In Leonhard Euler's original proof for Fermat's Last Theorem n=3, he made a mistake. For anyone interested in seeing the details of the proof, it is perhaps comforting to know that a genius like Euler was capable of getting the details wrong. The details of this blog are based on Harold M. Edwards's book,

*Fermat's Last Theorem*.

The mistake came when Euler tried to prove the key lemma of his proof. In his published proof, Euler attempted to prove this lemma using imaginary numbers. Euler is credited with the invention of the notation for imaginary numbers and one of the most fascinating equations ever derived which is known as Euler's Identity:

**e**

^{iπ}=-1So, it is not too surprising that he would apply this method to Fermat's proof. It is also very interesting to note that a less innovative method can be constructed based on Euler's other work. This is what I did in the proof that I presented earlier.

Also interesting is the fact that this technique turns out to be a very important innovation which makes possible additional proofs for n=5, n=7, and Kummer's proof for regular primes.

Euler tries to prove the following lemma:

**Lemma: Given that p**.

^{2}+ 3q^{2}is a cube, show that there exist a,b such that p = a^{3}- 9ab^{2}and q = 3a^{2}b - 3b^{3}Here's Euler's idea. What if we presuppose that there exists numbers of the form:

**a + b√-3**

If we allow that these numbers exist, then we have the following equation:

**p**)

^{2}+ 3q^{2}= (p + q√-3)(p - q√-3So far so good. Mathematicals in general will allow any extension of numbers as long as the extension is logically consistent. The rational numbers extend the integers. Real numbers extend the rational numbers and Complex numbers extend the real numbers.

Euler proceeds to show that the two complex numbers (

**p + q√-3**,

**p - q√-3**) are relatively prime to each other. His proof here also works. I will add the details for this proof later.

He then concludes that the two complex numbers must, therefore, be cubes based on the reasoning of the Relatively Prime Divisor Lemma. This is the mistake!

In modern notation, Euler assumed that Z[√-3] was characterized by unique factorization. This is not completely correct. In actual fact, Z[(-1 + √-3)/2] is characterized by unique factorization. I go into more detail on this subject in my blog about Euclidean integers. If you are not familiar with quadratic integers, start here.

This result would be greatly extended by Carl Friedrich Gauss.

## 7 comments:

p^2+3q^2 is also a prime!

Let p=500, q=511, ,,,

x^3+y^3=(a cube) times (a prime)

= (x+y)(x^2-xy+y^2)

Let s=x+y, then

= Q(s) = s P(s)

where P(s) is a prime polynomial.

I think the link is missing from the first line.

Rob

Hi Scouse Rob,

Thanks for noticing that. I added the link to the first line.

Cheers,

-Larry

Hi,

i want to correct Euler' s lemma by using the UFD Z[(-1 + √-3)/2].

but i have no idea how i can start the proof, if it has the same argument with the Euler's wrong proof, where will i use my Ufd Z[(-1 + √-3)/2].

fatifati,

Check out the Edwards book on Fermat's Last Theorem for more details or my coverage of the accepted proof.

Sorry, but i can not find the correct solution of FLT for n=3 by using the Ufd Z[(1+sqr(-3))] at Edwars book. Which page does it contain. There is only the wrong proof. Also, i wish i can understand your proof.

Hi Fatifati,

Edwards doesn't give the proof as much as he explains how to give a proof.

I took my proof from this book

It's also covered in Hardy's book: Introduction to Number Theory.

Good luck.

-Larry

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