It is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second into two like powers. I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
-Pierre de Fermat, 1637
Pierre de Fermat was a French lawyer and mathematician who lived from 1601 to 1665, and his work has formed the foundation of many advanced areas of mathematics. His work on maxima, minima, and tangents to curves formed the basis of differentiation. His work on reducing general power functions to the sums of geometric series was influential to Newton and Liebnitz in their development of the fundamental theorem of calculus. Fermat’s work with Blaise Pascal formed the basis of the theory of probability. His principle of least time lead to the principle of least action and the development of classical physics. Fermat was instrumental in establishing the basis upon which modern scientific and mathematical understanding is derived.
However, he was not always as rigorous as one might like. Although he dabbled in mathematics, he always considered himself an amateur. He communicated most of his work to his friends in letters, often with little or no proof of his assertions. Many mathematicians doubted several of his claims, especially given the difficulty of some of the problems he attempted to solve and the tools available to him with which to solve them.
His most famous assertion was made in a note he scribbled in the margin of the book Arithmetica by the Greek philosopher and mathematician Diophantus. The assertion concerns the equation a^n + b^n = c^n (where “^” means “to the power of”). While solutions for the case of n=2 were well known since the time of the ancient Greeks, Fermat asserted that for any other integer greater than 2 there were no numbers a, b, and c that could make this equation true. This assertion has come to be known as Fermat’s Last Theorem, as it was the last of his unproven assertions to remain unproven.
The most tantalizing part of his assertion was the quote at the end: I have discovered a truly marvelous proof of this, which this margin is too narrow to contain. For over 350 years, mathematicians searched in vain for this elusive proof. Fermat challenged his peers to prove the cases of n=3 and n=4, but he himself offered no proof of those cases. Euler proved the case for n=3 in 1770, n=5 was proved by others around 1825, and n=7 was proved in 1839.
Additional proofs were developed in the 1800s for 6 ,10, and 14. Sophie Germain proved that Fermat’s Last Theorem was true for any prime number p such that the expression 2p+1 was also prime (such values of p are called “Sophie Germain primes”). By 1993, computer-based proofs had been used to demonstrate the truth of Fermat’s Last Theorem for all values up to 4 million. But the proof of the general case of n>2 remained unproven.
Mathemetician Andrew Wiles was one of many to become enamored at a young age with the search for Fermat’s marvelous little proof. He described the problem this way: “Here was a problem, that I, a ten year old, could understand and I knew from that moment that I would never let it go. I had to solve it.”. Wiles worked on the proof off and on for years, with no success. Eventually he abandoned work on proving the theorem and went on to other areas of research, specifically into the advanced mathematical concept of modular elliptic curves. (Don’t ask me what they are – I’ve read about them a bunch and still don’t understand them.)
In 1985, mathematician Jean-Pierre Serre asserted that if a special case of the theorem concerning elliptic curves called the Taniyama-Shimura Conjecture was true, then Fermat’s Last Theorem would have to be true. In 1986, Ken Ribet proved this assertion which transformed the proof of Fermat’s Last Theorem into the proof of the Taniyama-Shimura Conjecture. This development meant that Andrew Wiles’ dream of proving Fermat’s Theorem was now a matter of proving one of the most important conjectures in his particular area of mathematics specialization.
Wiles’ search for a proof was renewed in earnest. Wiles worked for years on the proof in secret, starting in the summer of 1986. He dedicated all of his research time to proving Taniyama-Shimura. He compared the proof to walking into a giant mansion where all of the lights were turned out – you wander around in one room, feeling your way around the walls and the furniture, until eventually you find the light switch. Once you turn on the switch, the room is illuminated, and you can make your way to the next room. Repeat this process long enough, and eventually the entire mansion will be lit.
In a series of lectures presented at the Isaac Newton Institute for Mathematical Sciences over three days from June 21-23, 1993, Wiles presented his proof to the world. As he began lecturing on day one, word spread quickly through the mathematics community that his work might be the long-awaited proof of Fermat’s Last Theorem. On the third day, after finishing his proof and stating that his proof implied the correctness of Fermat’s Last Theorem, he concluded by saying, “I think I’ll stop here.” For his proof, delivered 354 years after the original theorem was posited, he not only received a standing ovation, but a host of other awards and accolades.
His proof is not at all the “marvelous proof” that Fermat had envisioned – it uses mathematical techniques that were not developed until the 20th century and beyond. At the time the proof was presented, there were only a handful of people in the world that had any potential of understanding it. Most likely, Fermat believed that he had found a proof using only 17th century methods, and the potential existence of such a proof is what has driven mathematicians for years to seek its solution.
Wiles journey has been documented in many books and television specials. His own description of his journey contains many words of advice not only for mathematicians, but for puzzle-solvers everywhere, including:
- I really believed that I was on the right track, but that did not mean that I would necessarily reach my goal.
