What is Fermat's Theorem?
140387
Actually, this is his last theorem. Fermat's last theorem states that for the equation xn + yn = zn (n is a superscript) there are no integers x, y, z that satisfy the equation if n is an integer greater than 2
926159
4^n = (3+1)^n\n\nexpand the same using binomial theorem r+1st  term\n\n= 3^r(nCr)\n\nyou get the LHS immediately\n\nIf you want to derive 1st principle\nNow consider the product of these four equal factors:\n\n(x + b)(x + b)(x + b)(x + b) where x = 3 and b= 1\n\nThat is, let us expand (x + b)4.\n\nAgain, if we were to actually multiply out, then before collecting like terms, we would find 24  or 16  terms; and again every term would consist of four factors (one from each).  There will be terms such as xxxx,  xxbx,  bbxx,  and so on.  That is, there will be term in x4, x³b, x²b², xb³, and b4.  What the binomial theorem does is to tell how many terms there are of each kind.  Those are the coefficients.\n\nNow, how is a term x4 produced?   By taking x from each factor.  xxxx = x4.  There is only one such term.  The coefficient of x4 is 1.\n\nHow will a term x³b be produced?\n\n(x + b)(x + b)(x + b)(x + b)\n\nBy taking b from any one factor, and x from the remaining three:\n\nbxxx + xbxx + xxbx + xxxb\n\nTherefore the number of terms equal to x³b will be the number of ways of taking 1 thing -- letter b -- from 4.  There will be 4C1 = 4 terms equal to x³b.  On adding those like terms, we will have 4x³b.\n\nA term x²b² will come from taking b from any two factors in every possible way, and x from the remaining two.\n\nbbxx + bxbx + bxxb + xbbx + xbxb + xxbb\n\nTherefore, the number of terms equal to x²b² will be the number of ways of taking 2 things -- letter b -- from 4.   We will have 4C2x²b², that is, 6x²b².\n\nA term xb³ will come from taking b from any three factors, and x from the remaining factor; therefore, there will be 4C3 terms equal to xb³; we will have 4xb³.\n\nAnd finally, a term b4 comes from taking b from each of the four factors; there is only 1 way to do this; we will have 4C4b4 = b4.\n\nThis explains the binomial coefficients for the expansion of (x + b)n when n = 4.  They are 1  4  6  4  1.\n\nAgain, the binomial coefficients are how many terms there are of each kind.  They are none other than the combinatorial numbers nCk , where k successively takes on the value 0, 1, 2, 3, 4.\n\nThis result is general.  The binomial theorem states that in the expansion of (a + b)n, the coefficients are the combinatorial numbers nCk , where k successively takes on the value 0, 1, 2, . . . , n.  Each term in the\n  expansion will have this form:          n!      \n(n − k)! k!   an − kbk  .9
1043613
This lesson will allow you to figure out the Pythagorean Theorem all by yourself.  Go ahead and click on the preface link.\n\nhttp://www.arcytech.org/java/pythagoras/index.html
51055
4Pi^2 + 5Pi^2 = 9Pi^2 is not a solution to Fermat - you're only dealing with powers of 2!\nFermat's theorem is x^n + y^n = z^n  - where n is greater 2\nCan you give us an example which makes this true where n = Pi?\n\nEdit:\nUsing Windows' own calculator-\n4^Pi = 77.880233648388115107020834778598\n5^Pi = 156.99254530886590757845919883265\n9^Pi = 995.04164489285525428416364538181
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?
699025
Fermats little theorem tells us x^65536 = 1 modulo 65537 for all x.\nThus, for x^65535 to also be 1 modulo 65537, we need x^65536 = x^65535 (mod 65537).\nNow, thats true for x=0, but x=0 doesn't work in the original. If x is not 0, we can divide both sides by x^65535, since that is coprime with 65537. That tells us that x = 1 (mod 65537), so there is only one possible value for x.
646347
This is a *VERY* famous math problem called "Fermats Last Theorem".  It took over 250 years to prove, but it has been proven to be true.\n\nHere is a snippet from the link below:\n"Using sophisticated tools from algebraic geometry (in particular elliptic curves and group schemes), and number theory (modular forms, Galois representations, Hecke algebras, and Iwasawa theory), the English-born mathematician Andrew Wiles, from Princeton University, with help from his former student Richard Taylor, devised a proof of Fermat's last theorem that was published in 1995 in the journal Annals of Mathematics."
806805
You can prove it by induction.\n\nIt is obviously true when n = 1.\n\nInduction pypothesis:  Suppose it is true for n\n\nWe need to show  is true for (n+1).\n\nDoes 24 divide (5^(2*(n+1))-1))?\n\nWe can rewrrite this as:\n\n25*(5^2n)-1  Subtract 24 from this to get 25*(5^2n)-25 \n\nThis is just equal to 25*(5^2n)-1).\n\nThis has to be divisible by 24 -- since one of the factors is (5^2n)-1) and our induction hypothesis says that it is.  \n\nAdd 24 to it and we get something that is also divisible by 24 -- but it is (5^(2*(n+1))-1)) -- so our induction is done.
45454
Do you mean the Pythagorean theorem?\n\nIn any right triangle, the area of the square whose side is the hypotenuse (the side of the triangle opposite the right angle) is equal to the sum of the areas of the squares of the other two sides.
594792
Famous Fermat contributed to the development of modern calculus. Fermat created the modern theory of numbers. Some of his results were not proved for over a century after his death, and one of them, the Last Theorem, took more than three centuries to prove.\n\nHe also contribute to mathematical physics.
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.
478213
Properties of mathematics makes it work. I'm not sure exactly what it has you do but essentially it does this.\n\nIf your number is X. You add 5 to it then subtract 5 from it and you get your original number. Of course the nice ones hide the fact that your doing exactly this. \n\nDavid Copperfield did a TV bit once where he had a 3X3 square. You start out at any square you want and move according to his directions. After each move he took away a square. In the end you always end up on one square. That was cool. Anyone heard of it?
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.
406361
well the theorem says that a^p ≡ a(modp) where p is prime.\nequivalent to this is a^(p-1) ≡ 1(modp) by just taking the a to the other side. \n\nso we know 5^6≡1(mod7)\n\nso\n5^74 = (5^6)^12 x 5^2 ≡ 1 x 5^2 ≡ 25 ≡ 4 (mod 7)
500368
The links below have good info. The first briefly addresses your first question, and the second is a good brief history of the problem which provides some insight as to the difficulties encountered. I actually played around with this problem a little on an elementary level and it is interesting. Cheers.\n\nhttp://www.sciam.com/print_version.cfm?articleID=000DBA55-B698-1C71-9EB7809EC588F2D7\n\nhttp://www.geocities.com/fermatnow/flt/
145253
Take a look at this link. You should find all you need about Fermat Numbers there.
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.