 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Power Sum Proposition (Posted on 2016-06-04) Does there exist an infinite number of positive integer values of N such that:
(1999.5)N + (2000.5)N is an integer?

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) solution Comment 1 of 1
Write (2000 - 1/2)^N + (2000 + 1/2)^N = [(4000-1)^N + (4000+1)^N]/2^N

For even N, each term in the expression in brackets is a multiple of 4000 and therefore 4, except the last which = 2.  So the bracketed value is twice an odd number and will not be divisible by 2^N.

For odd N, terms with even exponent cancel and terms with odd exponent double.  The smallest of these = 2*N*(4000) which is divisible by 2^6 and no greater power of 2.

So N is limited to 1,3,5.  Checking I find each to be a solution.

 Posted by xdog on 2016-06-04 13:59:35 Please log in:

 Search: Search body:
Forums (0)