Tuesday, May 17, 2005

Fermat's Last Theorem: n = 4

The easiest proof for Fermat's Last Theorem is the case n = 4. The proof for n=3 is a bit tricker. What is nice about this proof is that it arises quite naturally from the solution to Pythagorean Triples but it also proves Fermat's Theorem for all values n > 2 where n is even if we can prove Fermat's Last Theorem for all values n > 2 where n is odd. In fact, we can use the same time of reasoning to prove for all values n > 2 where n is not a prime if we can prove it true for all values n > 2 where n is prime.

Many textbooks say Fermat himself published this proof. This is not completely true. Fermat published a proof showing that a right triangle cannot have its area equal to a square. Fermat's proof does by implication show that there is no solution to n=4 but it is a bit more complicated than the proof that I am about to show.

Theorem for FLT/n = 4: There are no integer solutions to:
x4 + y4 = z2 where xyz ≠ 0

(1) We can assume that x2,y2,z are coprime. [From here since (x2)2 + (y2)2 = z2]

(2) From the solution to Pythagorean Triples, we know that there exist p,q such that:
x2 = 2pq
y2 = p2 - q2
z = p2 + q2

(3) Now from this, we have another Pythagorean Triple since y2 + q2 = p2.

(4) So, there exist a,b such that:
q = 2ab
y = a2 - b2
p = a2 + b2
a,b are relatively prime.

(5) Combining equations, we have:
x2 = 2pq = 2(a2 + b2)(2ab) = 4(ab)(a2 + b2)

(6) Since ab and a2 + b2 are relatively prime, we know that they are both squares. [See here for the proof.]

(7) So, there exists P such that P2 = a2 + b2.

(8) But now we have reached infinite descent since:
P2 = a2 + b2 = p which is less than p2 + q2 = z which is less than z2.

NOTE: if xyz=0, then this argument does not hold. This proof only works for xyz ≠ 0.

(9) So, the existence of a solution to the initial equation leads necessarily to the existence of another smaller square that has the same properties.


The case for n=4 is then stated as a corollary.

Corollary for FLT n = 4: There is no solution to:
x4 + y4 = z4 where xyz ≠ 0

Since this equation is equal to:
x4 + y4 = (z2)2

Corollary for FLT n divisible by 4: There is no solution to:
x4n' + y4n' = z4n' where xyz ≠ 0 and n' = (n/4)

Since this equation is equal to:
(xn')4 + (yn')4 = (z(2n'))2

Corollary for FLT n > 2: FLT is proven if FLT is proven for all cases where n is a prime.

The lesson learned from all this, is that we only need to proof that Fermat's Last Theorem is true for prime values of n.

If we prove that Fermat's Last Theorem is true for a given prime number, then it follows that it is true for any number which is divisible by that prime. For example, if we prove that there is no integer solution for x3 + y3 = z3, then we have likewise proven that there is no solution for (xi)3 + (yi)3 = (zi)3 = x3i + y3i = z3i.

This would also prove that x9 + y9 = z9 has no nontrivial integer solution.


Anonymous said...

Is it also true that u^2+x^4=y^4 has no solutions? I have found none up through {x,y} <= 1000, but the methods given here don't seem to result in a proof, nor do several variations I've tried.

Larry Freeman said...

u^2 + x^4 = y^4 is the same as:
u^2 = y^4 - x^4

Interestingly, this is the only proof that Fermat ever published.

You are right. There is no solution where xyz ≠ 0. Here's the proof:

Anonymous said...

I can't understand why you begin working with the equation x^4+y^4=z^2, why you don't begin directly with x^4+y^4=z^4?
If it isn't a typo, please, explain it to me.

Larry Freeman said...

Hi Alvaro,

Good question.

The main reason is that x^4 + y^4 = z^2 is more general than x^4 + y^4 = z^4. No solutions in x^4 + y^4 = z^2 implies no solutions in x^4 + y^4 = z^4 but not the other way around.

The theorem is therefore stated in the most general way.


stef said...

Hi Larry,

Love the blog but just wanted to clarify a point.

