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.