THE TWIN PRIMES CONJECTURE, a Proof by Contradiction by Brian S. McMillan If we assume that for some twin prime set (P1, P2) that these members constitute the largest or only pair of twin primes that could exist. IF: ((P1 * P2 + 1) /4)^(1/2) = N THEN: (P1 * P2 + 1) /4 ą N = (P1 + 1) = (P2 - 1) HENCE: P1 + 2 = P2 OR: P2 - 2 = P1 For example: IF: ((3 * 5 + 1) /4)^(1/2) = 2 THEN: (3 * 5 + 1) /4 + 2 = 6 = (5 + 1) = (7 - 1) IF: ((5 * 7 + 1) /4)^(1/2) = 3 THEN: (5 * 7 + 1) /4 + 3 = 12 = (11 + 1) = (13 - 1) IF: ((11 * 13 + 1) /4)^(1/2) = 6 THEN: (11 * 13 + 1) /4 + 6 = 42 = (41 + 1) = (43 - 1) IF: ((41 * 43 + 1) /4)^(1/2) = 21 THEN: (41 * 43 + 1) /4 + 21 = 462 = (461 + 1) = (463 - 1) IF: ((461 * 463 + 1) /4)^(1/2) = 231 THEN: (461 * 463 + 1) /4 + 231 = 53592 = (53591 + 1) = (53593 - 1) IF: ((53591 * 53593 + 1) /4)^(1/2) = 26796 THEN: (53591 * 53593 + 1) /4 + 26796 = 718052412 = (718052411 + 1) = (718052413 - 1) Which are NOT PRIME IF: ((53591 * 53593 + 1) /4)^(1/2) = 26796 THEN: (53591 + 53593 + 1) /4 - 26796 = 717998820 = (717998819 + 1) = (717998821 - 1) Which are NOT PRIME So ends our first RUN! So we have established that our first assumption that (3, 5) were the largest or only known twin primes that could exist. There must, therefore, be some twin prime sets that exist between our first set and our greatest set as yielded above. For example: IF: ((17 * 19 + 1) /4)^(1/2) = 9 THEN: (17 * 19 + 1) /4 - 9 = 72 = (71 + 1) = (73 - 1) IF: ((71 * 73 + 1) /4)^(1/2) = 36 THEN: (71 * 73 + 1) /4 - 36 = (1259 + 1) Which is the ONLY PRIME in this series. IF: ((101 * 103 + 1) /4)^(/2) = 51 THEN: (101 * 103 + 1) /4 - 51 = (2549 + 1) = (2551 - 1) IF: ((2549 * 2551 + 1) /4)^(1/2) = 1275 THEN: (2549 * 2551 + 1) /4 - 1275 = (1624349 + 1) = (1624351 - 1 ) Which, again, is sufficient proof of the falicy of our original assumption. So ends the extension of my proof by contradiction for the Twin Primes Conjecture. While this does not work for all Twin Prime Sets... it does however work for a sufficient number of them to constitute some what of a TWIN PRIMER of sorts. This is really shocking in the number of Primes and Twin Primes that it hits. Hope you like it. To see the rest of the proof, go to: www.godkings.com/twinprime.txt The Infinitude of Primes Proof by Contradiction After Euclid: IF: N = (P1 * P2 * P3 * ... * Pn + 1) WHERE: (P1 to Pn) is a collection of all the known primes thought to exist. THEN: N must be divisible by at least one prime which is not a member of our original set. HENCE: N - 2 = (P1 * P2 * P3 * ... * Pn - 1) THEN: N - 2 must be divisible by at least one prime which is not a member of our original set. After Fermat: IF: (2^(P-1) - 1) /P = Integer WHERE: P = PRIME or PSEUDO-PRIME After McMillan: FOR: Some Integer N IF: (2^((P-1)/2) ą 1) /P = N Integer THEN: (2^(P-1) - 1) /P = N (PN ą 2) After Euclid-Fermat IF: N - 2 = (2 * 2 * 2 * ... * 2 - 1) WHERE: 2 is a collection of all the known primes thought to exist. IF: (2^(3-1) - 1) /3 = 1 = (2^1 - 1) AND: (2^(5-1) - 1) /5 = 3 = (2^2 - 1) THEN: (2^(7-1) - 1) /7 = 9 = (2^3 + 1) After Euclid-McMillan FOR: Some Integer N IF: (2^((P-1)/2) ą 1) /P = N Integer THEN: N = (2 + 1) HENCE: (2^((3-1)/2) + 1)/ 3 = 1 IF: N = (2 * 2 + 1) HENCE: (2^((5-1)/2) + 1)/ 5 = 1 IF: N = (2 * 2 * 2 - 1) HENCE: (2^((7-1)/2) - 1)/ 7 = 1 HENCE: We can see that for some n multiples of 2, of which 2 is the largest prime thought to exist, that Fermat uses a special case for Euclid when the base is 2, for his prime number theorem. And that these two approaches are in fact identical. Furthermore, it must be noted that the Wilson Theorem also belongs to this group. But more on that at a later date. Some time during the middle of the 1600's, Pierre de Fermat discovered an induction format