- Certainly one thing that I’ve learned is that it is important to pick a problem based on how much you care about it.
- Always try the problem that matters most to you.
- Just because we can’t find a solution doesn’t mean that there isn’t one.
- When I got stuck and I didn’t know what to do next, I would go out for a walk. I’d often walk down by the lake. Walking has a very good effect in that you’re in this state of relaxation, but at the same time you’re allowing the sub-conscious to work on you.
- Well, some problems look simple, and you try them for a year or so, and then you try them for a hundred years, and it turns out that they’re extremely hard to solve.
- That particular odyssey is now over. My mind is now at rest.
Wikipedia has a marvelous description of the proof – the proof itself is over 200 pages long. It also has a number of references to other articles, books, and TV shows to learn more about the proof.
Of course, the best pop culture reference to the proof is in Tom Lehrer’s timeless classic That’s Mathematics (YouTube).
4^3=4(4+1)/2^2 – 3(3+1)/2^2=100 – 36=64
(b,c) are the integers .
b^3=[ b(b+1)/2]^2 – [ (b – 1)*b/2]^2
F(b) = [ b(b+1)/2 ]^2.
F(b – 1)= [ (b-1)*b/2 ]^2
b^3= F(b)-F(b-1).
c^3 = F(c)-F(c-1)
b^3+c^3= F(b)-F(b-1).+ F(c)-F(c-1)
b^3+c^3 = [ F(b)+F(c) ] – [ F(b-1)+F(c-1) ]
(A1)^3 = b^3+c^3
(A1)^3 = [ F(b)+F(c) ] – [ F(b-1)+F(c-1) ]
Attention; A1 have been identified and A1 possible be an integer or not
Assume have A2 and A2 is an integer
(A2)^3 = F(A2)-F(A2-1)
If A1 and A2 are identical
Have is an integer A
A^3=b^3+c^3
FLT is wrong.
I must prove A1 and A2 are not identical
FLT is right.
Example.
T = M – N
Z= X – Y
T coincide Z infer [ M coincide X infer N coincide Y ]
I prove [ M coincide X infer N coincide Y is wrong
I conclude T and Z are not identical
My form is.
L infer ( K infer U )
I shall prove ( K infer U ) is wrong
I conclude L is wrong.
Similar .
(A1)^3 = [ F(b)+F(c) ] – [ F(b-1)+F(c-1) ]
(A2)^3 = F(A2) – F(A2-1)
A1 coincide A2 infer [ F(b)+F(c) = F(A2) infer F(b-1)+F(c-1) = F(A2-1) ]
I prove this clause [ F(b)+F(c) = F(A2) infer F(b-1)+F(c-1) ] = F(A2-1) ] is wrong.
I conclude A1 and A2 are not identical
F(b)+F(c) = F(A2) infer F(b-1)+F(c-1) = F(A2-1)
in special case b=0
F(b)+F(c) = F(A2)
F(b)=0
F(c) = F(A2)
F(c-1)=F(A2-1)
F(b-1)+F(c-1) = F(A2-1)
F(c-1) = F(A2-1)
infer
F(b-1)=0
infer (b-1)=0 and ( b-1 )=-1 Because ( b-1 )=-1<0.
So
b=1
We see b=0 infer b=1
This is wrong.
So the clause
F(b)+F(c) = F(A2) infer F(b-1)+F(c-1) = F(A2-1)
is wrong.
I conclude A1 and A2 are not identical.
Never A is integer. A^3=b^3+c^3.
z^3 not =x^3+y^3
FLT.
Happy&Peace.
Cuong
Pierre de Fermat had said that:
2^3+8^3=8+512=520.
36^2-1^2-(3^3+4^3+5^3+6^3+7^3)=36^2-1^2-(27+64+125+216+343)=1296-776=520.
520=520.
2^3+8^3=36^2-1^2-(3^3+4^3+5^3+6^3+7^3)=[8(8+1)/2]^2 – [2(2-1)/2]^2 – [(2+1)^3+(2+2)^3+……..+(8-2)^3+(8-1)^3]
The formula on (N) .
x^3+y^3=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [ (x+1)^3+(x+2)^3+……..+a^3+……….+(y-2)^3+(y-1)^3 ]
(X+1)<a<(y-1).
Hypothetical Pierre de Fermat's equation.
z^3=x^3+y^3.
So
a^3=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [ (x+1)^3+(x+2)^3+…+(a-1)^3+(a+1)^3+…………+(y-2)^3+(y-1)^3 ] – z^3.
The trio (z,x,y) can not satisfy every (a).
Nam-mô Bổn sư Thích-ca Mâu-ni Phật.