Who here believes Wile Solved Fermat's Last Theorem?
492413
Fermat's last theorem has been solved and verified and Wiles already got the reward for it and that was some time ago.
271648
boy we carry those text books along with us\nif u want us to prove then provide the theorem plz
491741
you arguments are good, however i swear that proving FLT isn't so easy . the best will that you yourself prove FLT in a single page checking out every arguments.\nafter all it took 350 yrs to find a proof !!\nkeep on trying ,good luck.
253239
***********************Spoiler Alert***********************\n\nThe puzzle is solvable. I spent about 4 mind-racking hours on it. Click the link to see the puzzle together. :-)
1094642
Kurt Gödel, famous for his incompleteness theorems has shown that for any consistent formal theory that proves basic arithmetical truths, it is possible to construct an arithmetical statement that is true but not provable in the theory. That is, any consistent theory of a certain expressive strength is incomplete.\nhttp://en.wikipedia.org/wiki/Goedel\n\nThere are other limits to knowledge most of which do not lend themselves to short answers.\nAlonzo Church's undecidability theorem - there exists no one procedure for deciding the validity of any argument in the system.\nhttp://en.wikipedia.org/wiki/Alonzo_Church\nChaos theory (well never have accurate weather forecasts more than 2 weeks out)\nhttp://en.wikipedia.org/wiki/Chaos_theory\n\nThese limits affect not just science but all human learning. Beyond all of this, humans are simply not capable of knowing all there is to know. There will always be aspects of the universe unknown to us and inaccessible to science or religion because we’re not smart enough to ask the right question and incapable of understanding the answer.
859279
This would be a special case of the Fermat-Catalan Conjecture.\n\nThis conjecture is that there are only a finite number of solutions of x^p + y^q = z^r for 1/p + 1/q + 1/r < 1.  Ten solutions are known, but all of them have the three exponents different (except for 1^p + 2^3 = 3^2) and one exponent equal to 2.\n\nYour conjecture would be the special case that there are no solutions with p > 2 and q = r.  It is already known that there are no solutions with p = q and r > 2 (Darmon and Merel, 1997).
255722
It is discovered by Yang Hui, and there is no mistake.  It is identical to Pascal's triangle, except discovered 400 years earlier.
940136
Hahahahaha good luck on that one... it took Andrew Wiles seven years to prove :)\n\nFirst you need to prove the following theorem:\nSuppose E is an elliptic curve with integer coefficients in global minimal form. If it has discriminant Δ a product of primes p with exponents δp, and conductor N a product of primes p with exponents np, and if E is a modular curve (now known to be true), then we can perform a level descent modulo primes ℓ dividing one of the exponents δp of a prime dividing the discriminant. If pδp is an odd prime power factor of Δ and if p divides N only once, then we can descend to another elliptic curve E' with conductor N' = N/p, such that the L-series coefficients for L(s, E) and those for L(s, E') are congruent modulo ℓ.\n\nOnce you're done with that, prove this:\nAny elliptic curve over Q can be obtained via a rational map with integer coefficients from the classical modular curve\n\n    X0(N)\n\nfor some integer N; this is a curve with integer coefficients with an explicit definition. This mapping is called a modular parametrization of level N. If N is the smallest integer for which such a parametrization can be found (which by the modularity theorem itself is now known to be a number called the conductor), then the parametrization may be defined in terms of a mapping generated by a particular kind of modular form of weight two and level N, a normalized newform with integer q-expansion, followed if need be by an isogeny.\n\nFrom another point of view, given an elliptic curve E over Q we may define a corresponding L-series. The L-series is a Dirichlet series which we may write\n\n    L(s, E) = &#92;sum_{n=1}^&#92;infty &#92;frac{c_n}{n^s}.\n\nWe can take the same coefficients, and use them to define a function in powers of q\n\n    f(q, E) = &#92;sum_{n=1}^&#92;infty c_n q^n.\n\nIf we make the substitution q = exp(2πiτ), then the series becomes a Fourier series, and so the coefficients are sometimes called "q-series coefficients", but other times "Fourier coefficients". The function obtained in this way, remarkably, is a cusp form of weight two and level N and is also an eigenform (an eigenvector of all Hecke operators); this is the Hasse-Weil conjecture, which follows from the modularity theorem.\n\nSome modular forms of level two, in turn, correspond to holomorphic differentials for an elliptic curve. The Jacobian of the modular curve can (up to isogeny) be written as a product of irreducible abelian varieties, corresponding to Hecke eigenforms of weight 2. The 1-dimensional factors are elliptic curves (there can also be higher dimensional factors, so not all Hecke eigenforms correspond to rational elliptic curves). The curve we obtain by finding the corresponding cusp form, and then constructing a curve from it, is isogenous to the original curve (but not in general isomorphic to it).\n\n\nThat should get you started.
