Thursday, May 19, 2005

Leonhard Euler

The next mathematician in the story of Fermat's Last Theorem is Leonhard Euler, whose name is pronounced "Oiler". Euler was one of the most prolific mathematicians of all time. One of those interests was number theory and he was the first mathematician to make progress on Fermat's famous problem (Fermat, himself, provided a proof for n=4). Euler was responsible for finding a proof for n=3 and for finding a method of proof using imaginary numbers. I will go into detail on both of these points in future blogs.

Euler was born on April 15, 1707 in Basel Switzerland. His father was a Protestant minister with an interest in mathematics. In fact, his father attended lectures by the very famous Jakob Bernoulli. For those interested, the Bernoulli family is one of the most famous families in all the history of mathematics.

Euler entered the University of Basel when he was 14. He originally planned to study theology but his abilities in mathematics attracted the attention of Daniel and Nikolaus Bernoulli, sons of Jakob. The Bernoullis convinced Euler's father to let him study mathematics.

In 1727, he received second place in Grand Prize of the Paris Academy.

At 23, Euler became a professor of physics in St. Petersburg. In 1733, when he was 26, he married Katharina Gsell, the daughter of a Russian painter. They would have 13 children, only 5 of which survived infancy. Euler would later say that he made some of his greatest mathematical discoveries while he had a baby in his hands.

By 1740, he had lost the use of one of his eyes and was starting to lose sight in his other eye. By this time, he had a very strong reputation in mathematics. He had won the Grand Prize of the Paris Academy in 1738 and 1740.

In 1741, at the age of 34, he became Director of the Prussian Academy of Sciences in Berlin. He did not get along well with Frederick the Great and returned in 1766 to St. Petersburg where he stayed for the rest of his life. During this time, he wrote 380 articles, wrote books on the calculus of variations, the mathematics of planetary motions, on artillery and ballistics, on analysis, shipbuilding, and navigation among others. Frederick was apparently very upset when Euler left.

Euler's achievements are legendary. His total mathematical output fills over 70 large volumes. He was able to do all this even while he gradually lost his eyesight. For the last seventeen years of his life, he was completely blind. Almost half of all his output was accomplished during this time. He died in 1783 at the age of 76. Even after his death, it took 50 years to release his unpublished works.

It is really impossible to summarize Euler's achievements in the space of this blog. Instead, I will present just a glimpse of some of his very interesting results.

  • Euler invented the notation ƒ(x) for function (1734), e for natural logs (1727), i as the square root of -1 (1777), the notation π for pi, the notation Σ for summation (1755), and the notation for finite diferences ∇y and 2 y.
  • He was the first to apply Newton's calculus to physics.
  • He showed the usefulness of using imaginary numbers as exponents. Which includes one of the most amazing results in mathematics, known as Euler's identity: e=-1.

I go into detail about this equation here. It is used in the derivation of cyclotomic inteogers which are the basis for Ernst Kummer's proof for Fermat's Last Theorem.
  • He solved many long standing mathematical problems including the Basel Problem, Fermat's Last Theorem for n=3, and many other of Fermat's unproved theorems and conjectures.
  • He made many breakthroughs in physics deriving what are known today as Euler equations which are a set of laws of motion in fluid dynamics that are derived directly from Newton's laws of motions.
  • Along wtih Daniel Bernoulli, he established the law of torque on a thin elastic beam.
In summary, Euler made significant contributions to number theory, differential geometry, mechanics, theory of music, analysis of planetary orbits, differential calculus, lunar theory, differential equations, analytic mechanics, Fourier series, complex numbers, analytic functions of a cmplex variable, logarithm of complex numbers, ordinary and partial differential equations, calculus of variations, mathematical physics, Bessel functions, topology, hydrostatics, fluid mechanics, astronomy, and many other areas which I have not listed.

In 1976, his image appeared on the Swiss 10 franc note.




The contents of this blog were based on two sources:

1. Wikipedia.com
2. MacTutor Biography of Euler

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.

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.

Thursday, May 12, 2005

Fermat's One Proof

In his life, Pierre de Fermat only left one proof in relation to number theory. He used his method of infinite descent to show that the area of a right triangle cannot be a square within the domain of whole numbers.

