Given that a proof of Fermat's last theorem exists, is there a simpler one?
355573
Fermat claimed he had a proof 300 years ago but he\nnever demonstrated it. Andrew Wiles proved it in 1996\nbut had to use 20th century mathematics. There's an interesting book called Fermat's Enigma dedicated to\nDr. Wiles 7 year quest to prove it. There's also a \ndocumentary which has been aired on PBS which I've\nseen twice and wouldn't miss it if it came out again.
471450
weezy is on the right track, that glow1989 is on the right track.  For everyone who thinks that writing 1+1=2 is so obvious that it constitutes a self-contained proof of itself(!)... if you undertake a serious study of mathematical proof, you might find that 357 pages is a good length for a quick introduction to the ideas that are involved in considering questions in this area.  Of course, I am talking about actually UNDERSTANDING something - very different to simply passing an exam, regardless of the 'level' of maths you are enrolled in.\n\nYes, as you mentioned, this is a hard question.  'Proving' that 1+1=2 is not as simple as it sounds, because of the inherent question of what we mean by 'prove'.  You might have noticed that some proofs you have used depend on other proofs, which in turn depend on other proofs, ... which ultimately rest on a handful of axioms.  It is the nature of operations that are allowed by any axiomatic system, that determine the characteristics of mathematical problems that can be said to be SOLVABLE.  It is no small task to understand the axioms that underlie our systems of mathematics.  For example, maths students learn: that division by zero is undefined; that any object raised to a null power is equal to unity; some differences between rational and irrational numbers; what we mean when we consider objects to exist in 'sets'; and so on.  Those who consider themselves to be excellent students are usually able to list these rules in rapid-fire style.  But if we are asked to 'prove' why those rules are the way they are, the questions start to seem fuzzy, and you typically run out of convenient formulas to rely on.  You could easily write 357 pages just to explain what you actually MEAN by the numbers 1 or 2, without worrying about the definition of the addition operation!  As usual, people might do better to stop and consider what the question is asking before booting in a smug answer.\n\nLook into Hilbert's Problem (David Hilbert - German mathematician), which was a more workable and compact form of a proposal by Russell and Whitehead to formalise mathematics on a general basis, according to well-established axioms and procedures.\n\nOther notable mathematicians to work in this area were Alan Turing, Alonzo Church (look for Church's Lambda Calculus), Emil Post, and the brilliant Kurt Gödel.  In more recent decades, look for Gandy (1988) and of course many others.\n\nLol, molawby is spot-on.  I wish he'd posted b4 me... would've saved me a bit of typing.  If all you need to do is learn to sing the right song, to pass your exams and get the job that stipulates that you need to have completed such-and-such a course, then you don't have to worry about any of this.  Of course, that doesn't mean you have to dismiss it either.  But if you are chasing a deeper understanding, it probably does take the 20 yrs, and you won't necessarily get paid for it, heheheh.\n\nFor bedside reading, I suggest Roger Penrose etc...
176676
check this site\n\nhttp://scienceworld.wolfram.com/biography/Fermat.html
10761
this is a common "proof" that is often bandied about, even though it's wrong.  \n\nLet a=b. \n\nmultiply by a:\na^2 = ab\n\nsubtract b^2:\na^2 - b^2 = ab - b^2\n\nfactor both sides:\n(a+b)(a-b) = (a-b)b\n\ncancel (a-b)\n(a+b) = b\n\nsubstitute a=b from the first step:\n2b = b\n\ndivide by b:\n2=1\n\nsubtract 1 from both sides\n1=0\n\n\nfortunately sanity is saved when you realize that the step where you cancel (b-a) from both sides is actually dividing by zero, since you specify beforehand that b=a.  this is not so much a proof as a mathematical fallacy.  see the link below for another, more complicated version of the same thing.
301417
I'll try. I gave this proof to THREE people already:\n\nTake any two ordinary constants. (e.g. x and y)\nMake them alike, as in x = y = any non-zero number (in this case, 2)\nMultiply each by y, but x works equally well (y²=xy).\nSubtract x² from each equation (y²-x² = xy-x²).\nFactor y²-x² into (y+x)(y-x) to make ((y-x)(y+x) = xy-x²).\nFactor xy-x² into x(y-x) to make ((y-x)(y+x) = x(y-x)).\nDivide by (y-x) to get (y+x=x).\nSubtract x from both sides of this equation to get y=0.\nThis means that y=0.\n\nbut since, y equaled 2 in the first case, that means 2 = 0.\nIf you multiply this equation by 2.5, you get 5 = 0.\n\nSince 0 + 0 = 0, you can plug in 2 for the first two zeroes, and five for the last one, and then, TA DA! 2 + 2 = 5.\n\nP.S., this proof has a little falsity in it. you can't divide by (y-x) because that is zero!
182195
Fermat's Last Theorem states that there are no non-zero integers x, y, z where \nx^n + y^n = z^n\nwhere n is an integer greater than 2.\nThe rest of your question really doesn't make any sense. Did you mistype it or leave something out?
