Prove that Goldbach's conjecture that every even integer greater than 2 is the sum of two primes is equivalent to the statement that every integer greater than 5 is the sum of three primes.
125660
You r wrong, Not for every positive number... Such as your referance is not true for 11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53...its a long list... you cant sum them up using 2 prims...\nbut its true for all even number(because no such even number has found yet) and Goldbach's Conjecture says so..\n**Goldbach's Conjecture: For any even number n greater than or equal to 4, there exists at least one pair of prime numbers p1 and p2 such that n = p1 + p2.\n\noh now I see u also ment even Integer.. I am sorry..\n\nand 2nd part...\nif all even number greater than 3 is sum of 2 primes..\nthen all even number greater then 6(not 5)is sum of 3 primes..\nthat means n = 5+p;\nbut 5 = 2+3;\nso, n = 2+3+p
130367
The subject is very complicated. On this theme there is a avery good divulgation book published by a great mathematician, Marcus Du Sautoy, "The Music of the Primes". It is a pure delice!
370958
To see what conjecture they are interested in, it really is necessary to consider a number of examples, exactly as they said. So lets set up a table. We will call the product in question N.\nRemember that we are only interested in prime numbers for p.\n\np               N                N/p\n---------------------------------\n2                    2               1\n3                    3               1\n5                  25               5\n7                721           103\n11      3628801     329891\n13  479001601 36846277\n\nNotice that the numbers N so far have all been divisible by p. So, if we represented  them on a p-clock, they would all be p o'clock (or 0 o'clock if you prefer). That is the conjecture they are interested in. You are supposed to conjecture that the value of 1*2*3*...*(p-1)+1 in p-hour arithmetic is always p o'clock.
560388
Every even integer greater than 2 can be written as the sum of two (not necessarily distinct) primes.
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:D i think the person asking this question is not stupid, as most people here make him out to be, but rather is having fun, and seeing how far ppl will go to be picked :D ... anyways\nHere's my version of the proof, i'm going to do it in a step by step manner\n\nGiven: 5*5\nProof: the value is equvalent to 25\n\nStatement                |         Reason\n1)5*5                    |  1)Given\n2)=25                    |  2)Multiplicative Property of                      \n                         |     Arithmetic
420791
It is not known to be a fact.  However, it would be a special case of the Fermat-Catalan conjecture about integer solutions of x^p + y^q = z^r, and it has been proved that a similar special case to yours has no solutions.
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(A)  Show base case is true:\n\n(2*1)^2 = (2*1)(2)(2*1 + 1) / 3\n4 = 4\n\nAssume the equality holds for n = k\nShow that it holds for n = k+1\n\n2^2 + 4^2 + 6^2 + ... + (2k)^2 + (2(k+1))^2\n= 2k (k+1) (2k + 1) / 3 + (2k+2)^2\n= 2k (k+1) (2k + 1) / 3 + (4k^2 + 8k + 4)\n= (2k^2 + 2k) (2k + 1) / 3 + (12k^2 + 24k + 12) / 3\n= (4k^3 + 6k^2 + 2k + 12k^2 + 24k + 12) / 3\n= (4k^3 + 18k^2 + 26 k + 12) / 3\n\n2(k+1)(k+2)(2k+3) / 3 \n= 2 (k^2 + 3k + 2)(2k + 3) / 3\n= 2 (2k^3 + 3k^2 + 6k^2 + 9k + 4k + 6) / 3\n= (4k^3 + 18k^2 + 26k + 12) / 3\n\n(B)  Show base case is true\n5 (2!) ≥ 3^2\n10 ≥ 9\n\nAssume it is true for all n < k, Show it is true for n = k+1\n\n5(k+1)! ≥  5 (k+1) k! ≥  5 * 3 * 3^k ≥ 5 * 3^(k+1)
143801
Wow this is a long question.... hehe.\n\n1) Firstly consider any integer divided by 6, we must get one of the residues ( remainders ), r = 0,1,2,3,4,5. If the remainder of y divided by 6 is r, thus we can write \n y = 6 x "quotient" + "remainder" = 6k + r, where let k denote the value of the quotient. Since r = 0,1,2,3,4,5, we have just shown that any integer y ( we may assume positive ), can be written in one of the "forms"\n y = 6k , 6k + 1, 6k + 2, 6k + 3, 6k + 4, 6k + 5.\n\nNow among these, the ones that are composite ( NOT prime ) \nfor any integer k, are\n      6k , Even\n    6k + 2, Even\n    6k + 3 = 3(2k + 1) , a multiple of 3. \n    6k + 4, Even\n\nRemembering that the only even prime number is 2.\nThus, the only possibilities for forms for primes in here is either 6k+1 or 6k + 5 . All other forms denotes a composite integer. Remember that a composite integer is an integer with at least two distinct prime factors.\n\nIf p is form 6k+1, Let p = 6k+1,\nThus, p^2 + 2 = ( 6k + 1 )^2 + 2\n              = 36k^2 + 12k + 3\n              = 3 ( 12k^2 + 4k + 1 )\n              = A multiple of 3 > 3 \n              = A Composite number.  \n\nIf p is form 6k+5, Let p = 6k+5,\nThus, p^2 + 2 = ( 6k + 5 )^2 + 2\n              = 36k^2 + 60k + 27\n              = 3 ( 12k^2 + 20k + 9 )\n              = A multiple of 3 > 3 \n              = A Composite number.\n\nThus, in both cases, if p is prime then p^2 + 2 is always composite.\n\n2)\nAssume that the number of primes are finite.\nNow the integer N is NOT divisible by any of the primes in the list p_1, p_2, ..., p_n.\nWhy? because say p_k divides N. Then, p_k must divide the following integer : p_1*...*p_n where it does NOT contain p_k. Which is a contradiction .\nThus, primes  must be infinite for if not, they produce the contradiction shown above.\n\n3)\n\nI will not use bertrand's conjecture because its NOT proven yet... So, i'll skip this one... its redundant. I will not  use something that has not been proven yet... i.e, a conjecture.