The proof itself was found after his death in his notes on Diophantus's Arithmetica.

In what follows, I will go through each sentence of Fermat's proof and provide details to make his step clear. My goal is to shed light on Fermat's thinking in his one existent proof.

I will show how this proof can also be used to prove the case of Fermat's Last Theorem where n = 4. In my next blog, I go over a simpler proof for n=4.

My analysis is based on the work by Harold M. Edwards called Fermat's Last Theorem. The translation is taken from T. L. Heath's translation.

Fermat writes:

(1) Fermat: "If the area of a right-angled triangle were a square, there would exist two biquadrates the difference of which would be a square number."

A biquadrate is a value to the fourth-power. So, the biquadrate of 2 is 24 = 16.

This corresponds to the following equation:
p4 - q4 = z2.

When I first read this, it was surprising to me that Fermat assumed that this equation was obvious from the problem of showing that right triangle's area cannot be equal to a square.

This equation proves n = 4 since if x4 + y4 = z4, then:
x4 = (x2)2 = z4 - y4

So, proving there is no right triangle that has an area equal to a square will also prove Fermat's Last Theorem for n=4.

The steps to this equation can be traced as follows:

(a) A right triangle is characterized by the Pythagorean Theorem:
a2 + b2 = c2.

(b) The area of a rectangle is base x height.

(c) A right triangle of a given base and height is created by dividing the same rectangle across the diagonal.

(d) Therefore, the area of a right triangle is base * height / 2.

(e) Or, using a,b,c from above: area = ab/2

(f) From the solution to Pythagorean Triples, we know that:
a = (2pq)d
b = (p2 - q2)d
where we know that p,q are relatively prime.

(g)So, saying the area of a right triangle is equal to a square comes down to proving that there are no solutions for:
z2 = ((2pq)d[p2 - q2]d)/2 = (pq)(d2)[p2 - q2]

(h) From a previous result, we know that d will divide z so, we are left with showing no solutions for:
(z/d)2 = (pq)[p2 - q2]

(i) Since p,q are relatively prime, it follows that (pq) and [p2 - q2] are relatively prime [see here for details]

(j) From this we conclude, that pq is a square and p2 - q2 are squares.

(k) And since p,q are relatively prime, p,q are themselves squares.

(l) So, there exist P,Q such that p = P2 and q = Q2

(m) Since pq is a square, there exists a value k such that k2 = pq.

(n) And from (m), k divides (z/d), so we get:

(z/dk)2 = p2 - q2 = P4 - Q4

(2) Fermat: "Consequently there would exist two square numbers the sum and difference of which would both be squares."

This one is a bit easier to derive
z2 = P4 - Q4 = (P2 + Q2)(P2 - Q2)

Now, all, we need to show is that (P2 + Q2) is relatively prime to (P2 - Q2) which means that p + q is realtively prime to p - q. [ The trick is to remember that p,q are relatively prime and they are of different parity (see 13(b) here). See here for details]

In other words, there exist two square numbers (P2,Q2) the sum (P2 + Q2) and difference (P2 - Q2) are both squares.

(3) Fermat: "Therefore we should have a square number which would be equal to the sum of a square and the double of another square..."

We know that P2 + Q2 is equal to a square. Let's say S2.
We also know that P2 - Q2 is equal to a square. Let's say T2
So that P2 = Q2 + T2
Which combined with the other equation gives us:
Q2 + T2 + Q2 = T2 + 2Q2 = S2

We can assume that S,P, Q are relatively prime and also that P,T,Q are relatively prime with S,P,T odd and Q even. [See here for details]

We can also assume that Q,S,T are relatively prime. [See here for details]

(4) Fermat: "...while the squares of which this sum is made up would themselves have a square number for their sum."

And as stated before: P2 = Q2 + T2.

(5) Fermat: "But if a square is made up of a square and the double of another square, its side, as I can very easily prove, is also similarly made up of a square and the double of another square."

So, he is saying this:
T2 + 2Q2 = S2 --> S = t2 + 2q2.

