
want to do now is to show you how mathematicians would go about analyzing
whether equations like this one or this one, or this one have integer solutions.
Of course we already know that the second equation has tons of solutions
but the first equation doesn't have any although we have not seen a proof that
the first is impossible. To build a bit of suspense let's analyze this equation
here which is a mix of both. We still have all even exponents but there are
4th powers on the left and a square on the right. Does this equation have
integer solutions? Well to build some intuition and to perhaps luck out I'd
start with a bit of trial and error, just substitute some small numbers for X and
Y and let's see whether this sum becomes an integer square. In fact, once you do
this you immediately notice that we do get solutions if one or both of X and Y
are equal to 0 for example 0 to the power of 4 plus 1 to the power of 4 is equal
to, obviously, 1 squared. Actually this works for all the Fermat like
equations that we are considering today. Of course these solutions aren't terribly
exciting and we usually exclude them by requiring solutions in positive integers.
What's next? Well this is the 21st century and so let's do a computer search.
What we ask the computer to do is basically the same as what we just did
but more systematically. Let's solve for Z, so square roots on both sides. Now
systematically try all possible choices for X and Y starting with small numbers.
First, okay not an integer so X is equal to 1 and Y is equal to 1 are not
part of a solution. Next, doesn't work either. Next, next, next, next, next, and so
on. Obviously if there exists an integer solution, then this procedure will
eventually find it, though it may take a zillion years or so. In fact, as soon as
computers became available people tried this sort of computer search for many of
the equations that Fermat and Euler were interested in, and they struck
gold. Here's one of the shortest papers in the history of mathematics, a counter-
example to one of the cases that make up Euler's conjecture, appearing about 200
years after Euler began pondering. Now let's see where it fits in. Okay right
there in the lower right corner four 5th powers adding to a 5th power and so
the full Euler conjecture is dead it really only takes one counterexample to
kill a conjecture. And how about the handsome hero who discovered it? Here he
is the first supercomputer, the CDC6600.
Those were the times :) However, as far as I know the CDC 6600 and it's successors
only managed to kill off one more instance of Euler's conjecture and so
there's still plenty of room for the next Andrew Wiles to stake their claim
to greatness. Anyway, back to our analysis. So let's suppose we've had the computer
running for a year or two and still no solution has appeared. Perhaps it's time
to start suspecting that maybe just maybe there may not be a
solution after all. Now if you are a mathematical consultant for the Simpsons
you're probably beginning to panic what with the deadline for the episode looming
and still no solution to our equation. In which case the best alternative is to go
for a prank solution like the ones I showed you. So amongst all the computer
printouts you find a convincing near-miss solution for this equation and
attempt to pass it off as the real thing. Okay
here is my attempt at helping out the Simpsons in this respect. I'll leave you
to power up your calculators and see how close I got. On the other hand, a
mathematician who has a whole lifetime to play with this problem might have a
serious go at showing that there is no solution, and there's a common strategy
for trying to piece together a proof by contradiction. We begin by assuming that
there is at least one solution to our equation and that the specific numbers
blue X, blue Y and blues Z form the first of the solutions that our computer
program would eventually discover. Starting with this supposedly smallest
solution we now follow our mathematical nose and explore the properties of such
a solution and also what noteworthy consequences follow from this solution. In
the case of this particular equation the mathematician John Cassels came up with
this very short and very elegant chain of implications. So here we go.
Right. Now this chain culminates in a second, smaller solution to our equation
down there. However, and this is the punchline, on closer inspection it also
becomes clear that our computer search would have revealed this second smaller r s u
solution before the supposedly first X Y Z solution. Obviously this is completely
impossible and the only way to resolve this contradiction is to conclude that
there's no solution to our equation to begin with. Tada, proof by contradiction
complete! Well complete except for justifying the seven scary implication
in the proof. We'll get to that but first it's time for a
little confession :) The fact that my building-suspense equation has no positive
integer solutions is actually exactly what Fermat proved, although he followed
a much windier path. Fermat also noticed that proving this
impossibility is just one baby step away from what we really want to prove namely
that X to the power 4 plus Y to the power 4 is equal to Z to the power 4 has no
positive integer solutions. Before returning to the long chain of
implications, let me indicate the final baby step to show that this equation has
no positive integer solutions and we'll round off the easier part of the video.