Your simple proof for x^4 + y^4 = z² proves all cases for x^n + y^n = z^n when n≡0mod4

i.e. You give as a corollary "Corollary for FLT n > 2 and even: There is no solution to:
x^2n + y^2n; = z^2n; where xyz ≠ 0" .((why couldn't you have just said Corollary for FLT : There is no solution to:
x^4n + y^4n = z^2n where xyz ≠ 0" as this is equal to (x^n)^4 + (y^n)^4 = (z^n)^2 ?))

It does not prove the case for n≡2mod4. (You say at the top of the page "it also proves Fermat's Theorem for all values n > 2 where n is even."



Larry Freeman said...

Hi Rhuaidhri,

You are correct that the argument here does not prove that FLT is true for n > 2 if n is even but only if n ≡ 0 (mod 4).

I have modified the wording of this article to make this point clear.

Thanks very much for noticing the incomplete sentence.



Rhuaidhri said...

Hi Larry,

I reread your proof and it leaves out a number of steps (they may not be obvious to everyone). Shouldn't it complete a loop by showing the assumption x^4 + y^4 = z^2 leads to a smaller triple of the form a^4 + b^4 = c^2 not just a smaller triple of form a^2 + b^2 = c^2 ?

I have a number of proofs for various things to do with FLT and the Pythagorean triples, most of them are pretty short but rule out alot of possible solutions, if you want them posted let me know.



Larry Freeman said...

Hi Rhuaidhri,

I think that the proof stands as it is without any extra steps.

The main idea is that if x^4 + y^4 = z^2 has nontrivial integer solutions, then there exists integer q,p such that y^2 + q^2 = p^2.

But this is imposible since q,p,y have certain properties that integers cannot have. That is, they lead to an infinite descent.

Since q,p,y cannot exist, it follows directly that x^4 + y^4 = z^2 cannot have any nontrivial solutions.

I hope that this comment makes the main idea more clear.



Ralph said...

Your proof is incorrect. You use the parametrization for PRIMITIVE Pythagorean triples. It is not general.

Larry Freeman said...

Hi Ralph,

The proof is correct. If there are any general solutions, then there must be primitive solutions (see here).

In the proof, I show that there are no solutions for a primitive Pythagorean triple. Using the logic in the link above, it follows that if there is no primitive solution, then there cannot be a general solution.

Please let me know if I am misunderstanding your question.


Scouse Rob said...

It took me a long while to figure that last corollary out.

To make it clearer you could perhaps state that the the N=4 case solves N for all powers of 2 (that are greater than 2) due to N then being divisible by 4.

This only leaves primes greater than 3 and composites composed of them. So only the primes need to be solved.

Enjoying the blog, although I'm only at the begining.


Larry Freeman said...

Hi Rob,

Thanks very much for your comment. I have updated the last corollary in order to make it more clear.

Please let me know if you still have questions.



Scouse Rob said...

I think this sentence is wrong:

"but it also proves Fermat's Theorem for all values n > 2 where n is even"

Is it not proved only for multiples of 4?

(ie N=10 needs a proof for N=5 etc)


Larry Freeman said...

Hi Rob,

Thanks for the comment.

I did not provide enough details for the corollary presented. I have updated the corollary in order to keep it simple.



Ravi Fernando said...

Thanks for the proof! There are a few parts that took me a while to figure out though:

First, a and b must be relatively prime because p and q must be relatively prime, which is in turn because x^2, y^2, and z are relatively prime. (This may have been what Ralph was talking about: step 4 uses the parametrization for primitive Pythagorean triples without specifically noting that p, q, and y must be coprime.)

Then, after step 6, it should be noted that a and b must be squares, by the same theorem that step 6 uses, because a and b are relatively prime and ab is square.

Then (sqrt[a])^4 + (sqrt[b])^4 = a^2 + b^2 = P^2 leads to infinite descent, as desired.


alvareztorres said...

Hello, what about solutions of the p^20+q^20+r^20=t^4?

Nørresundby said...