419322
It was believed that he saw the shadows of the pillars on the tiled temple floor, that looked like a right angled triangle with squares attached. \nFrom there he developed his theory that the square drawn on the hypotenuse of a right angled triangle, is equal in area to the sum of the squares drawn on the other two sides.\nPythagoras put forward a proof, but said proof contained errors to do with transient numbers, however it seemed the theory was sound.\nSome two centuries later the Great Euclid himself proved it, with his brilliant proof using congruent triangles, true poetry of geometry.
382403
Strictly speaking, you have decided NOT to embrace the "undecidable" and NOT the "unverifiable".  Unverifieable, inherently implies the potential applicability of the scientific and/or logical verification process, which in turn could be a function of the Uncertainly Principle, time, knowledge, instrumentation, etc.  Whereas the situation at hand is by far much more hopeless than that; the very notion of "verifiability" does NOT even apply in this case.  Yes my friend those ancient desert-dwellers really knew how to screw us all with their creation of an undecidable god that can only be attained through the pure fairytale process of "faith".   And hence started the greatest conspiracy ever plotted against the human race, and which, in its many mutated forms, continues to this very day.   ... As for the pity part - well, how can I pity someone who has a great family, enjoys his life, and has otherwise had the courage to construct his own reality in accords with his moral will?    Besides, we do not much care for twice-removed guests like you DOWN HERE anyway. :-)
1178235
It's probably a mistyping of FLT which is shorthand for Fermat's Last Theorem. \n\nFLT states that threre are no integer values a, b, c that are solutions to the equation a^n + b^n = c^n when n > 2.
525104
Fermat, I think.  He wrote he had found a proof for this, but put a note in the margin that there wasn't enough room there for the proof.  And no-one could find the proof anywhere else in his notes and he died.  A fellow named Andrew Wiles worked for a long time to find the proof, and after some ill-fated public announcements I think the proof was finally completed.\n\nI hope this helps.
103563
Curiouscat is way kewl.....good answer on her part.\n\nE = mc^2 has been proven valid over and over again. Altho, in science, everything is theory, because someone may come along and find a flaw or correction.\n\nHowever up the present, this equation is rock solid.
379145
Look up Gauss. Story is he added up 1..100 in grade school and invented the n*(N+1)/2.
849522
Pythagorean Theorem\n \nLet's build up squares on the sides of a right triangle. Pythagoras' Theorem then claims that the sum of (the areas of) two small squares equals (the area of) the big one.\n\nIn algebraic terms, a2 + b2 = c2 where c is the hypotenuse while a and b are the legs of the triangle.\n\nThe theorem is of fundamental importance in the Euclidean Geometry where it serves as a basis for the definition of distance between two points. It's so basic and well known that, I believe, anyone who took geometry classes in high school couldn't fail to remember it long after other math notions got thoroughly forgotten.\n\nBelow is a collection of 69 approaches to proving the theorem. Some of the proofs are accompanied by Java illustrations, but most have been written in plain HTML.\n\nRemark\nThe statement of the Theorem was discovered on a Babylonian tablet circa 1900-1600 B.C. Whether Pythagoras (c.560-c.480 B.C.) or someone else from his School was the first to discover its proof can't be claimed with any degree of credibility. Euclid's (c 300 B.C.) Elements furnish the first and, later, the standard reference in Geometry. In fact Euclid supplied two very different proofs: the Proposition I.47 (First Book, Proposition 47) and VI.31. The Theorem is reversible which means that a triangle whose sides satisfy a2 + b2 = c2 is necessarily right angled. Euclid was the first (I.48) to mention and prove this fact.\n\nW. Dunham [Mathematical Universe] cites a book The Pythagorean Proposition by an early 20th century professor Elisha Scott Loomis. The book is a collection of 367 proofs of the Pythagorean Theorem and has been republished by NCTM in 1968.\n\nIn all likelihood, Loomis drew inspiration from a series of short articles in The American Mathematical Monthly published by B. F. Yanney and J. A. Calderhead in 1896-1899. Counting possible variations in calculations derived from the same geometric configurations, the potential number of proofs there grew into thousands. For example, the authors counted 45 proofs based on the diagram of proof #6 and virtually as many based on the diagram of #19 below. I'll give an example of their approach in proof #56. (In all, there were 100 "shorthand" proofs.)