Okay, if there was a solution to this equation, that one, then we could rewrite
it like this, right? Rename the Z squared in the brackets
U and in this way we would have produced a solution to my building-suspense
equation with no solutions and this means that the hypothetical solution of
X to the power of 4 plus Y to the power 4 equals Z to the power 4 that we
started with cannot exist either. In other words this special case of
Fermat's Last Theorem cannot have any positive integer solutions. I'll leave it
as an easy puzzle for you to show, using the same trick, that the same is true for
all equations with exponents that are multiples of 4. In general, once you
succeed in showing Fermat's Last Theorem for some particular number in the
exponent, all the multiples of this exponent are also taken care of. So, for
example, Euler proved Fermat's Last Theorem for the exponent 3 and in this
way also disposed of the cases 6, 9, 12, 15, etc. This also means that for a complete
proof of Fermat's Last Theorem it's enough to take care of all the odd prime
power exponents and exponent 4. In fact at the time that Wiles announced his
proof, all prime number exponents up to 125000 had already been taken care
of. So quite a bit had been done. Anyway, as
promised, I did show you the simplest proof of this impossibility by showing
you THIS :):)
But, obviously, if this was really all I did, I'd be in for lots of downvotes, hell
I'd downvote the video myself. Anyway to give this video some real substance, let
me now explain this argument to you as best as I can. If that all looks too
scary a trip that's cool. Just push the YouTube eject button now and find an easier
maths video to watch for the moment and I'll see you next time. ++++++Intermission+++++++++
For you brave souls willing to take the trip, good on you :)
Buckle your mathematical seatbelts and let's get going. Ready? To begin let me
introduce you to the three main weapons that will help us with the proof. First
we'll be using our Simpson's fact that after dividing by 4 any even power of
an odd number must have remainder 1. The second weapon is an easy one. Suppose
you have a three-term equation like this. Then if any two of the terms have a
common factor, the third term must also have the same common factor. For example,
if both D and E on the left are divisible by 3 then F also has to be
divisible by 3. Pretty obvious, right? Our third and final weapon is tricker to
explain. Let's begin by carefully considering
some fourth power such as this one here. Then it's pretty easy to see that in its
prime factorization all the powers of the individual primes have to be fourth
powers themselves. In this example 2 to the power of 4 and 5 to the power
of 4 are obviously numbers that are fourth powers themselves. 3 to the
power of 8 in the middle is, too because we can write it like this. Now
suppose that somebody gives you two mystery numbers whose product is equal
to our fourth power and very important those two numbers are relatively prime,
so have no common factor apart from 1. Then we can conclude that our two
mystery numbers are also both fourth powers. Why because the prime factors of
one kind, say all the 2s in our original number must all go into just
one of our mystery numbers, for example, like this, or like that. Of course, this in
turn implies that both P and R are fourth powers themselves. Unfortunately we will
need to use a slightly trickier version of this principle, where the highest
common factor of P and R is exactly 2. What can we then say about P and R? Well,
the highest common factor being 2 implies that the odd primes must split
up just as before, with all the primes of one type in either P or R, for example
like this. Now what about the 2s. Hmm because the highest common factor is 2
both P and R must have a factor of 2 but then one of P and R must have exactly one
2, otherwise the highest common factor would be at least 4. Now it follows
that one of our factors let's say P must be of the form 2 times a fourth power
and the other factor R is 8 times a fourth power and this is true in general.
Oh and there are two more things that we can say for certain about these new
fourth powers, the solid red and the solid green bits. First, the two new
fourth powers are relatively prime. Second the red fourth power that goes
with the 2 must be odd. So to summarize, if an integer fourth power is the
product of two integers with highest common factor 2, then one of the
numbers is of the form 2 times an odd fourth power and the other is of the
form 8 times a fourth power. Furthermore both new 4th powers are
relatively prime. Okay, so that was the hardest bit. We just have to carefully
apply all three weapons now to construct our proof by contradiction. The first
thing that follows from our hypothetical blue solution being the smallest
solution is that any two of the numbers X, Y and Z are relatively prime, so X
and Y are relatively prime, X and Z are relatively prime and Y and Z are relatively prime. Why?