Hey Larry...

Could you please tell my why x^4+y^4=z^2 is equal to x^4+y^4=z^2

Nørresundby said...

Hey Larry..

Can you please tell my why, x^4+y^4=z^4 is equal to x^4+y^4=z^2 ??


Larry Freeman said...

Hi Louise,

Thanks for your question. To be clear, the two forms are related and you are right, the two values are not (unless z=+/-1 or z=0 in which case z^2 = z^4).

Let's assume that there are no integers such that x^4 + y^4 = u^2.

Then this proves that there are no integers such that x^4+y^4 = z^4.

Why? Assume that there is an x,y,z such that x^4 + y^4 = z^4.

Let u=z^2.

Then it follows that there is also an x,y,u such that x^4 + y^4 = u^2.

But this is impossible by assumption so we have a contradiction and we have shown that the two forms are related.

I hope that helps.


Luiza said...

Hi Larry,
I'm happy to discover this blog.
I hope you will help me to present another solution of Fermat last theorem, made in 1989. It is not a joke.
The person who find it was not promoted, for unknown (by me) reasons.

Larry Freeman said...

Hi Luiza,

The only accepted solution to Fermat's Last Theorem is the one made by Andrew Wiles.

I'm sure that the 1989 proposed proof that you are referring to has a flaw.

Feel free to post the link in your comment.


Luiza said...

it's about a proof witch was not officialy verified. The person who writed, published the resume of the proof in 1989 and died in 1990, in Romania.
I believe he had not many instruments to communicate with those interested about his solution.
The fact that Andrew Wiles has the only accepted solution prove only that it's notorious. He made a "show" around his work, in order to capt the attention. So, he had the neccessary support for the later acceptance.
The other one liked another way to do that. Discreetly. Maybe too discreet, but he was afraid to talk to much, at those moment.
Now I try to rebuild his solution, in his memory.
It's not easy, because it's not my professional occupation right now.
The author of this solution is Ion Melcu, and he published it in France, in french.

Larry Freeman said...

Hi Luiza,

Good luck in putting together the solution. It is very sad when a person dies so early and it would be a great tribute if there was some mathematical insight that could be found in the work already done.

In fairness to Professor Wiles, his solution is brilliant. In my view, it is one of the mathematical highpoints of the last century.


Michael Ejercito said...

It can be proven that the sum of an even perfect fourth power and an odd perfect square can not be a perfect fourth power.

The sum of a perfect square and a perfect fourth power can be a perfect square. (The smallest example is 3^2 + 2^4=5^2.)

Can the sum of an odd perfect fourth power and an even perfect square be a perfect fourth power?

Dirk said...

I must have missed something, but I don't know what. Teh theorem states : If the area of a right-angled triangle were a square... . I have a right angled triangle with height 4 and base 2, it's area is 4*2/2 = 4 a square ! The same holds for 16*2/2 = 16 a square. What did I miss ?

Michael Ejercito said...

I must have missed something, but I don't know what. Teh theorem states : If the area of a right-angled triangle were a square... . I have a right angled triangle with height 4 and base 2, it's area is 4*2/2 = 4 a square ! The same holds for 16*2/2 = 16 a square. What did I miss ?You are right; I wonder how Fermat missed that.

Dirk said...

@Michael Ejercito : This can't be!!! There is an infinite set of examples. Therefore something must be wrong in the wording of this theorem. It's impossible that Fermat and the rest of the mathematical world overlooked it. Where is then the error in the proof of this theorem p^4 - q^4 = z^2 ?

Larry Freeman said...

Hi Michael, Dirk,

Remember, the issue is integers.

Fermat is saying that if a square has the same area as a right triangle, they can't both have integer sides.

For example, there is no problem solving Fermat's Last Theorem for real numbers. :-)


Dirk said...

@Larry,thanks but how is the phrasing in the beginning of your proof ? "In his life, Pierre de Fermat only left one proof in relation to number theory. He used his method of infinite descend to show that the area of a right triangle CANNOT BE SQUARE WITHIN THE DOMAIN OF WHOLE NUMBERS." There must be a problem in the wording of the theorem, I think.