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Proving irreducibility over Q is rather straightforward based on what the form of a nontrivial factorization would have to look like.\n\nReduction modulo every prime follows from X^4 - 10x^2 + 1 being equal to all three of the following:\n\nx^4 - 2x^2 + 1 - 8x^2 = (x^2 - 1)^2 - 2(2x)^2\nx^4 + 2x^2 + 1 - 12x^2 = (x^2 + 1)^2 - 3(2x)^2\nx^4 - 10x^2 + 25 - 24 = (x^2 - 5)^2 - 6(2^2)\n\nIf 2 is a quadratic residue modulo p, then the first is a difference of two squares.\n\nIf 3 is a quadratic residue modulo p, then the second is a difference of two squares.\n\nIf neither 2 nor 3 is a quadratic residue modulo p, then their product 6 is a quadratic residue modulo p, and so the third is a difference of two squares.
344009
No, because it doesn't.\nThere's a number of "proofs" out there that rely on dividing by 0, even when it's not obvious that's what's going on.\nDivide by zero, and you can prove anything!
456169
There are many proofs but with fallacies. Please visit:\nhttp://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html\n\nhttp://www.math.toronto.edu/mathnet/falseProofs/second1eq2.html
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this is probably not a perfect answer but this site may give you some clues.\n\nhttp://www.gomath.com/Questions/question.php?question=4751\n\naltrnatively type 'prove v*v = 2gr*r/s+v0*v0-2gr' into a search engine and see what comes up.
1147854
Fermat's Last Theorem states that no integer solution exists for the equation\n\na^n = (b^n) + (c^n)  \n\nwhere n is greater than 2.\n\nYour equation has n = 2, so it does not contradict (or prove) Fermat's last theorem - it is one example of it.
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well i dont know how much there would be in the line of proofs steming from Fermats Last Theorem.\n\nbut i can tell you that Andrew Wiles (the man who proved fermats last theorem) developed a fair amount of new mathematics in the process of the proof, it is speculated that the new math he developed will be studied for hundreds of years.  his new math may lead to new lines of thinkings and ways to do proofs\n\nLook at it this way fermats last theorem was proved using the shimura taniyama conjecture...basically he proved that conjecture and said therefore Fermats Last theorem is true.  But he was able to say that anythign that happened in one world happened in another.  the math is way above my head...for now.\n\nSo yes, a new proof would be important.  It has been said that the answer to a question isnt nearly as imporant as the means to get there.  Implying the new developments made by proving something stem to new ideas and proofs...such as the new math used in the proof of Fermats Last Theorem
580657
Fermat's Last Theorem states that, for an integer n > 2, there are no non-zero integers x,y,z such that x^n + y^n = z^n.  What you wrote above does not prove this.  Yes, we know that 3^2 + 4^2 = 5^2, but there are no integers such that x^3 + y^3 = z^3.  If you still think your explanation actually proves this, please explain more.\n\n"There lies a commmon factor between a,b,c when a=b+c".  For your example, 9+16=25, the common factor is 1.  This does not, in any way, prove FLT.  You cannot just end a proof with a vague statement and not show how it proves what you're trying to prove.  How do the two facts 9+16=25 and "common factor when a=b+c" relate to prove FLT?\n\nUpdate:  I think you are confused as to what FLT is.  I stated it correctly above and the link below has a great explanation.  I believe you're trying to prove that there are an infinite number of integers x,y,z such that x^2 = y^2 + z^2.  Of course this is true, and has been known for a long time.  They're called Pythagorean triples.  FLT excludes the case n=2.  Read the link below for more information.
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Ask your friends if they can find the error in the following proof.\n(it's obviously not right, but it looks cool!)\n\na=b\na+1 = b+1\n(a+1)^2 = (b+1)^2\na^2 + 2a +1 = b^2 + 2b +1\ncancel the ones and rearrange\na^2 - b^2 = 2(b-a)\n(a+b)(a-b) = 2(b-a)\ntherefore a+b = -2\nPick numbers for a and b\nand look!! 5 + 7 = -2\nThus shown.\nLOL
759262
The rules for simplifying integer exponents are pretty simple:\n\nhttp://en.wikipedia.org/wiki/Exponentiation
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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.
222795
It is an easy problem.\n\nPut it as the\n\nProposition. There does not exist a positive integer n such that 2^n+1 is divisible by 7.\nProof. Re-write the comparison 2^n+1=0 mod 7 as 2^n=-1 mod 7 or else 2^n=6 mod 7.\n\nFurther, it is clear that 2^0=1 mod 7, 2^1=2 mod 7, 2^2=4 mod 7, 2^3=1 mod 7, hence, one observes the periodicity of remainders (periodicity with respect to n of length 3). Consequently it suffices to consider only 3 types of powers of 2: namely, for\n\n1) n=3*k, 2) n=3*k+1, 3) n= 3*k+2 (k is a positive integer).\n\nThese cases imply respectively:\n\n1) 2^(3*k)=1 mod 7;\n2) 2^(3*k+1)=2 mod 7;\n3) 2^(3*k+2)=4 mod 7;\n\nThese cases exhaust all possibilities. In neither of the cases the remainder 6 is attained (or, what*s the same, -1). The proposition has thereby been proved.