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.
QED
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.
Tuesday, May 17, 2005
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39 comments:
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.
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:
http://fermatslasttheorem.blogspot.com/2005/05/fermats-one-proof.html
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.
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.
-Larry
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."
Regards,
Rhuaidhri
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.
Cheers,
-Larry
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.
Thanks,
Rhuaidhri
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.
Cheers,
-Larry
Your proof is incorrect. You use the parametrization for PRIMITIVE Pythagorean triples. It is not general.
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.
-Larry
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.
Rob
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.
Regards,
-Larry
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)
Rob
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.
Cheers,
-Larry
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.
-Ravi
Hello, what about solutions of the p^20+q^20+r^20=t^4?
Hey Larry...
Could you please tell my why x^4+y^4=z^2 is equal to x^4+y^4=z^2
Hey Larry..
Can you please tell my why, x^4+y^4=z^4 is equal to x^4+y^4=z^2 ??
Louise
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.
-Larry
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.
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.
-Larry
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.
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.
-Larry
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?
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 ?
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.
@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 ?
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. :-)
-Larry
@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.
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!
Larry, sorry I forgot the hypotenusa of the triangle. Nevertheless, the phrasing of the theorema is not 100 pct correct!
Hi Dirk,
I agree that the title is unclear.
I used the historical title.
I'll think about changing the title.
Cheers,
-Larry
If a right triangle and a square have the same area, at least one side of either figure is irrational.
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,
w^2=-3s^2+-2sqrt(2s^4+2k^4)
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.
Regards
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.
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.
-Larry
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
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.
Cheers,
-Larry
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