This means that

2Q2 = S2 - T2 = (S - T)(S + T)

And:

Q2 = (1/2)(S-T)(S+T)

Now Q,S,T are relatively prime. We know that Q, S-T, S+T are even. [see above for details]
Let S-T = 2u, S + T = 2v
So that Q2 = (1/2)(2u)(2v) = 2uv
Now, u,v are relatively prime [see here for details]
And, either u or v is even, let's assume u
So that, u = 2w and Q2 = 2(2w)v
And w,v are relatively prime so w,v are squares.
Let w = W2, v = V2
And S + T + S - T = 2S = 2u + 2v = 2(2W2 + V2)
So that S = 2W2 + V2

(6) Fermat: "From this we conclude that the said side is the sum of the sides about the right angle in a right-angled triangle and that the simple square contained in the sum is the base and the double of the other square is the perpendicular."

So, he is saying this:
S = 2W2 + V2 -> (2W2 )2 + ( V2)2 is a square.

(2W2 )2 + ( V2)2 = (2w)2 + ( v)2 = (u)2 + ( v)2 =
[(1/2)(S-T)]2 + [(1/2)(S+T)]2 =
[S2 - 2ST + T2]/4 + [S2 + 2ST + T2]/4 =
= [2S2 + 2T2]/4 = (S2 + T2)/2.

Now, since S2 = 2Q2 + T2
We have:
(2Q2 + T2 + T2)/2 = Q2 + T2

Which equals P2.

(7) Fermat: "This right-angled triangle will thus be formed from two squares, the sum and differences of which will be squares."

We can once again apply the Solution to the Pythagorean Triples so that:

2W2= 2mn
V2 = m2 - n2
P= m2 + n2

So we've proven that the difference is a square. Now, we need to prove that the sum is a square.

Since Q2 = 2(2w)v = 4W2V2.

Q2= 2(2mn)(m2 - n2)=4mn(m2 - n2).

And

(Q/2)2 = mn(m2 - n2)

Since m,n and m2 - n2 are relatively prime, then m,n,m2-n2 are all squares.

Letting m = M2 and n = N2

We get:
V2 = M4 - N4 = (M2 - N2)(M2 + N2)

Since M2 - N2, M2 + N2 are relatively prime, both the sum and differences are squares.

(8) Fermat: "But both these squares can be shown to be smaller than the squares originally assumed to be such that both their sum and differences are squares."

So M2 + N2 P2 + Q2.

Since M2 + N2 ≤ m + n is less than P which is less than P2 + Q2.

(9) Fermat: "Thus if there exist two squares such that their sum and differences are both squares, there will also exist two other integer squares which have the same property but a smaller sum"

We have proven that:
if P2 + Q2 is a square and P2 - Q2 is a square, then there exists M,N such that:
(a) M2+N2 is less than P2 + Q2
(b) M2 + N2 is a square
(c) M2 - N2 is a square.

(10) Fermat: "By the same reasoning we find a sum still smaller than the last found, and we can go on ad infinitum finding integer square numbers smaller and smaller which have the same property."

This leads to an infinite number of smaller solutions.

(11) Fermat: "This is, however, impossible because there cannot be an infinite series of numbers smaller than any given integer we please."

So that we have a proof by Infinite Descent.

(12) Fermat: "The margin is too small to enable me to give the proof completely and with all detail."

At least, this time we were able to reconstruct the proof. :-)

-Larry

Monday, May 09, 2005

Infinite Descent

Pierre de Fermat was very proud of his technique known as Infinite Descent. He wrote that this method "will lead to marvelous advancement in the theory of numbers" (quoted from: http://fermat.workjoke.com/flt2.htm)

As he often did, Fermat left very few details on how to apply the method. He did provide one example of this method in his proof that the area of a right triangle cannot be equal to a square number. An elegant application of this proof is found in the case of FLT: n=4 where the proof rests on the method of infinite descent and the solution to Pythagorean Triples.

The basic method is very straight forward. One proves that if an assumption is true, it means that the assumption must also be true for an element that is smaller. In other words, an assumption leads to the proposition of an infinite number of cases where it is true.

This technique is especially useful in the domain of positive integers. In this case, infinite descent is impossible since it contradicts the Well Ordering Principle. In other words, there must always be a smallest element in any set of positive integers. But if there is a smallest element, then, there cannot be an infinite descent. In this way, the method can be used as way to prove by contradiction certain negative assumptions. It can also be used to prove a positive conclusion as I will show below.