Dirk said...

Larry, how about this: A right angled triangle with height 18 and base 4 and a square with side 6.Both have the same surface : 36!

Dirk said...

Larry, sorry I forgot the hypotenusa of the triangle. Nevertheless, the phrasing of the theorema is not 100 pct correct!

Larry Freeman said...

Hi Dirk,

I agree that the title is unclear.

I used the historical title.

I'll think about changing the title.



Michael Ejercito said...

If a right triangle and a square have the same area, at least one side of either figure is irrational.

Stangerzv said...

Dear Larry

Is there a way to proof for 2s^4+2k^4=a^2(perfect square), k can never be a rational number, where "s" and "a" are integers.

This equation, I got from the solution of polynomial w^4+6(ws)^2+s^4-8k^4=0. Where,


Where "w" is an integer. I would appreciate it if anybody could help me as this equation leads to similiar form of Fermat's Last Theorem. I am writing up a paper on this but stumbled on this part.


John said...
This comment has been removed by the author.
John said...

Excuse a simple question, but with Fermat's theorem, if a solution were to be found, would the numbers xyz necessarily have to be the sides of a right triangle? Or, put another way, if you could prove the theorem is true for the values of the sides of all right triangles, would you have proved the entire theory?

I don't quite understand to what extent Fermat's theorem deals, if at all, with right triangles.

Larry Freeman said...

Hi John,

Fermat's Last Theorem is true. This was proven by Andrew Wiles.

FLT is about the case where n >= 3. Right Triangles involve n=2. These are known as Pythagorean Triples and there are an infinite number of them. This solution was famously discussed by the Ancient Greek Diophantus.


Dirk said...

Hi Larry,
I have the following, most probably, philosophical, question:
Since it is proven that the theorem of Fermat is true, it follows that any theorem that would claim the opposite is untrue.
Since one of the characteristics of mathematics is that it consists of a number of statements which are 1. proven to be valid starting from a limited number of hypothesis (the so-called axiomas); 2. up to now not proven to be inconsistent with each other; is it right to state the following theorem :
"The square root of 2 is a rational number is equivalent to the statement that the Theorem of Fermat is not true" ? You will immediately notice that I could have replaced the second part of my statement by any other theorem of which we know that it is true.
In more general terms, is it true that, up to our knowledge in mathematics, each inconsistency in mathematics can be reduced to another known inconsistency ?
Of course, an obvious answer would be: you are right, if you can prove it; but I feel that this answer is the same as the one Gödel gave about inconsistencies in mathematics. Just to have your feeling about the question. Thanks

Larry Freeman said...

Hi Dirk,

I appreciate the question.

From a mathematical viewpoint, your reasoning would not be valid.

Mathematical proofs focus on assumptions and conclusions that are directly derived.

Saying FLT is true is really saying that given a certain set of assumptions, FLT is true.

This is what happens for example with NonEuclidean Geometry where certain Euclidean theorems still hold up but others (such as the theorem that a triangle's angles add up to 180) do not.

For these reasons, even if we know that sqrt(2) is not rational and FLT is true, it may turn out that there is an interesting proof that shows that if FLT is false, sqrt(2) is rational.

My point is that the statement of two true theorems is not enough. The relationship requires its own proof in order to be valid.



john said...

can you explain line 7 ? i dont understand where the p^2 came from

Larry Freeman said...

Hi John,

This follows from step #6 where I show that a^2 + b^2 is itself a square.

That is, there is a number P such that P^2 = a^2 + b^2.


Matt said...


In step 8, why does it matter that all those numbers are smaller than each other? arent they suppossed to me? Just curious

Michael Ejercito said...

Fermat's Last Theorem is true. This was proven by Andrew Wiles.

to be more precise, he finished the proof by proving that semistable elliptical curves can not be modular. (It had already been shown that an integer solution to n>=3 would imply the existence of modular semistable elliptic curve.)

Larry Freeman said...