contextual proof for grouping prime numbers into 2 of 4 categories. The four categories are as follows: WHERE: n = some integer Even integer = 4n Odd integer = 4n + 1 Even integer = 4n + 2 Odd integer = 4n + 3 FOR: Some prime number P = 4n + 1 FOR: Some prime number P = 4n + 3 This means that all prime numbers must belong to either the (4n + 1) or (4n + 3) category. This also means that any two odd numbers which are separated by 2 can never be of the same category. All (4n + 1) series Primes are only once the sum of two perfect squares. While all (4n + 3) series Primes are never the sum of two perfect squares. The induction format is the most efficient method for any type of mathematical proof that there can ever be. If the first odd number equals 1 = (4*0 + 1) If the second odd number is prime equals 3 = (4*0 + 3) If the third odd number is prime equals 5 = (4*1 + 1) If the fourth odd number is prime equals 7 = (4*1 + 3) If the fifth odd number equals 9 = (4*2 + 1) If the sixth odd number is prime equals 11 = (4*2 + 3) If the seventh odd number prime equals 13 = (4*3 + 1) I could go on... however, the point is that the first odd twin prime series and only odd numbers separated by 2, begins with (4n + 3) and ends with 4(n+1) + 1... and the second series immediately adjacent to the first, begins with 4(n+1) + 1... which is coincidentally the end of the previous series and ends with 4(n+1) + 3... of which these first three odd primes comprising the first two twin prime series also constitute the first and only prime triplet in the set of real integers. However, it is not necessary to prove this for the proof at hand. It is only necessary to illustrate a primary beginning to the two odd number series of (4n + 1) and (4n + 3). One last note: This also indicates that there is a sub-series embedded within the the two series which are the topic of this discussion. Now, as we know, n can be any integer for the series as long as it satisfies the equivalent for some prime such that: IF: (3, 5) comprise the first odd twin prime set. AND: (5, 7) comprise the second twin prime set. Which means that the third twin prime set must begin with either (4n + 1) or (4n + 3). It just so happens that.... 11 = (4*2 + 3) And its twin equals........ 13 = (4*3 + 1) IF: (11 - 3) = 8 = 2^3 AND: (13 - 5) = 8 = 2^3 THEN: 3 + 8 = 11 SO: 5 + 8 = 13 The next twin prime set and all others thereafter will always be separated from the first two odd prime sets by an even number which is, therefore, a multiple of 2 and beginning with one of only two possible series in the first two sets. For example: IF: (2657 - 1) /4 = 664 THEN: 2657 and 1 must both be members of the (4n + 1) series AND (2659 - 3) /4 = 664 THEN: 2659 and 3 must both be members of the (4n + 3) series OR: (7877 - 1) /4 = 1969 THEN: 7877 and 1 must both be members of the (4n + 1) series SO: (7879 - 3) /4 = 1969 AGAIN: 7879 and 3 must both be members of the (4n + 3) series IF: (4517 - 1) /4 = 1129 AND: (4519 - 3) /4 = 1129 OR: (9857 - 1) /4 = 2464 THEN: (9859 - 3) /4 = 2464 Since our very first twin prime set does not share the same remainder n, and we know that twin prime sets must contain primes separated by 2... then this is our first contradiction. So either the number 1 must be considered a prime or our contradiction holds. Because 0 cannot be divided by 0 and therefore, the prime number 2 also cannot be a member of the set. _________________________________________________________ PROOF: IF: (3 - 3) /4 = 0 AND: (5 - 1) /4 = 1 IF: (5 - 1) /4 = 1 AND: (7 - 3) /4 = 1 ALSO: (11 - 3) /4 = 2 AGAIN: (13 - 1) /4 = 3 For some (4n + 3) series prime there must be some prime number greater than (4n + 3) which is a 4(n+1) + 1 series prime... and a member of a set of twin primes. IF: (3167 - 3) /4 = 791 = n THEN: (3169 - 1) /4 = 792 = (n+1) _________________________________________________________ The above example completes the proof. Not to beat a dead horse or anything... but this also means that beginning with our first twin prime set of (3, 5) there is some even integer that when added to 3 will correspond with the next greater twin prime in the set of infinite primes. Now to quote a phrase that William Dunham uses in his book "The Mathematical Universe". "To respond to the accusation that the related examples for twin primes were contrived... I plead guilty! They were contrived." And "admittedly this strategy may seem strangely indirect and unnecessarily devious." However, it is quite valid in support of the argument. But mathematically speaking, I owe it all to Pierre de Fermat for reasoning the series', without which this proof would not have been possible. The next proof for this same theorem, the one and only at the beginning of this page, I owe to Euclid... yep! You got it... the man himself. PYTHAGORAS, FERMAT and GAUSS Pythagorean Induction Theorem by Brian S. McMillan AFTER Gauss, Karl Friedrich: For any number n Ascending series sum; IF: 1+2+3...+ n = n (n+1)/2 WHERE n = Integer AFTER Pythagoras: For any odd number n odd Ascending series sum; IF: 1+3+5...+ n odd = (n odd +1)^2 /4 AFTER McMillan: For any even number n even IF: 2+4+6...+ n even = ((n even+1)^2-1)/4 AND: (n odd+1)^2/4 + ((n even+1)^2-1)/4 = n (n+1)/2 WHERE: n odd+1 = n even = n FOR ANY NUMBER n IF: n^2 + (n+1)^2 - 1 = 2n (n+1) THEN: n^2 + (n+1)^2 = 2n (n+1) + 1 SO: (n-1)^2 + n^2 = (n+n) (n-1) + 1 HENCE: a^2 + b^2 = c^2 THE IDENTITY FOR THE EQUATION To be added later GAUSS-FERMAT QUADRATIC INDUCTION THEOREM IF: (2^P -2) = 2 PN (PNą2) AND: (2 PN (PN+2) - (PN+2)^2 + 2^2)^(1/2) = PN THEN: (2 PN (PN-2) - (PN-2)^2 + 2^2)^(1/2) = PN OR: (2^P - (PNą2)^2 + 2)^(1/2) = PN ALSO: (2^P - (2^((P-1)/2)ą1)^2 + 2)^(1/2) = PN In the last two equations above the quadratic is in it's simplest terms for the non-hyperbolic or non-parabolic form of the "Little Theorem of Fermat"... analogous to the simplest long form or format "FOR ANY NUMBER n" at the beginning of this page. Contrary to the last two examples above which always have a positive integer as the fourth term, the next two examples below, obviously not in their simplest terms, define through their symmetry a break in the terms which give the "Fermat Little Theorem" it's character or functionality. WHERE: PN is defined HENCE: (2^((P-1)/2) +1) = (PN+2) ALSO: (2^((P-1)/2) -1) = (PN-2) HYPERBOLIC QUADRATIC EQUATION FOR THE GAUSS-FERMAT INDUCTION THEOREM WHERE: PN is defined IF: ((2^P / (PNą2)^2 - 1) * (PNą2)^2 + 2)^(1/2) = PN AND: ((2^P+2)/(PNą2)^2 - 1) * (PNą2)^2)^(1/2) = PN GENERAL HYPERBOLIC QUADRATIC FOR THE GAUSS INDUCTION THEOREM WHERE: n = Integer IF: ((2n (n+1)/(n+1)^2 -1)(n+1)^2 + 1)^(1/2) = n THEN: ((2n /(n+1) -1)(n+1)^2 + 1)^(1/2) = n OR: ((n^2 /(n+1) -1)(n+1) + (n+1))^(1/2) = n HENCE: n^2 + 2n + 1 - (n+1)^2 = 0 WHERE: n = Integer THEN: ((n (n+1)/(n+1) -1)(n+1) + 1)^(1/2) = n OR: ((n-1)(n+1) +1)^(1/2) = n HENCE: n (n+1) - (n+1)^2 + n + 1 = 0 PROGRESSIVE SERIES HYPERBOLIC QUADRATIC IF: ((2 (n+1)/(n+1) -1)(n+1)^2 +1)^(1/2) = (n+1) AND: ((2 (n+2)/(n+1) -1)(n+1)^2 +1)^(1/2) = (n+2) AND: ((2 (n+2)(n+1) - (n+1)^2 + 1)^(1/2) = (n+2) THEN: (n+1) (n+2) - (n+1)^2 - n - 1 = 0 OR: 2 (n+1)^2 - (n+2)^2 - n^2 + 2 = 0 ((2 (2n+n/2)/(n+1) -1)(n+1)^2 - 2 (n+n/2) + 1)^(1/2) = 2n FOR: Some X and n WHERE: X and n are defined THEN: ((2 (X+n)/(n+1) -1)(n+1)^2 +1)^(1/2) = (X-n) The final equation listed immediately above, is when this begins to get interesting... do you know why? Signed: Brian S. McMillan Copyright (1996-2005) Except quotes, Copyright 1996-2006, Brian S. McMillan