prove FLT on a single  page mathematicaly if a^n=c^2-b^2.for given a and n there are infinite sets of c and b.
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.
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.
292309
The Pythagorean Theorem? It's a^2 + b^2 = c^2.\n\nIf there's a right triangle, and a and b are the two "legs" (the shorter sides), and c is the hypotenuse (the longest side).
464622
Use any recursive algorithm.\n\nFor instance:\n\nf(x) = f(x) + 1
248592
Yes. There is not a continuous one, though. It is tricky to come up with an explicit example, but showing that one exists isn't too bad. First show that R is in one-to-one correspondence with 2^N. Then since Nx{0,1) is equivalent to N, we get R^2 is equivalent to (2^N)^{0,1} which is equivalent to 2^N which is equivalent to R.\n\nIn general, if X is an infinite set, there is a bijection from X to X^2. The axiom of choice is needed for this, though.
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."
203992
a) [1,2] is a sub set of domain of g so for every x in [1,2]\nwe have a unique value of g.\n\nb) if the sequence converge, in the limit Xn+1 must be equal to X, so if the limit is L we have:\n\nL=(3L^2+3)^1/4\n\nL^4=3L^2+3\n\nL^4-3L^2-3=0\n\ndelta=9+12=21\n\nL^2=(3+√21)/2  (L^2>0)\n\nL = √[(3+√21)/2]  (L>0 because X0>0)\n\n4<√21<5  ---> 7<3+√21<8\n\n---> 3<(3+√21)/2<4  ---> 1<L<2\n\nc) we have (Xn+1)^4=3Xn^2+3 let Yn be Xn^2\n\n---> (Yn+1)^2=3Yn+3 ---> Yn
634046
If f(n) = f(n + 11) = f(n + 18) then\nf(n) = f(n + c1 * 11) = f(n + c2 * 18) is also true,\nbecause f(n) = f(n + 11) = f(n + 11 + 11) = f(n + 11 + 11 + 11) etc.\nso f(n + 1) = f(n + 1 + c1 * 11)\nfor c1 = 13 we get:\nf(n + 1) = f(n + 1 + 13 * 11)\nf(n + 1) = f(n + 144) = f(n + 8 * 18) =>\nf(n + 1) = f(n) for every n in Z, so f(n) has to be a constant.
963823
LHS = (2n)!\nRHS = 2^n( n!)^2\n\nfor n=1,\n(2n)! = (2)! = 1.2 = 2\n2^n( n!)^2 = 2^1( 1!)^2 = 2\nSo LHS = RHS, condition is not satisfied.\n\nfor n = 2,\n(2n)! = (4)! = 1.2.3.4 = 24\n2^n( n!)^2 = 2^2( 2!)^2 = 16\nSo LHS > RHS\n\nYour problem has errors!!!\n\nI'll try to prove:\n (2n)! >= 2^n( n!)^2 .. .... (ii)\n\nfor n = 1, \n\n(2n)! = (2)! = 1.2 = 2\n2^n( n!)^2 = 2^1( 1!)^2 = 2\nSo LHS = RHS, condition is satisfied.\n\nlet, for any arbitrary integer m, (ii) is satisfied.\n\nSo, \n (2m)! >= 2^m(m!)^2 \n\nnow, if n = m+1, (ii) becomes,\n(2(m+1))! >=2^(m+1) ((m+1)!)^2  \n\nnow,\n(2(m+1))! = (2m+2))! = (2m+2)(2m+1) (2m)! \n= 2(m+1)(2m+1) (2m)! \n\n2^(m+1) ((m+1)!)^2  = 2.2^m  (m+1)^2 (m!)^2 \n\n\nwe have assumed, (2m)! >= 2^m(m!)^2 \nnow, multiplying by 2(m+1)(2m+1) we get,\n2(m+1)(2m+1) (2m)! >= 2^m 2(m+1)(2m+1) (m!)^2 \n\nnow, 2^m 2(m+1)(2m+1) (m!)^2  = 2.2^m (m+1)(2m+1) (m!)^2 \n\nagain 2m+1 > m+1\n\nso, 2(m+1)(2m+1) (2m)! >= 2^m 2(m+1)(m+1) (m!)^2 \n\nor,\n(2(m+1))! >=2^(m+1) ((m+1)!)^2  \n\n(ii) is true for n = m+1\n(Proved)
577215
This again is a direct theorem from the field axioms: http://mathworld.wolfram.com/FieldAxioms.html . \nSuppose a*b = 0. If a is non-zero, multiply both sides by a^(-1) to get a^(-1)ab = a^(-1)0, and thus b = 0, since the product of 0 and any number is 0.\nIf b is non-zero, multiply both sides by b^(-1) to get the similar result a = 0. Thus, either b or a must be 0. \nThe final case is that both a and b are 0, which is trivial.\nFurther discussion of this theorem can be found in the source.\nThis theorem can be used to solve equations by getting 0 on one side of the equation and factoring the other side completely, then using the theorem to imply that at least one of the factors must be zero. For example, x^2 = -3x - 2 can be rearranged to be x^2 + 3x + 2 = 0. We factor the left side to read (x + 1)(x + 2) = 0. From the theorem we know that either x + 1 = 0 or x + 2 = 0 or both. Thus we have two solutions, x = -1 or x = -2. It is obvious that in this case, only one factor could be 0.