Hi Matt,

Step #8 is the critical step which completes the proof. If a positive integer solution always implies that a smaller integer solution also exists, then there can be no integer solution.

The reason is that integers cannot descend infinitely. That's the principle behind proof by infinite descent.


Larry Freeman said...


You are correct about Wile's proof.

In this blog, I am showing historical results that preceded Wile's master proof.


Scouse Rob said...

Morning Larry.
(Well it is morning here.)

In step (7), to begin the infinite descent, don't we need an equation of the following form?


I finally figured out that in step (6) ab is proved to be a square.

Now because a and b are relatively prime this means that a and b are both squares, so:

a=G^2, b=H^2

Which gives us in step (7):


Which is the same form as the original equation and the infinite descent now clearly works.

I love this proof. It is so simple and elegant.
(Once you can get your head around the concept of infinite descent.)

Love the blog as well. :-)


Scouse Rob said...


Could I suggest that you expand on step (6) a little to explain that ab and a2 + b2 are relatively prime because p and q are relatively prime?

Could you also state that p and q are relatively prime in step (2), as you do in step (4)?

Sorry if it seems as though I want to be led fully down the path without having to think too much for myself on these proofs.

But I do. ;-)


Phil said...

Hi Larry,
Your correspondents want to discuss aspects of your proof of n = 4, but I would like to submit a proof for FLT which may have been Fermat's own version.
After playing around with Diophantus' solution to Q8 Bk II around the early 1630's, Fermat knew that Pythagorean triples could be described algebraically in terms of a power 2 raised to a unit power by the equation (x^2)^1 + (y^2)^1 = (z^2)^1, when x,y,z are integers. From this he found the equation (x^n)^k + (y^n)^k = (z^n)^k described any power separated into two powers, when power n is raised to power k. Your first conjectural reconstruction of "The easiest proof for Fermat's Last Theorem is the case n = 4...." concerns Fermat's marginal note to Q29 Bk V of the Arithmeticorum annotated around the mid 1630's, saying that he had proved x^4 + y^4 = z^2 had no solutions. Because values of z^4 can be written as z^2, it followed that (x^4)^1 + (y^4)^1 = (z^4)^1 had no solutions.
Fermat vigorously proved his theorem by applying the logic of mathematical proof introduced by the ancient Greeks:
Let A=(x^4)^1+(y^4)^1=(z^4)^1
Let B=(x^4)^k>2+(y^4)^k>2=(z^4)^k>2
Let C=(x^k>2)^4+(y^k>2)^4=(z^k>2)^4
Let D=(x^k>2)^1+(y^k>2)^1=(z^k>2)^1
-As A had no solutions, and A implied B,
B had no solutions.
- As B equalled C, C had no solutions.
- As D implied C, D had no solutions.
Replace (k>2) with any integer >2 in equations B, C, and D, and FLT (D) had no solutions when n>2 and k=1.
In the early 1650's, Fermat derived his second proof of n = 4 by proving Pythagorean triangles do not have square areas, as shown in your reconstruction.
The lesson not learned by all later mathematicians from Euler to Wiles, and even now in 2010, is that Fermat proved his theorem true for all n>2 by extending the case for n=4, not by taking different approaches to odd prime values of n.
Phil Cutmore

Graeme said...

Regarding Phil's "proof" of FLT...
I will recap Phil's argument here:
Let A=(x^4)^1+(y^4)^1=(z^4)^1
Let B=(x^4)^k>2+(y^4)^k>2=(z^4)^k>2
Let C=(x^k>2)^4+(y^k>2)^4=(z^k>2)^4
Let D=(x^k>2)^1+(y^k>2)^1=(z^k>2)^1
As A had no solutions, and A implied B, B had no solutions.
As B equalled C, C had no solutions.
As D implied C, D had no solutions.

