The first step will be stated as a lemma:

Lemma: x

^{3}+ y

^{3}= z

^{3}has a solution only if there exists p,q such that:

(a) gcd(p,q)=1

(b) p,q are positive

(c) p,q have opposite parity (that is, one is even, one is odd)

(d) 2p*(p

^{2}+ 3q

^{2}) is a cube.

(1) We can assume that

**x,y,z**are coprime.

(2) Since,

**x,y,z**are coprime, we know that at most one of them is even.

(3) We also know that at least one of them is even since if

**x**and

**y**are odd, then

**z**must be even.

(4) So, we now divide this proof up into

**Case I: z is even**and

**Case II: x is even**. Since

**x,y**are symmetrical,

**Case II**will also cover the case where

**y**is even.

**Case I: z is even**

(1) Then

**x,y**are odd.

(2)

**x + y**and

**x - y**are even.

(3) Let

**2p = x + y**and

**2q = x - y**

(4) Then:

**x = (1/2)[x + y + ( x - y )] = p + q**

**y = (1/2)[x + y - ( x - y )] = p - q**

(5) Now

**gcd(p,q) = 1**

(a) Assumef = gcd(p,q)is greater than1

(b) Then there existsP,Qsuch thatp = fP, q = fQ

(c) But thenfdivides bothx,ysincex = f(P + Q), y = f(P - Q)

(d) This is a contradiction sincex,yare coprime.

(6) We can assume that

**p,q**are positive

(a) From beforep = (1/2)(x + y), q = (1/2)(x - y)

(b)xcannot equalysince they are coprime.

(c) ifx + yis negative, then substitute(-x),(-y)sinceximplies that^{3}+ y^{3}= z^{3}(-x)^{3}+ (-y)^{3}= (-z)^{3}

(d) ifyis greater thanxthen flipx,ysince they are symmetric.

(e) This covers all cases.

(7)

**p,q**have to have different parities since

**x,y**are both odd.

(8) Finally,

**2p*(p**is a cube.

^{2}+ 3q^{2})**z**

= (x + y)(x

= (p + q + p - q)[(p+q)

= 2p(p

^{3}= x^{3}+ y^{3}== (x + y)(x

^{2}- xy + y^{2}) == (p + q + p - q)[(p+q)

^{2}- (p+q)(p-q) + (p-q)^{2}] == 2p(p

^{2}+ 3q^{2})**Case II: x is even**

(1) Then

**z,y**are odd since they are coprime to

**x**.

(2)

**z+y, z-y**are both even.

(3) There exists

**p,q**such that

**2p = (z - y), 2q = (z + y)**

(4) And

**z = (1/2)[(z - y) + (z + y)] = (1/2)(2p + 2q) = p + q**,

**y = (1/2)[(z + y) - (z - y)] = (1/2)(2q - 2p) = q-p**

(5)

**p,q**have opposite parity since

**z,y**are odd.

(6)

**gcd(p,q)= 1**[Same argument as Case I]

(7)

**p,q**are positive [Same argument as Case I]

(8)

**2p*(p**is a cube.

^{2}+ 3q^{2})**x**

= (z - y)(z

= [q + p - (q - p)][(q+p)

= 2p*(p

QED

^{3}= z^{3}- y^{3}== (z - y)(z

^{2}+ zy + y^{2}) == [q + p - (q - p)][(q+p)

^{2}+ (q+p)(q-p) + (q-p)^{2}] == 2p*(p

^{2}+ 3q^{2})
## 26 comments:

Just a small observation, but I thought I'd say it anyway. In the statement of the lemma, you state that 2p*(p2 + 3q2) is a square, though as I see it, it should be a cube. I'm very much enjoying the rest of the proof. Thanks for trying to make it so accesible for starting mathematicians.

Hi Sam,

Thanks for noticing the typo. I fixed the lemma.

I am very glad to hear that you are enjoying this blog! :-)

Cheers,

-Larry

Tiny Nitpick:

No end square bracket in the 4th line of (8) (z is even)

Rob

Rob,

Thanks for pointing out the typo. Just fixed it.

-Larry

Hello. I'm don't understanding the proof of p and q are positive, item 6, case z is even. Can you help me? Thanks.

Hi Juli,

p = (1/2)(x+y)

If x+y is positive, we are done. x+y ≠ 0 since x ≠ y (from #6b). If x+y are negative, then (-x + -y) are positive and we can substitute (-x)^3 + (-y)^3 = (-z)^3 so we can still assume that (x + y) are positive.

q = (1/2)(x-y)

From the arguments above, we only need to show that x-y is positive.

This is easy to do since x^3 + y^3 are symmetric and we can swap x,y without changing the equation.

In other words, we can assume that x is greater than y and therefore (x-y) is positive.

Please let me know if you have still have questions.

-Larry

Ok, Larry. I think I understood :) Thanks!

