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.
Question about Euler's conjecture. Understanding that the most general version of it has been disproved, does anything remain of it?
It looks like from Wikipedia that counterexamples have been shown for k equals 3, 4, 5, 7 and 8, but has it been shown that counterexamples exist for all k? Or is there some new modified conjecture stating that this limitations may exist for certain classes of k?