Fermat is known to have used this technique to prove that there is no square equal to a right triangle. He left a proof using this technique for the case of n = 4 which I will go over in a future blog. He also wrote about its use in the proof for the case of n = 3 in a letter to Christian Huygens.

If Fermat had really found a proof for his theorem, it was without doubt based on this method.

To show the technique in action, I will use it to prove the following theorem:

Thereom: Relatively Prime Divisors of an n-power are themselves n-powers.

This theorem says that if gcd(v,w) = 1 and vw = zn
Then, there exists x,y such that v = xn, w = yn

(1) So, we start with gcd(v,w) = 1, vw = zn
(2) Assume that v is not equal to any number xn
(3) v ≠ 1 since 1 is an xn power
(4) Now, v is divisible by a prime number p. [Fundamental Theorem of Arithmetic]
(5) So, there exists k such that v = pk
(6) p divides z since zn = vw = pkw [By applying Euclid's Lemma]
(7) So, there exists m such that z=pm
(8) So, zn = vw = pkw = (pm)n = pnmn
(9) Dividing p from both sides gives us:
kw = p(n-1)mn
(10) From Euclid's Lemma, p divides k or w.
(11) It can't divide w since it already divides v and gcd(v,w)=1. Therefore, it divides k
(12) We can apply this same argument for each p in p(n-1)
(13) So, we can conclude that p(n-1) divides k.
(14) So, there exists V such that k = p(n-1)*V
(15) So, kw = p(n-1)mn = p(n-1)*V*w
(16) Dividing p(n-1) from both sides gives us:
Vw = mn
(17) Now, gcd(V,w)=1 since V is a divisor of v and gcd(v,w) = 1
(18) Likewise, V cannot be an n-power. If it were, then v = pnV would make v an n-power which goes against our assumption.
(19) Finally, V is less than v since p(n-1) > 1.
(20) Thus, we have a contradiction by infinite descent.

QED

We proved that assuming that a relatively prime divisor of an n-power is not itself an n-power means that there must necessarily be a smaller relatively prime divisor that is also not an n-power and so on and so on.

Here is what Fermat wrote about Infinite Descent in a letter to another mathematician:

"As ordinary methods, such as are found in books, are inadequate
to proving such difficult propositions, I discovered at last
a most singular method...which I called the infinite descent.
At first I used it to prove only negative assertions, such as
'There is no right angled triangle in numbers whose area is a
square'... To apply it to affirmative questions is much harder,
so when I had to prove 'Every prime of the form 4n+1 is a sum
of two squares" I found myself in a sorry plight (en belle
peine). But at last such questions proved amenable to my methods."
-Quoted from Andre Weil's Number Theory

Saturday, May 07, 2005

Pythagorean Triples: Solution

OK, here's the solution for Diophantus's problem of determining the solution for:
x2 + y2 = z2

(1) We know that we can assume that x,y,z are coprime. [See my previous blog for details]

(2) The second important insight is that z has to be odd.

(a) Assume the opposite that z is even.

(b) Then, there exists another value Z such that z = 2 * Z

(c) Also, z2 is then divisible by 4 since:

z2 = (2 * Z)2 = 4 * (Z2)

(d) We know that x,y must both be odd because of (1).

(e) Since they are odd, there must also exist values X,Y such that:

x = 2 * X + 1
y = 2 * Y + 1

(f) But x2 + y2 cannot be divisible by 4 since:

x2 + y2 = (2 * X + 1)2 + (2 * Y + 1)2 =
= 4X2 + 4X + 1 + 4Y2 + 4Y + 1 =
= 4[ X2 + X + Y2 + Y ] + 2


(g) So, we have a contradiction and we reject our assumption.

(3) Since z is odd, either x or y must be even since an odd number is always the sum of an odd and an even number.

(4) Let's assume x is even. The same argument will also work if y is even.

(5) Now, we know that:

x2 = z2 - y2 = (z - y)(z + y)

(6) And, z - y and z + y must be even since z,y are odd.

(7) So, we know that there must exist u,v,w such that:
x = (2u)
z + y = (2v)
z - y = (2w)


(8) Which means that:
(2u)2 = (2v)(2w) [From (5) and (7)]

(9) Dividing both sides by 4 gives us:
u2 = v * w

(10) We need 1 more insight before the solution. Here it is: v,w are coprime

(a) Assume that v,w are not coprime.