Hi Larry,

I read your blog for n=3 case of FLT and hand copied it on my note book: it needs over a dozen pages!

It works very well, of course :-)

But I have a question.

The heart of proof, I think, is the application of Fermat's Infinite Descent method. In your blog (May 22nd, 2005) you say that:

if a solution exists of the Fermat equation x^3 + y^3 = z^3

then

another solution will exist (e.g. A^3 + B^3 = C^3), and that solution will be smaller than the former.

Well.....

Inside the proof you say that the second solution will be smaller AT LEAST in one element [eg. (A less than x) OR (B less than y) OR (C less than z)].

This will work ok for the Infinite Descent method.

But, I say, what about oscillating triples?

Let me do an example.

(35,21,7) was a solution of the equation;

(31,28,19) is a smaller solution with respect to the first element (31<35);

(25,29,22) is the next solution, and it's also smaller than the second solution, with respect to the first element, 25<31;

(26,28,25) is the next solution, and it's also smaller than the third, with respect to the 2nd element 28<29.

So, I can write down as many triples I want: (40,25,30), (38, 27, 32), (36, 28, 35), ...

Note that every triple has ONE element smaller than the corrispondent element of the previous triple.

But this descent can really be infinite!

I don't understand....

Where is my mistake?

Thank you

Luigi (ITALY)

PS this blog is great!

Hi Luigi,

I hope I understand your question correctly.

I think the answer comes from understanding the modern equivalent of infinite descent.

In modern set theory classes, this idea is called the Well Ordering Principle.

The idea is that for any finite or infinite set of positive integers, you can always find the least one (for sets of negative values, you can always find the highest one)

In your example, if I understand you correctly, you are using negative integers as possible solutions (for example, it is possible to find an infinite number of negative answers to the pythagorean triples)

Infinite Descent really applies to positive values. If you can find an infinite number of smaller positive integers, then it implies that the set violates the well-ordering principle property of integers.

Real numbers and fractions, don't have this property.

I wrote a blog on it here.

I hope that answers your question.

-Larry

Hi Larry,

I can't explain my problem so well, I'm sorry. I didn't think to negative numbers.

I thought this:

what happened if the second triple has JUST ONE element smaller than the corresponding element in the previous triple (triple=solution of Fermat equation)?

Can the remaining two elements be larger than the corresponding elements in the previous solution?

For example, if (10,20,30) was a triple.

Can (15,25,28) be the next solution if I follow the "step 3" lemma?

In fact ONE element in the final triple is smaller than the corresponding element in the starting triple (28<30).

The next triple can be, I say, (20,30,10). And also ONE element in this solution is smallar than the corresponding element in the previous triple (10<28)

So, the next triple can be (21,22,15). And here also ONE element is smaller (22<30).

If I follow this method, I can write as many triples I want. Every triple has one element smaller (and some other elements larger) than the corresponding elements in the previous triples.

Without negative numbers I can write infinite triples with oscillating elements within them.

Luigi

PS I wander for your speed in replay! Thanks ;-)

Hi Luigi,

If you are not including negative numbers, then I am confused on how you can get an infinite number of solutions.

If in each case, at least one of the three is less than the previous one, then there is only a finite number of steps before you reach 0.

When you get to (x,y,0), it is not possible to find a solution that is smaller without going to negative numbers.

Am I missing something in your logic?

Thanks,

-Larry

Hi Larry,

I understand your answer.

But what about if:

in the first step the smaller element is the 2-nd: (x,y,z)->(A,B,C) with B<y;

in the next step the smaller element is (I say) the 3-rd: (A,B,C)->(A',B',C') with C'<C;

in the next step the smaller element is the 1st: (A',B',C')->(A'',B'',C'') with A''<A';

in the next step the smaller element is again the 3rd: (A'',B'',C'')->(A''',B''',C''') with C'''<C'' and so on

Hi Luigi,

If I understand your question, then the confusion regards the meaning of infinite descent.

At each point, you need to show that the least positive integer is decreasing.

If you have three values, you can't just pick A,B, or C, you would have to pick the smallest positive integer.

If you can show that the smallest positive integer in a set will decrease infinitely, then you have proven that there is no integer solution.

-Larry

Hi Larry,

Again, this is not the question.

I said that "step 3" lemma stated that if a solution exists, then a smaller solution will exist. And this solution will be smaller than previous only for ONE ELEMENT (at least).

Recursively, another smaller solution will exist. Again this solution will be smaller than previous only for one element. And so on.

My question is this:

1) The presence of a smaller element in a newer solution doesn't require that the other two elements will be in turn smaller. So they can be both larger than correspondig elements in previous solution.

2) In the next recursion step, the solution can be smaller than previous only in one element that can be other than the previous recursion step. For example: if in the first recursion step the 1st triple element was smaller than 1st element in the previous triple, in the second recursion step the 2nd triple element can be smaller than 2nd element in the previous triple (and the 1st element can be larger).