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Prime numbers are numbers that are divisible by itself and produce only two factors..\n\nThe two factors are 1 and the number itself(other than 1)...\n\nBetween 1 to 10.. There are only 4 prime numbers..\n\n2 = 2 x 1\n3 = 3 x 1\n5 = 5 x 1\n7 = 7 x 1\n\nThe remaining numbers are not prime numbers as they do not fit the criteria...\n\n1 = 1 x 1\n4 = 2 x 2 = 4 x 1\n6 = 3 x 2 = 6 x 1\n8 = 4 x 2 = 8 x 1\n9 = 3 x 3 = 9 x 1\n10 = 5 x 2 = 10 x 1\n\n\nHope it helps to clear your suspicions...\n\n\nCheers.... (",)
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1) (I had had another proof but I like this one better)\nlet x = 4k+a for k = 0,1,2,3,... and a = 0,1,2, or 3\nS = 1^x + 2^x + 3^x + 4^x = 1 + 2^(4k+a) + 3^(4k+a) + 4^(4k+a)\n= 1 + 2^a(2^4)^k + 3^a(3^4)^k + 4^a(4^4)^k \n≡ 1 + 2^a(1)^k + 3^a(1)^k + 4^a(1)^k (mod 5) by Fermat's theorem\n≡ 1 + 2^a + 3^a + 4^a (mod 5)\n= 10, 30, 100 ≡ 0 (mod 5) i.e., S divisible by 5 when a = 1,2,3 \n≡ 1+1+1+1 ≡  4 (mod 5) i.e., S not divisible by 5 when a=0, i.e, x divisible by 4\n \n2)\nx^2 - y^2 = 12^2 = 144\n(x+y)(x-y) = 144\nThe 15 factors of 144 are 1,144; 2,72; 3,48; 4,36; 6,24; 8,18; 9,16; 12\nx+y and x-y will be integers only when cofactors are both even or both odd.\nThis leads to (37+35)(37-35); (20+16)(20-16); (15+9)(15-9);\n(13+5)(13-5)...[i.e.,4 cases (x,y) = (37,35), (20,16), (15,9), (13,5)]\n\n3)\nIf x is even (x=2k), x^4 + 4^x = (2k)^4 + 2^(2x) and is divisible by 2, so not prime.\nIf x is odd (x=2k+1), then\nx^4 ≡ 1 (mod5) for any x by Fermat's theorem,\nand 4^(2k+1) ≡ 4*4^2k ≡ 4*2^4k ≡ 4*(2^4)^k ≡ 4*1^k ≡ 4 (mod 5)\nx^4 + 4^x ≡ 1+4 ≡ 5 ≡ 0 (mod 5) and is thus divisible by 5 , so not prime.\n\nI hope you understand these and are not just handing it in.\nIf there's a next time please ask three questions.
122239
>> Prove that for each positive integer n\n>> there exist n consecutive positive integers\n>> none of which is an integral power of a\n>> prime number.\n\nFor example, I challenge you with n=5 and\nyou respond 16, 17, 18, 19, 20 and I win\nbecause 16 = 2^4, 17 = 17^1, 19 = 19^1\nare all integral powers of a prime.  But if\nI challenge you with n=3 and you reply with\n33, 34, 35 then you win because none of\nthose are prime powers.\n\nA proof (or winning strategy for the second\nplayer) is that given n, you can choose 2n\nprimes greater than n and use the CRT\n(Chinese Remainder Theorem) to find a\npositive integer x that simultaneously\nsatisfies\n\nx == -1 modulo prime_1\nx == -1 modulo prime_2\nx == -2 modulo prime_3\nx == -2 modulo prime_4\n[...]\nx == -n modulo prime_(2n-1)\nx == -n modulo prime_(2n)\n\nThen each of the n consecutive positive\nintegers x + 1, x + 2, ..., x + n is\ndivisible by at least two primes and\nso is not a power of a prime.