Rather than point out flaws, I will restate it with k>1 rather than k>2:

Let A=(x^4)^1+(y^4)^1=(z^4)^1
Let B=(x^4)^k>1+(y^4)^k>1=(z^4)^k>1
Let C=(x^k>1)^4+(y^k>1)^4=(z^k>1)^4
Let D=(x^k>1)^1+(y^k>1)^1=(z^k>1)^1
As A had no solutions, and A implied B, B had no solutions.
As B equalled C, C had no solutions.
As D implied C, D had no solutions.

This second version of the proof is equally valid, as I'm sure Phil will agree, since the implications are the same: A->B, B=C, and D->C. Yet, it proves something which is clearly false. Therefore, there must be a flaw in both versions of the proof.

Graeme said...

Larry, I'm troubled by the selection of p,q in step 2 such that q is even and q<p. If you slapped a pair of absolute value bars around the second equation in that step, I would be less troubled.

Graeme said...

After more thought, I'm still very troubled. Slapping absolute value bars around the right hand side of equation

(2b) y^2 = p^2 - q^2

makes the next step problematic. In order for this equation to yield a new Pythagorean triple, p must be greater than q. In order to use q=2ab later, q must be even.

Interchanging p and q won't work, because the larger of the two must be odd.

I'm sure I'm missing a way to cope with the smaller of p,q being odd, but I can't see it. The infinite descent escalator has ground to a halt.

Graeme said...

Problem solved... I'm no longer troubled.

With the help of "lebesgue", on the NRICH forum, I now understand that equation (2b) y^2=p^2-q^2 is enough to establish that q must be even. This is a consequence of y^2 being odd, and thus one more than a multiple of 4.

Larry, since your goal is for your proof to be accessible to amateur mathematicians, maybe a little comment just after step 2 is in order to the effect that p is odd and q is even as a consequence of the 2nd equation in step 2.

Also, you might point out at the beginning that x is chosen to be even, and y is chosen to be odd WLOG.

Also, if you were to number your equations, then some of the text can be made shorter and clearer.

An example of all of these changes can be found here: http://mathhelp.wikia.com/wiki/Proof_x%5E4_%2B_y%5E4_%3D_z%5E2_has_no_solutions

Phil said...

Hi Larry, in reply to Graeme, my/Fermat's version demonstrates why k > 2 has to be 3 minimum, which is proved by implication that D = (x^3)^1 + (y^3)^1 = (z^3)^1 has no solutions. To prove FLT all we do is let k be any integer > 2 (not > 1) in B and C, and by implication D will never have any solutions. I agree that no proof of FLT occurs when k > 1 is 2 minimum, because D reduces to Pythagoras theorem. It is explained below why this "flaw" is a paradox, which Fermat resolves when working through the details of his proof.
My proof is part of a much larger conjectural reconstruction of FLT. To understand some of Fermat's possible thought processes when playing around with powers of numbers, including his "Eureka" moment "...which this marvellous proposition truly explains", the following should be added in my version after "...had no solutions".
"Because values of z^4 can be written as z^2, when k = 1, (x^4)^1 + (y^4)^1 = (z^4)^1 has no solutions. It follows that the multiples of n = 4 also have no solutions, i.e. when n = 8, 12, 16, 20, ...etc. because each of these power can be written as a fourth power, i.e. 2^8 = 4^4, 3^12 = 27^4. Now, the rules of powers allow them to be manipulated in different ways. If integer a raised to power n is a^n, then a^n raised to power k is (a^n)^k. This can be expressed as (a^n.k)^1, or (a^k.n)^1, or (a^k)^n. This means that each power can be written as a product of two powers n.k. Therefore the n.k multiples of n = 4 are 4 = 4.1, 8 = 4.2, 12 = 4.3, 16 = 4.4, 20 = 4.5, ...etc.
From here Fermat simply played around with a few of the lower powers and found no solutions were possible. He asked himself how and why Pythagorean triples existed, and found that PT rearranged to (z^2)^1 - (y^2)^1 – (x^2)^1 always gave a zero answer. When applied to higher powers he found triples that only came close to, but not zero, i.e. when n = 3, triples (7,6,5), (9,8,6), (12,10,9); when n = 4, triples (3,2,1) and (9,8,7); and when n = 5 and higher, triples (3,2,1). Surprisingly, this proof by 'scientific experiment' demonstrated that no power greater than squares could be separated into two powers. But this was not a 'vigorous' proof demanded by the mathematical community.
With this in view, Fermat went on to list all the powers n and multiples k of n = 3, 5, 6,...etc. that had no solutions. He included the multiples of n = 2 because something did not seem right:
n n.k multiples
4 4.1 4.2 4.3 4.4 4.5...
2 2.1? 2.2 2.3 2.4 2.5...
3 3.1 3.2 3.3 3.4 3.5...
5 5.1 5.2 5.3 5.4 5.5...
6 6.1 6.2 6.3 6.4 6.5...etc
He was right because there is a paradox. If all powers (n = 3,4,5,6,... etc., including n = 2) do not have solutions, then neither do their multiples (n.k). But the first multiple of n = 2, (2.1) has solutions (as proved by Pythagoras theorem) but the second, third etc multiples, 2.2, 2.3, 2.4,...etc. do not. He resolved this paradox by equating multiples 2.2 to 4.1, and 2.3 to 3.2... etc. all of which had no solutions. But it was something he had to be careful of later on.
His "Eureka" moment came when he noticed that n.k: 12.1 is the lowest common multiple (LCM) of n.k: 4.3 and n.k: 3.4, This meant that as the first multiple of n = 12 and the third multiple of n = 4 had no solutions, then the fourth multiple of n = 3 also had no solutions. Then by implication the first multiple of n = 3 had no solutions. From the way powers can manipulated Fermat realized that as any k > 2 can be a multiple of n = 4, then it would imply that (x^k)^1 + (y^k)^1 = (z^k)^1 had no solutions, and did not need to calculate the LCM at all.
Continue: "Fermat vigorously proved his theorem..."
regards Phil