(b) Then, there exists d such that d > 1 and d divides both v,w

(c) Then d divides both v + w and v - w

(d) But:

z + y + z - y =
2v + 2w

So 2z = 2v + 2w which means that z = v + w
So d divides z

(e) And:

z + y - (z - y) = 2v - 2w
So 2y = 2v - 2w which means that y = v - w
So d divides y

(f) Which is a contradiction since z,y are coprime [by (1)].

(g) So, we reject our assumption.

(11) By the properties of coprimes, we know from (9),(10) that v,w are themselves squares (see here for proof). [For those who need a review of coprimes, here is a link.]

(12) So, there exists p,q such that:

v = p2
w = q2

(13) And, we have our solution since:

z = v + w = p2 + q2
y = v - w = p2 - q2
x = 2u = 2pq [Since u2 = vw means u = pq]

We also know that:

(a) p,q are relatively prime. [Otherwise, z,x,y would not be relatively prime]

(b) p,q are opposite parity (that is, one is odd and one is even) [Since z is odd]

(14) For sure enough:

(p2 + q2)2 = (2pq)2 + (p2 - q2)2

(15) Now, to generate our answer, we can pick any p,q we want so long as they are integers.

For example, if p = 2 and q = 1, we get:

z = (2)2 + (1)2 = 5.
y = (2)2 - (1)2 = 3.
x = 2pq = 2*2*1 = 4

(16) We can do even better than this because we know that for each x,y,z, if they have common factors, the relation still holds.

z = d[p2 + q2]
y = d[p2 - q2]
x = d[2pq]

For example, if p = 2 and q = 1 and d = 2, we get:
z = (2)(5) = 10
y = (2)(3) = 6
x = (2)(4) = 8

And sure enough, 62 + 82 = 36 + 64 = 100.

QED

What's also nice about this result is that it is not too difficult to apply Fermat's method of infinite descent and prove Fermat's Last Theorem for n=4.

Thursday, May 05, 2005

Coprime Numbers

In the last blog, I spoke about Diophantus's problem: to divide a square into the sum of two smaller squares.

In other words, to find solutions for x,y,z where:
x2 + y2 = z2.

The first step in solving this problem is to realize that we can assume that x,y,z are coprime (or another way to say it, relatively prime). That is, no two of these values are divisible by the same prime. So, if p is a prime that is a factor of x, then we know that it is not a factor of y and not a factor of z.

When we have a situation where the three numbers are not coprime (for example, 6,8,10), we will be able to divide out common factors and end up with three numbers that are.

In the case of 6,8,10, the three numbers share the prime 2. If we divide out 2, then we are left with 3,4,5 which are coprime.

This assumption is important because it greatly simplifies the task of analyzing the conditions for when a solution exists. In my next blog, I will show how this assumption gives us the solution to Diophantus's problem.

Interestingly, we can apply this same assumption to Fermat's Last Theorem. From this point on, we will only need to consider the case where x,y,z are relatively prime.

One of my goals in this project is to provide complete proofs each of the conclusions presented. This blog relies on one lemma. A lemma is an intermediate statement that requires proof and is used in a larger theorem.

Lemma: All solutions to xn + yn = zn can be reduced to a form where x,y,z are coprime. [Here is the proof.]

Wednesday, May 04, 2005

Pythagorean Triples

When Fermat wrote his note in the margin, he was making a comment on the problem of determining Pythagorean Triples.

In Book II, Problem 8 of the Arithmetica, Diophantus poses the problem of how to divide a given square number into the sum of two smaller squares.

In other words, solve the problem:
x2 + y2 = z2.

Any three numbers that satisfy this equation are called Pythagorean Triples. They are called Pythagorean Triples since this is the same equation as the Pythagorean Theorem.

The Pythagorean Theorem is so well known that I refer people to this link if you would like to see a proof for it.

An example of a Pythagorean Triple is 3, 4, and 5 since 32 + 42 = 52.

I encourage everyone who has not already seen the solution to Diophantus's problem to try and solve it. This is without doubt what Fermat did and in solving this problem, he stumbled upon his famous generalization.