If you go on with this two arguments, you can build an infinitely long triple series. In each triple you can find an element that is smaller than the corresponding element in the previous triple (as required by your "step 3" lemma), but the other two elements will be larger.

Indeed your "step 3" lemma doesn't require that the smaller element in each triple is always the same (in each recursion step). It only states that in a single recursion step ONE ELEMENT is smaller. In the next recursion step ANOTHER ELEMENT can be the smaller.

I hope I explained my question.

Thanks for answers!!!

Luigi

Hi Larry,

thanks for your patience :-)

Read together the "step 1" lemma.

We have (x,y,z) that comply with the equation x^3 + y^3 = z^3.

In the proof, we make that triple primitive (i.e. pairwise coprime).

Then, we separate the proof into tree directions:

I) x:odd, y:odd, z:even, and then we prove that z^3= 2p(p^2+3q^2).

II) x:odd, z:odd, y:even, and then we prove that y^3= 2p(p^2+3q^2).

III) y:odd, z:odd, x:even, and then we prove that x^3= 2p(p^2+3q^2).

In "step 3" lemma we also have a triple (x,y,z) and we prove that exists another triple (A,B,C) that comply with the equation A^3 + B^3 = C^3. Moreover, this triple (A,B,C) is smaller than (x,y,z).

In details, we also have tree cases:

I) x:odd, y:odd, z:even, then z^3=2p(p^2+3q^2), then z^3>A^3, then z>A.

II) x:odd, z:odd, y:even, then y^3=2p(p^2+3q^2), then y^3>C^3, then y>C.

III) y:odd, z:odd, x:even, then x^3=2p(p^2+3q^2), then x^3>B^3, then x>B.

So we see that only one element in the triple (A,B,C) is smaller than the corresponding element in the (x,y,z) triple.

That element is NOT chosen by me, but by parity of the element in the (x,y,z) triple.

Suppose that we are in the case II, so y>C.

In the next recursion step we start from the (A,B,C) triple whose parity isn't the same of the (x,y,z) parity.

So, we suppose that, this time, we are in the case III, so B>B'.

Here is my problem.

B>B' doesn't imply that A>A' and C>C' too.

So A' and C' can be larger than A and C.

In the next recursion step (of the Infinite Descent algorithm), we can find, A''<A', but B''>B' and B''>B'.

And so on...

That's all. I think nothing say to me that the recursion steps will be these:

y>C

C>C'

C'>C''

C''>C'''

and so on.

What do you think?

Thanks

Luigi

Hi Larry,

there are you?

What do you think about my last post?

Luigi

Hi Luigi,

I'm sorry that I haven't been able to respond. I've gotten busy at work.

I'll try to post a response in the next day or two.

Cheers,

-Larry

Hi Larry,

Sorry for my hurry....

I'm waiting for you

Luigi

Hi Larry,

Here I am! (I think)

My mistake was this:

Let's suppose that (x,y,z) is the former triple and (A,B,C) is the next triple;

I thought that ONLY ONE of the A,B,C elements had to be smaller than ONE of the x,y,z elements.

Indeed, this is what really happens:

ALL OF THE NEXT TRIPLE'S ELEMENTS (A,B,C) are (all together) smaller than ONE of the x,y,z elements in the previous triple.

So, infinite descent goes!

I read more and more 1 and 3 lemmas to understand this, it was so subtle...

What do you think? I'm so satisfied! :-) :-)

Luigi

Sounds good, Luigi.

I'm glad it's clear now.

-Larry

Eh, eh, eh, dear Larry,

I love Number Theory. I wanted to know Euler ;-) What a great brain!

You know, Larry, your blog is the most complete resource I found about FLT in the web.

Thanks again, Luigi

hello sir ,

I have a silly doubt in the proof of this lemma which is when you are proving the gcd(p,q)=1,

you have contradicted the fact that x and y are co prime . Can you tell me how x and y are coprime

@Akshay,

Assume that x,y are not coprime so that there exist k > 1 and x=ak and y=bk.

It follows that k divides z since z^3 = x^3 + y^3 = k^3(a^3 + b^3) so that there exists c such that z = ck.

Then it follows that:

(ck)^3 = k^3(a^3 + b^3). Dividing k^3, we get: c^3 = a^3 + b^3.

We can keep on removing common factors until there is x',y',z' such that:

(z')^3 = (x')^3 + (y')^3 and z',y',x' have no common factors.

Since x,y,z only exists if x',y',z' exists, we can work with x',y',z' and assume that they have no common factors.

Hello:

Just saw your wonderful Blog. I have a question.

(x, y, z) is the starting triple.

What is the next triple? I see two new variables p, and q. But I do not see a third variable??

Thanks.

Guy.

Euler's proof of FLT for n = 3 is incorrect, please see paper with the same title which is available in Vixra.org/abs/1605.0123.author Quang Nguyen Van

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