Prove that for each positive integer n there exist n consecutive positive integers none of which is an integral power of a prime number.
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.
25462
101 (b) is a prime number because it can't be divided (without leaving a remainder) by anything, except itself and 1.
667802
Your claim that there are infinitely many solutions for x and y for a given x and n does not hold if we consider only integer solutions.\n\nConsider that we have:\nx^n = z^2-y^2\nx^n = (z-y)(z+y)\nSince z, y are integers, |z-y| >= 1 and |z+y| >= 1\nAlso, |z-y| <= x^n and |z+y| <= x^n\nWe can assume that z,y > 0, since if there is a solution with one of them less than 0, then switching the sign will still be a solution.\nSo, now we know that |z+y| <= x^n.  For z,y >=0, this has x^n + 1 solutions (not all of these will solve the original equation, but there are no others that do).\n\nThus, there are at most 4x^n +4 unique solutions to this equation in the integers.  The others are found by taking a positive pair (z,y) and noticing that (-z,y) (z,-y), (-y,-z) are also solutions.\n\nHowever, if you are talking about non-integer solutions, then there are obviously infinitely many, and that result is not useful.
368830
Looks like it's 50 numbers. For example , let the numbers be\n51 thru 100 inclusive. Since even the lowest one can't divide\ninto the highest, none are divisible by the other.  Now lets' suppose there is another number less than 51 in this set. Then\nthat number times 2 will be some number less than or equal to\n100 so by contradiction, there is no other number less than 51\nin the set.
1015
A to the power of n is:\n\nA * A * A ... n times.\n\nor\n\n1 * A * A * A ... n times.\n\nsince n is 0, w're left with\n1.\n\n--\n\nbut really it's defined to be 1.
520800
Let the first integer be x.\n\nThe next consecutive integer is x+1.\n\nNow setup the equation.\n\n8(x) + 7(x+1) = 127.\n\n8x + 7x + 7 = 127.\n\n15x + 7 = 127.\n\n15x = 120\n\nx = 8.\n\nThis means that the integers are 8 and 9.\n\nTo check, 8 times 8 is 64. 7 times 9 is 63. 64 + 63 = 127.\n\nIt checks out!
265759
lets take your 3 integers as x, y and z consecutively, then y = x+1, and z= x+2\nsetting up the equations\nxy = -6z\nsubstituting for y and z\nx(x+1) = -6(x+2)\nopening up\nx^2 + x = -6x - 12\nrearranging\nx^2 + 7x + 12 = 0\nfactorising\n(x+3)(x+4) = 0\nso either x = -3 or x = -4\ntherefore ur consecutive numbers are -3, -2, -1 or -4,-3,-2
406836
If n is contains an odd prime factor, this is part of an unsolved problem called Beal's conjecture. This says that if any\nsuch solution exists, then a and b cannot be coprime.\nIf n is a power of 2, it is a multiple of 4, since n> 2.\nA glance at Wikipedia also shows that this\ncase is also part of Beal's conjecture.\nAt the present time no counterexamples to\nBeal's conjecture are known.\nIncidentally, there is a $100000 prize for the\nfirst proof or counterexample to Beal's conjecture.
138177
n, n+1 for every n>=2\n\nif 2 <= 3 then either 2 = 3 or 2^2 = 4 <=3, both of which are false.\nif 3 <= 2 then either 3 = 2 or 3^2 = 9 <=2, both of which are false.\n...\n\nif n <= n+1 then either n = n+1 or n^2 <= n+1, both of which are false for all n>=2.\nif n+1 <= n then either n+1 = n or (n+1)^2 <= n, both of which are false for all n>=1.\n\nThus, n and n+1 are incompartible for all n>=2.
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This is Goldbach's conjecture and has NOT been proven\ntrue or false.  Good luck solving it!!\n\nI looked at the above answers, adn they dont seem to understand the question.  You shouldl have said:\n\nAny even number greater than 2 CAN be written as the sum of two primes. Prove true or false.
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we need to prove this by induction...\n\nfor n belonging to Z\n\nf(n)=n*(n+1)*(2n+1)/6\n\nfor n=1\n\nLHS:\n f(1)= 1^2 =1\nRHS\n1*(1+1)8(2*1+1)/6=6/6=1\n\nThus its true when n=1;\n\nAssume f(k) is true i.e. \n1^2+2^2+......+k^2=k(k+1)(2k+1).. Let this be equation 1\n\nNow we need to prove f(k+1) is true...\nie\nf(k+1)=(k+1)(k+2)(2(k+1)+1)/6\n          =(k+1)(k+2)(2k+3)/6\nL.h.s\n\nf(k+1)=1^2+2^2........+k^2+(k+1)^2\n          =k(k+1)(2K+1)/6 + (k+1)^2 from eq 1\n          =(k+1)(k(2k+1)/6 +k+1)\n          =(1/6)(k+1)(2k^2+k+6k+6)\n          =(1/6)(k+1)(2k^2+7k+6)\n          =(1/6)(k+1)(k+2)(2k+3)\n\nHence f(k+1) is true if f(k) is true....\n\nThus its true for all positive integers.......
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All primes (ignoring 2) are odd.\n\nDivide odd numbers into 3 groups: those of the form 6n - 1, those of the form 6n + 1, and those of the form 6n + 3.\n\nIgnoring 3 itself, all primes are of the form 6n + 1 or 6n - 1, as all numbers of the form 6n + 3 or 6n - 3 are divisible by 3 and not prime. \n\nYour proposition follows from that fact.
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Having researched this topic as far as possible online, I found very few helpful resources.\n\nWhile the result you mention, using the subsequence 6(5)^n, is intriguing, I suspect that other sequences will yield vastly different bounds on the index.  In particular, is it not true that any prime number p always has a corresponding Fibonacci number F(p+1) of which it is a factor?\n\nGood luck with your own research!
511344
Is there an infinite number of twin primes? (i.e. N and N+2 are both prime).\n\nFor other problems, see:\nhttp://mathworld.wolfram.com/UnsolvedProblems.html
601657
three consecutive even integers are\nx, x+2 and x+4\n\nEquation is\n2((x+2)+(x+4)) = 6x - 12\n\nSolving for x\nx = 12\nSo, 12, 14 and 16
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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.
622260
differences between consecutive primes\n\n3-2 = 1\n5-3 = 2\n7-5 = 2\n11-7= 4\n13-11 = 2\n17-13 = 4\n19-17 = 2\n23-19 = 4\netc.