If you solve the problem, you should be able to prove there an infinite number of Pythagorean Triples and find a method for listing them out.

You can find the solution here.

This blog is based on the following sources:

Tuesday, May 03, 2005

Diophantus of Alexandria

When Fermat made his famous note in the margin, he was making a comment on a problem from Diophantus of Alexandria. While Fermat is today considered the father of number theory, he would probably have given this title to Diophantus.

Not much is known about Diophantus's life. He was a Hellenized Babylonian who lived in Alexandria, Egypt. He was born around 200 AD and he died around 284 AD.

His classic work is the Arithmetica which consists of 130 problems in 13 books. Today, only 6 of the books survive. It was written around 240 AD.

He focused on problems that have a positive rational number as a solution (a rational number is any fraction made from two whole numbers). He considered negative numbers and irrational numbers to be "useless," "meaningless," and "absurd."

When he spoke about quadratic equations, he only offered one solution. It is not clear if he knew that all quadratic equations have two solutions.

Diophantus is considered to be the father of algebra. Diophantine equations are named after him.

Much of what we know about him comes from the words written on his tombstone, which is itself an algebra problem for determining his age:

This tomb hold Diophantus. Ah, what a marvel! And the tomb tells scientifically the measure of his life. God vouchsafed that he should be a boy for the sixth part of his life; when a twelfth was added, his cheeks acquired a beard; He kindled for him the light of marriage after a seventh, and in the fifth year after his marriage He granted him a son. Alas! late-begotten and miserable child, when he had reached the measure of half his father's life, the chill grave took him. After consoling his grief by this science of numbers forfour years, he reached the end of his life.
This blog is based on the following sources:

Monday, May 02, 2005

Fermat's Achievements

Pierre de Fermat published very little in his lifetime, and yet by most accounts, he is considered to be one of the greatest mathematicians of all time.

I was very surprised when I learned that Fermat's Last Theorem was published after Fermat's death. It is pretty amazing that the mathematical world took a note in the margin so seriously.

Fermat was very precise about his mathematical pronouncements. When he was not certain about a finding, he labeled it a conjecture. Indeed, shortly after Fermat's death, all but one of his theorems had a proof just as Fermat had claimed.

Fermat was born in Toulouse, France on August 17, 1601. He was born into a wealthy family. He studied law at the university in Orleans and became a councillor at the parliament in 1631. He quickly moved up the ranks and in 1652, he became the chief magistrate of the criminal court.

In 17th century France, magistrates were required to spend large amounts of time in isolation. It was during this time, that he worked on mathematics.

His achievements in mathematics are incredible.

Sir Isaac Newton said that his invention of calculus was based a large part on Fermat's method of tangents.

In 1654, Blaise Pascal wrote a letter asking about Fermat's views on probability. Their series of correspondences became the foundation of probability theory.

Rene Descartes is perhaps most famous for his invention of Cartesian coordinates and his classic work La Geometrie. Fermat independently came up with a three-dimensional geometry. While Fermat's version of analytic geometry was more complicated and advanced, Descarte's classic work became more popular because its notation was more convenient. Today, both men are seen as the fathers of analytic geometry.

He made advances in optics. He came up with Fermat's Principle, the idea that light always traverses the path that takes the least time.

With Wiles' amazing proof, we are still left with a question. Did Fermat really have a proof?

Wiles' proof rests on twentieth century mathematics including the theories of elliptic curves, modular forms, and Galois Representations.

Wile's proof goes way beyond the mathematics that Fermat helped to create. It is very possible that Fermat's proposed proof made an incorrect assumption about cyclotomic factorization. This was a mistake that got Gabriel Lame into falsely believing that he had found a proof of Fermat's famous problem (I will be talking more about this in a later blog).

It is also possible that Fermat, in his lifetime, figured out that he had made a mistake. If he had truly found such a marvelous proof, why did he never tell anyone about it while he was alive? Perhaps, he had figured that he had made a mistake but forgot to correct his notes.

Still, the possibility that Fermat had a proof raises an interesting question: is there a way to simplify Wiles' result? Is it possible to derive a result from Wiles' proof that corresponds to the mathematics of Fermat?

The information in this blog was taken from the following sources:

Sunday, May 01, 2005

The Problem

For this blog, I assume that everyone is familiar with exponents. If you are not, here is an introduction.