Because if this was not the case and two of the numbers had a non-trivial common
factor, say 3, then all three terms would have to be divisible by 3 to the power
of 4. Then we could divide the whole equation
by this common factor to the power of 4 like this. Okay, it's all pretty
obvious, right? And this would give a new solution that is smaller than the
smallest solution we started with, which of course is impossible. Anyway
let's say it again, any two of X, Y, and Z in the hypothetical solution we started
with have to be relatively prime. Okay, step two. Notice that not all three terms
can be odd. Why? Because if all three were odd
we have odd plus odd is even on the left and odd on the right which is impossible
of course. And this means that at least one of the terms has to be even. In fact,
only one of the terms can be even since the three terms are relatively prime.
Okay, so conceivably there are three different distributions of odd and even
among these terms. The first distribution is exactly the same as that in the
second Simpsons pseudo counterexample but then we can conclude in the same way
as earlier, using our division by a 4 weapon, that this distribution of odd and
even is impossible. This leaves us with two possibilities. Are you already regretting
following me on this number theory quest? Well it's too late now, you
can't leave :) In the following I'll assume that X is even. If Y is even, we just have
to exchange the roles of X and Y in everything that follows. Okay, anyway
let's color code X, Y and Z. So yellow for even and orange for odd. Now let's solve for
X to the power of 4 and, like we learned in high school, write the
difference of squares on the left side as a product like this. Now I want to
show that the two components Z minus Y squared and Z plus Y squared have
highest common factor 2 to be able to apply the fancy weapon that we forged
earlier. It's clear that 2 is a common factor since the orange Z and Y are
both odd. Now to show that there is no higher common factor remember that any
common factor of two numbers is also a factor of their sum. In this case the sum
of our two components is this. So the Y squared's cancel and the Z plus Z is
2 Z. This means that any common factor of our two components also has to
be a factor of 2 Z. However any common factor of the two components also has to
be a common factor of their product X to the power 4 on the right. But at the
same time X and Z are relatively prime which means that the highest common
factor of the two components is 2. pretty tricky ! :) Okay and this is
exactly the situation we prepared for. We are looking at a fourth power which is a
product of two integers with highest common factor 2. So one of the factors
is 2 times the fourth power and the second factor is 8 times another
fourth power or the other way around. Our weapon also guarantees that U and V are
relatively prime and that u the power following the 2 is odd, so let's color
u orange for odd. Summarizing we have the following two cases. Fancy transition :)
Okay let's focus on the first box. Subtracting the second equation from the
first the Zs disappear and we're left with this. Then dividing by 2 gives this
here. Alright, nice. If you do the same with the other box we get that one here.
Now it's easy to see that the second equation is impossible. Why, well remember
since Y is odd, then after division by 4, Y squared leaves a remainder of 1
and the same is true for u to the power of 4. On the other hand, 4 times v to the
power 4 in the middle is divisible by 4 and so it gives a remainder of 0. And so
in total we get remainder 1 on division by 4 is equal to remainder minus 1 which
is impossible. The only way to avoid a contradiction is for the first equation
to be true. Again pretty much following our nose, we solve
for 4v to the power of 4, write the left side as a product again.
Now since u and v are relatively prime, remember that one, we can argue as before
that the two factors in the product on the left have highest common factor 2,
and then that both factors necessarily have to be of the form 2 times a
fourth power. Okay consider nutting out the details to be your, don't know,
don't remember, probably your third or fourth puzzle :) Anyway great! And believe it or
not we're almost there. Add the bottom equation to the top. This gets rid of the
Y's. Divide by 2 and we've got our second equation and so our hypothetical
first solution on top implies the existence of a second solution. It's also
not hard to see that both r and s have to be smaller than the X at the top. And
that's the punchline. Our computer program would have come across the
second solution before the first solution. That isn't possible and
completes the proof by contradiction. And this is real sweat, you know :) So, anyway
this is very slick, very pretty and also very tricky, right? Now this was the
simplest proof of the simplest case of Fermat Last Theorem, explained as simply as
possible. Still very tricky to understand. Now multiply this by at least a
million and you get an idea of the trickiness of Wiles's proof.
Well maybe I'll do that next week. No, just kidding, maybe in a million years. Anyway
I hope you enjoyed this video as much as I enjoyed putting it together. If there's
anything you did not understand please leave a comment. Even if I don't get
around to answering, there are a lot of very cluey people roaming the comment sections
of these videos who will be very happy to help you. Anyway this is it for today.