All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Integers and Prime Relationship (Posted on 2015-09-27)
Find all integers k such that : gcd(4*M +1, k*M+1) = 1 for every integer value of M.

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 The pattern is clear (spoiler) | Comment 1 of 2
After playing with this a little, the pattern becomes clear.

Let M = 2.  Then we need gcd(9,2k+1) = 1.  k cannot be 4, or 4 + 3n where n is any integer

Let M = 1.  Then we need gcd(5,k+1) = 1.  k cannot be 4, or 4 + 5n where n is any integer

Let M = 5.  Then we need gcd(21,5k+1) = 1.  k cannot be 4 + 7n where n is any integer (or 4 + 3n, but we knew that already)

Let M = 8.  Then we need gcd(33,8k+1) = .1  k cannot be 4, or 4+ 11n where n is any integer (or 4 + 3n, but we knew that already).

In fact, k cannot be 4 + ab where a is odd and n is any integer

So, k can only be 4 +/- 2^n where n is any non-negative integer

Possible values include
n = 0   k = 4 +/-  1 = 3 or 5
n = 1   k = 4 +/-  2 = 2 or 6
n = 2   k = 4 +/-  4 = 0 or 8
n = 3   k = 4 +/-  8 = -4 or 12
n = 4   k = 4 +/- 16 = -12 or 20
n = 5   k = 4 +/- 32 = -28 or 36
etc.

possible values therefore include
... -28, -12, -4, 0, 2, 3, 5, 6, 8, 12, 20, 36 ...

 Posted by Steve Herman on 2015-09-27 19:25:15

 Search: Search body:
Forums (0)