Donec Visus said...

OK, Fermat proved that the equation raised to the fourth power could not generate whole solutions using his theory of limitless descent, but it was Andrew Wiles who proved the remainder of the theorem in 1986. His proof is final and superior to all others which attempt to prove the theorem. It shows that there are no valid integer solutions for x*y*z<> 0 for all equations in x^n+y^n=z^n for all values greater than 3 and truly verifies Fermat's theorem. If you wish to learn more, read his theorem yourselves.

Dorthee Berman said...

If I wanted to prove something for, let's say a n of 8, or 16...how could I go about that argument??

I.B said...

how about this,,, I counted and then conclude
a^4 + b^4 = C^4 - 2(ab)^2
what this equation same as (xn ') 4 + (y') ​​4 = (z (2n ')) 2?

Jez said...

Step ( 9 ) isn't clear for the infinite descent in step (8) as it is not explained how the smaller square - P^2, has the same properties.
This is, I believe the missing bit :-
In step (6) it states ab and a^2+b^2 are both squares. In the same way, since a and b are relatively prime, it follows that a and b are both squares. So a=Q^2 and b=R^2 for integers Q, R.
Therefore Q^4 + R^4 = P^2
This is what step (9) refers to. I hope this has helped clarify the solution.

Jez said...

I don't know what school of Mathematical logic you attended but your argument falls down on the last line, ie D implied C so D had no solutions is nonsense as the implication is the wrong way.
C would have to imply D to prove that.

S. G. Saunders said...

Larry, I believe that your infinite descent argument may be flawed, as stated (or at the very least, incomplete). The flaw, as I see it, is that in step 5, when you combine equations, you are able to compare to x2, which is obviously a square; however, when you try to repeat this argument in step 8, you would only be able to compare to q, which CANNOT be a square, because of the factor of 2 in q=2ab, and that a and b are squares, so q/2 is a square, hence q is not(also, we know p is a square and so is x2, so from x2=2pq, 2q must be a square, so again q cannot be). I can only assume that the "same properties" you are referring to in step 9 is that you have come across another Pythagorean triple (P2=a2+b2) which is smaller than (y2+q2=p2). However, it does not indeed possess the same properties, as you suggest.