576001
By the transitive property, a=c
958370
We wish to show that:\n[k=1, n]∑kk! = (n+1)! - 1\n\nClearly, this is true for n=1, since 1*1! = 2! - 1 = 1. Suppose it is true for some n. Then for n+1:\n[k=1, n+1]∑kk!\n(n+1)*(n+1)! + [k=1, n]∑kk!\n(n+1)*(n+1)! + (n+1)! - 1\n(n+2)*(n+1)! - 1\n(n+2)! - 1\n\nWhich is what we would get if we substituted n+1 for n in our formula. So by induction, the formula holds for all n. Q.E.D.
509288
if a=b+c then there exists an integer q and another integer r s.t. a=bq+r  q=1 r not = 0 => a and b do not share a common factor.  furhermore this line of reasoning seems to have little to do with fermat's last th.
970176
Not without a fallacious proof like the one above.  (Notice how the "proof" includes a division by (a-b) which is a division by zero.  This is not allowed and leads to the seeming contradiction that 0 = 1.\n\nZero does *not* equal One.\n\nYou could use similar fallacious methods:\n\n1^0 = 1\n1^1 = 1\n\nTake the log base 1 of both sides:\n0 = log 1 (1)\n1 = log 1 (1)\n\nTherefore 0 = 1.
727089
Suppose it is true for some n . I do not know because you say 22 I can compute and check\n\nSo 1st step is right\n\nlet it be true for some n\n\nwe need to prove for n+1\n\n60*5^n < 6^n\nmultiply both sides by 5\n60*5^(n+1) < 6^n . 5 ...1\n\nbut 6^n. 5 < 6^n . 6 as 6^n > 0\nso 6^n. 5 < 6^(n+1) ....2\n\nfrom 1 and 2\n60.5^(n+1) < 6^(n+1)\n\n\n QED
134713
It can be proved with the use of the integral, but you don't have to.\n\n1)The first some is n(n+1)/2\n\nTo prove that you can add all the consecutive numbers twice, but rearraging them, like that: 1+n; 2+(n-1); 3+(n-3)\n...n+1. Then you have n times the number n+1, which means that this sum equals n(n+1). But since you added twice the sum you wanted, to get the desired result, divide by 2.\nThus you obtain n(n+1)/2\n2)Prove (with the use of mathematical induction) that the sum of all cubes of consecutive integers from 1 to n is  \n[n^2*(n+1)^2]/4\n3)Elevate the first sum, n(n+1)/2, to the second power as requested. See that you get \n[n(n+1)/2]^2= [n^2*(n+1)^2]/4\n4)Observe that the results of 2) and 3) are the same.\n\nQ.E.D\n\nGood Luck.
16153
this is a mathematical trick that is done as a teaching method on commonf fallacies of math.\n\nLet a=b\n\na^2=a*a\n\nsince a=b, a^2=ab\n\na^2+a^2=a^2+a^2\n\nsince a^2=ab, a^2+a^2=a^2+ab or 2a^2=a^2+ab\n\nsubtract 2ab from each side to get 2a^2-2ab=a^2+ab-2ab or 2a^2-2ab=a^2-ab\n\nthis can be rewritten as 2(a^2-ab)=1(a^2-ab)\n\ndivide both sides by a^2-ab to get 2=1\n\n\ncheck it over, mathematically its sound.. except for one thing... from what we already described, a^2=ab! therefore a^2-ab=0!  and if you all remember, you can not divide a number by 0!  \n\n2(a^2-ab)=1(a^2-ab) is actually saying that 2*0=1*0 which is a correct statement!\n\nthus no issues