Fermat's Last Theorem states for the equation: xn + yn= zn, there are no whole number solutions where x * y * z ≠ 0 and n > 2.

If x * y * z = 0, then it is easy to find a solution. For example (5)n + (0)n = (5)n.

Likewise, if we consider real numbers, then the solution is straight-forward algebra:
z = (xn + yn)(1/n).

Finally, if n = 2, then we have the Pythagorean Theorem a2 + b2 = c2. This is solveable by any Pythagorean Triple such as 3,4,5 (32 + 42 = 52) .

I think that this is the real appeal of the problem. It is easily stated and on its surface looks like it shouldn't be too difficult to resolve one way or the other.

Pierre de Fermat rarely published any of his results. He prefered to describe the problem and claim that he had found a solution. This has made the problem even more appealing: did Fermat actually have a proof?

The theorem itself became public without proof in 1670 when Fermat's son, Clement-Samuel published his father's notes. Unfortunately, Fermat was not around to explain his famous theorem because he had died in 1665. Instead, the reader was left with the famous statement of the problem:
"It is impossible for a cube to be written as a sum of two cubes or a fourth power to be written as the sum of two fourth powers or, in general, for any number which is a power greater than the second to be written as a sum oftwo like powers."
And this very mysterious statement about the proof:
"I have a truly marvelous demonstration of this proposition which this margin is too narrow to contain." (both quotes are from Fermat's Engima)
For over 350 years, this problem remained unsolved. Many of the greatest mathematicians were able to make progress on the problem including Leonhard Euler, Carl Friedrich Gauss, and Ernst Kummer but none of these great minds offered a solution.

The solution had to wait until 1995.

Saturday, April 30, 2005

Popular Books on Fermat's Last Theorem

Fermat's Last Theorem is one of the most famous math problems of all time. When Andrew Wiles proposed that he had a solution, it was front page news.

There are numerous books which provide a general introduction to the problem and the history of its solution. The most popular one seems to be Simon Singh's Fermat's Enigma. Simon Singh also has a web site.

Additionally, PBS dedicated an entire episode of Nova to the investigation of the theorem and its solution by Andrew Wiles. The Nova episode is called The Proof. PBS has a web site.

The popular books and the Nova episode provide great background on the story and the significance of Wiles' proof. Unfortunately, they provide only a high level coverage of the proof's major ideas.

Many books have claimed to provide details for amateurs. I find all of these books to be very difficult. They are great for more advanced math students, but for most others, they can be quite a challenge.

In this blog, I will try to present details that bridge the gap between these works and Wiles' proof.

Friday, April 29, 2005

Purpose

I have long been fascinated by Fermat's Last Theorem and greatly excited by Andrew Wiles' proof. I have wanted for a long time to explore the history of the problem and also the amazing proof by Wiles.

Just this week, for example, I learned that another mathematician is about to provide a very significant result that builds on the work done by Wiles. For those interested this mathematician, Chandrashekhar Khare, has a home page here: http://www.math.utah.edu/~shekhar/. After I get to Wiles' proof, I will attempt to also analyze Khare's result.

This blog will be an effort to trace the history of Fermat's Last Theorem and work through the details of the second version of Wiles's proof. Fermat has been called the Prince of Amateurs (Bell, Men of Mathematics) so I hope it is appropriate that this blog is also run by a mathematical amateur.

My goal in this blog is to present a set of proofs in a style very much like Euclid. Each proof presented will either rely on a previous blog or rely on a set of identified postulates. I am not a professional mathematician so I may make mistakes. I am hopeful that other participants will provide useful comments to keep the quality of this blog up.

The solution to Fermat's Last Theorem embodies a large part of the history of mathematics including many of its major results and many of its most famous persons. This blog will start with Fermat. For the most part, it will follow a historical flow, going from Fermat to Euler to Gauss to Kummer, etc.

It would be a shame to trace the history of the proof without also looking at the lives of the mathematicians involved. Some blog topics will explore the life and achievements of the participants of this story.

I look forward to people adding their comments and corrections to the biographies and the proofs. I think that this is one of the most exciting stories in all of mathematics and a blog is a great way to explore it.

Cheers,

-Larry