However, as Scouse Rob pointed out, since ab is a square and a and b are relatively prime, a and b are also squares, hence the infinite descent can quickly be salvaged, as follows:

(6) Since ab and a2+b2 are relatively prime, we know that they are both squares. Further, since ab is a square and a and b are relatively prime, both a and b are squares as well.

(7) So there exists P, A, and B such that P2=p and A2=a and B2=b, respectively.

(8)Now we have reached infinite descent since: P2=A4+B4...


JulieL said...

Could you please help me understand why if there is not a solution to x^4 + y^4 =z^4, then there is no solution to x^4 - y^4 =z^2 ?

Larry Freeman said...

@JulieL, I believe it works the other way. If there is no solution to x^4 - y^4 = z^2, then, there is no solution to x^4 + y^4 = z^4.

If z is not a square, then it is possible (logically) that there is a solution to x^4 - y^4 = z^2 but no solution to x^4 + y^4 = z^4.

Unknown said...

Anyone who can devise a simple proof of Fermat's Last Theorem and publishes it on the Unsolved Problems web site at http://unsolvedproblems.org/ is eligible for a prize of US$2500 plus a bottle of champagne!

Reliable Home Inspection said...

Here is a simple proof

To recap, Michael has agreed that if n is an even number, particularly when n is 4, there4 exists a second Pythagorean triple with h^2,x^2, y^2 where h^2 = x^2 + y^2

Because this is a Pythagorean triple, there exist some whole numbers r,s such that y = 2rs, and say s is even. Squaring both sides y^2 = 4 r^2 y^2.

Using Euclid's formula  for the larger Pythagorean triple set, z^2, x^2, y^2

x^2 = h^2 - y^2

or substituting r,s for  h,y and collecting terms

x^2 = 4r^2s^2 - r^2


x = ( 4 r^2 s^2  - r^2 )^1/2


x = r(4s^2 - 1)^1/2

since s is even, values of s are 2,4, 6, 8, etc.

x must always be a irrational number.
This is one method.

After I made the last post, I realized that even though you could not extend Fermat's proof for n=4, you could use it as it is to prove the case for n being all the other even numbers higher than 2.

Here is what I said previously.

Once the case for n  =4 is established it is rather easy to see how to extend the proof for the n being even above 2. And the reason is rather simple,since x^8 is just some other number j, where j = x^2.

So now you have 23 different proofs of FLT for the evens.

You tell me which one proof you agree with and I will demonstrate the case with then use that agreement to demonstrate the csase for n being odd.
A Response by Kevin :
2016-11-24 at 09:39GMT

And how do you prove other cases (n=3,5,7,9,...) ?

What you're showing is already well known (only the way is funny, the result is obvious since two centuries).


Here is the equation from the funny proof.

x = r(4s^2 - 1)^1/2

From this equation you know that the part that determines whether x is rational or irrational is (4s^2 - 1)^1/2 . Let us call that part k.

Therefore k = (4s^2 - 1)^1/2

So x^h = r^h k^h where h is some odd number.

While we cannot say anything about this number when h is odd, we can say something about x when we are talking about x^(h-1)

x^(h+1) ) is an irrational number from the proof that FLT when n=4.


x = r(4s^2 - 1)^1/2

the part of x  that is irrational is

(4s^2 - 1)^1/2

Looking at this portion , we can say something about the   position on the number line.

If we say that we look at some arbitrary odd number, j, we can look at the 2 values of x when the power of of n is j-1 and j+1. We know that x is irrational when power of n is even.
Let us call the irrational part of x when n = j+1, h. And let us call the irrational part of x when n = m.

So now if we look at the number line between (4s^2)1/2 and (4s^-2)^2. Placing h,x, and m on this number line looks like this

---(4s^2)^1/2 -1------- (4s^2 - 2)^1/2  ------ m ------- x -------- h --------   (4s^2)^1/2 ---- (4s^2)^1/2

x is sandwiched between 2 irrational numbers that are sandwiched between 2 consecutive whole numbers. There fore x is irrational

to see the full argument on the fake math site go to