\n\nPythagorean Theorem generalizes to spaces of higher dimensions. Some of the generalizations are far from obvious.\n\nLarry Hoehn came up with a plane generalization which is related to the law of cosines but is shorter and looks nicer.\n\nThe Theorem whose formulation leads to the notion of Euclidean distance and Euclidean and Hilbert spaces, plays an important role in Mathematics as a whole. I began collecting math facts whose proof may be based on the Pythagorean Theorem.\n\nWherever all three sides of a right triangle are integers, their lengths form a Pythagorean triple (or Pythagorean numbers). There is a general formula for obtaining all such numbers.\n\nMy first math droodle was also related to the Pythagorean theorem. Unlike a proof without words, a droodle may suggest a statement, not just a proof.\n\nSeveral false proofs of the theorem have also been published. I shall collect some in a separate page. It is better to learn from mistakes of others than to commit one's own.\n\nIt is known that the Pythagorean Theorem is Equivalent to Parallel Postulate.\n\nThe Pythagorean configuration is known under many names, the Bride's Chair being probably the most popular. Besides the statement of the Pythagorean theorem, Bride's chair has many interesting properties, many quite elementary.\n\nProfessor Edsger W. Dijkstra found an absolutely stunning generalization of the Pythagorean theorem. If, in a triangle, angles a, b, g lie opposite the sides of length a, b, c, then\n\n(EWD) sign(a + b - g) = sign(a2 + b2 - c2),  \n\nwhere sign(t) is the signum function:\n\n  sign(t) = -1, for t < 0, \nsign(0) =  0, \nsign(t) =  1, for t > 0. \n \n\nThe theorem this page is devoted to is trea
624004
Martin Gardner, for many years the recreational maths columnist in Scientific American magazine,\n\nHe was a great populariser. A puzzle he set interested me so much that it prompted me to discover the trinary tree of Pythagorean Triples, some ten years ago, in order to solve the problem.\n\nThe problem was, there is a right-angled field with one side 48, there are ten different integer solutions, what are they?\n\nThe first 8 are fairly easy being multiples of \n\n(3, 4, 5) (yields two solutions: 12 x the 4 and 16 times the 3)  \n(5, 12, 13) x 4\n(7, 24, 25) x 2 \n(8, 15, 17) x 6\n(12, 35, 37) x 4\n(16, 63, 65) x 3\n(24, 143, 145) x 2\n\nThe ninth solution is further down that series and is a primitive triple\n(48, 575, 577)\n\nBut the tenth eluded me. I therefore started to research the topic. I reasoned there might be a clue in the intervals between the 3 numbers in a triple. I took the triple (5, 12, 13) and wrote it as\n\n(-12) ^2 + (-5) ^2 = 13^2 and wondered if there was another triple with intervals 7 18 and 25 that wouid be three positive numbers if I solved the quadratic equation created by putting x, x+7, and x+25 into Pythagoras' Theorem:\n\nx^2 + x^2 + 14x + 49 = x^2 + 50x + 625\nSimplifying\nx^2 - 36x - 576 = 0\n(x - 48) (x + 12) = 0\n\nSo one triple is (48, 55, 73) and the other is (-12, -5, 13) which is what we started from.\n\nThe triple (48, 55, 73) gave me the tenth solution to Gardner's problem but also a method of generating all possible primitive triples from (3, 4, 5).\n\nBy writing 5, 12, 13 as\n(-5)^2 + 12^2 = 13^2, a similar search for a triple with intervals 17, 1, 18 leads to seeking a triple (x, x+17, x+18) and the quadratic equation\n\nx^2 + x^2 + 34x + 289 = x^2 + 36x + 324\nx^2 - 2x - 35 = 0\n(x -7) (x + 5) = 0\nwith triples (7, 24, 25) and (-5, 12, 13) with which we started\n\nSimilarly \nBy writing 5, 12, 13 as\n(-12)^2 + 5^2 = 13^2, a similar search for a triple with intervals 17, 8, 25 leads to seeking a triple (x, x+17, x+25) and the quadratic equation\n\nx^2 + x^2 + 34x + 289 = x^2 + 50x + 625\nx^2 - 16x - 336 = 0\n(x - 28) (x + 12) = 0\nwith triples (28, 45, 53) and (-12, 5, 13) with which we \n\nThese three techniques generate three new triples, from any triple and can be expressed as matrices e.g. \n\n(2 1 2\n 1 2 2    X (5, 12, 13) = (48, 55, 73)\n 2 2 3)
998897
Fermat lived 300 years ago and is best remembered for his last theorem which says that there are no two integers, excluding 1, that can be raised to any common power greater than 2 and summed such that the total equals the value of another integer raised to the same power ie x^n+y^n can never equal z^n where x,y,z are integers greater than 1 and n is an index greater than 2. The thing that made this famous was that though sort of obvious no one could prove it and doing so became one of the great challenges in maths for 3 centuries. It wasn't until 1994 that Andrew Wiles finally had a proof accepted as being valid.