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One question in number theory most likely :\n\nfor  all integers a, b, c and n,\n\nprove that :  \n\na^n + b^ n does not equal c^n for n > =3
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Consider the numbers\nA1,\nA1+A2,\nA1+A2+A3\n..\nA1+A2+A3+...+An.\n\nDo the computations modulo n. If none of these is divisible by n, then we have n numbers and n-1 non-zero residue classes (mod n), so there are two of these numbers which are equivalent (mod n) by the pigeon hole principle. But then subtracting gives a sum of the required form that is 0 (mod n), in other words, so that n divides that sum.\n\nLooking at the proof, there is no need for the sequence to be monotonic.
338539
There is a proof contained in my link, using the Fourier series.
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Since [L : K] = [L : E][E : K] and [L : K] is prime, \neither [L : E] = 1 or [E : K] = 1.\n\nThus either E = L or E = K
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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).
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64 proofs\nexplanation\nProof #1\nThis is probably the most famous of all proofs of the Pythagorean proposition. It's the first of Euclid's two proofs (I.47). The underlying configuration became known under a variety of names, the Bride's Chair likely being the most popular.\n\n    \n\nThe proof has been illustrated by an award winning Java applet written by Jim Morey. I include it on a separate page with Jim's kind permission.\n\nFirst of all, ABF = AEC by SAS. This is because, AE = AB, AF = AC, and\n\n  BAF = BAC + CAF = CAB + BAE = CAE.\n \n\nABF has base AF and the altitude from B equal to AC. Its area therefore equals half that of square on the side AC. On the other hand, AEC has AE and the altitude from C equal to AM, where M is the point of intersection of AB with the line CL parallel to AE. Thus the area of AEC equals half that of the rectangle AELM. Which says that the area AC2 of the square on side AC equals the area of the rectangle AELM.\n\nSimilarly, the are BC2 of the square on side BC equals that of rectangle BMLD. Finally, the two rectangles AELM and BMLD make up the square on the hypotenuse AB.\n\nThe configuration at hand admits numerous variations. B. F. Yanney and J. A. Calderhead (Am Math Monthly, v.4, n 6/7, (1987), 168-170 published several proofs based on the following diagrams\n\n   \n\n\n\nProof #2\n    \n\nWe start with two squares with sides a and b, respectively, placed side by side. The total area of the two squares is a2+b2.\n\n    \n\nThe construction did not start with a triangle but now we draw two of them, both with sides a and b and hypotenuse c. Note that the segment common to the two squares has been removed. At this point we therefore have two triangles and a strange looking shape.\n\n    \n\nAs a last step, we rotate the triangles 90o, each around its top vertex. The right one is rotated clockwise whereas the left triangle is rotated counterclockwise. Obviously the resulting shape is a square with the side c and area c2.\n\n(A variant of this proof is found in an extant manuscript by Thâbit ibn Qurra located in the library of Aya Sofya Musium in Turkey, registered under the number 4832. [R. Shloming, Thâbit ibn Qurra and the Pythagorean Theorem, Mathematics Teacher 63 (Oct., 1970), 519-528]. ibn Qurra's diagram is similar to that in proof #27. The proof itself starts with noting the presence of four equal right triangles surrounding a strangenly looking shape as in the current proof #2. These four triangles correspond in pairs to the starting and ending positions of the rotated triangles in the current proof. This same configuration could be observed in a proof by tesselation.)\n\n\n\nProof #3\n    \n\nNow we start with four copies of the same triangle. Three of these have been rotated 90o, 180o, and 270o, respectively. Each has area ab/2. Let's put them together without additional rotations so that they form a square with side c.\n\n    \n\nThe square has a square hole with the side (a-b). Summing up its area (a-b)2 and 2ab, the area of the four triangles (4·ab/2), we get\n\n  c2 = (a-b)2+2ab \n  = a2-2ab+b2+2ab \n  = a2+b2 \n \n\n\n\nProof #4\n    \n\nThe fourth approach starts with the same four triangles, except that, this time, they combine to form a square with the side (a+b) and a hole with the side c. We can compute the area of the big square in two ways. Thus\n\n  (a + b)2 = 4·ab/2 + c2  \n\nsimplifying which we get the needed identity.\n\n\n\nProof #5\n\nThis proof, discovered by President J.A. Garfield in 1876 [Pappas], is a variation on the previous one. But this time we draw no squares at all. The key now is the formula for the area of a trapezoid - half sum of the bases times the altitude - (a+b)/2·(a+b). Looking at the picture another way, this also can be computed as the sum of areas of the three triangles - ab/2 + ab/2 + c·c/2. As before, simplifications yield a2+b2=c2.\n\nTwo copies of the same trapezoid can be combined in two ways by attac