A prime and a cube (Posted on 2020-07-13)

	Submitted by Brian Smith
Rating: 4.0000 (3 votes)
Solution:	(Hide)
Let the cube be q^3. Then 16p+1 = q^3. Subtract 1 from each side and factor into 16p = (q-1)*(q^2+q+1). The second factor can be written as q^2+q+1 = q*(q+1)+1. Since this is one greater than the product of consecutive integers it must be odd. p is prime thus only odd factors of 16p are p and 1. This implies two cases. Case 1: q^2+q+1=1 and 16p=q-1 In this case q=0 or -1, neither of which creates an integer p and then are not solutions. Case 2: 16=q-1 and q^2+q+1=p In this case q=17 which makes p=307 which is prime. Therefore the solution to the problem is p=307 which is the only prime to make 16p+1 equal a perfect cube.

	Subject	Author	Date
	Explanation to Puzzle Answer: Method II	K Sengupta	2022-05-28 03:59:52
	Explanation to Puzzle Answer	K Sengupta	2022-05-26 09:52:52
	Puzzle Answer	K Sengupta	2022-05-26 02:50:08
	No Subject	selena goz	2020-07-15 20:17:58
	re(3): Just an observation	broll	2020-07-14 23:00:55
	re(2): Just an observation (on an interesting program line)	Charlie	2020-07-14 07:35:28
	solution	xdog	2020-07-14 07:33:30
	Possible solution	broll	2020-07-13 22:45:58
	re(2): Just an observation	Charlie	2020-07-13 17:32:39
	re: Just an observation	Charlie	2020-07-13 15:44:09
	Just an observation	Charlie	2020-07-13 15:23:45