2005 a square? Baseless! (Posted on 2005-05-30)

	Submitted by McWorter
Rating: 3.6000 (5 votes)
Solution:	(Hide)
Suppose 2b^3+5=x^2, with b and x integers. The left side is the sum of an even and an odd integer, so the left side is odd. Hence x is odd. Let x=2k+1. Then 2b^3+5=(2k+1)^2=4k^2+4k+1, or 2b^3=4k^2+4k-4=4(k^2+k-1), or b^3=2(k^2+k-1). Since the right side of the last equation above is even, b is even. But then the left side is divisible by 4. However, k^2+k is always even, so k^2+k-1 is odd. Hence the right side is not divisible by 4, contradiction. Therefore, 2005 base b cannot be a square.

	Subject	Author	Date
	Puzzle Solution	K Sengupta	2023-09-11 07:49:22
	re: Since it was not tacitily stated...	McWorter	2005-06-04 23:53:17
	Since it was not tacitily stated...	pcbouhid	2005-06-04 23:16:40
	Modulo solution	Nick Hobson	2005-06-01 20:10:16
	re(2): Solution	Robby Goetschalckx	2005-06-01 07:52:01
	A Modulus Based Solution !?!	brianjn	2005-06-01 07:30:34
	re(3): Solution	pcbouhid	2005-06-01 00:35:35
	re(3): Solution	pcbouhid	2005-06-01 00:32:00
	re(2): Solution	Bruno	2005-05-31 23:31:09
	re: Solution	pcbouhid	2005-05-31 19:35:47
	re: Possibly?	McWorter	2005-05-31 17:32:27
	Solution	Robby Goetschalckx	2005-05-31 09:01:23
	Possibly?	Michael S. McCracken	2005-05-31 04:38:22
	radical solution	Larry	2005-05-31 00:12:53
	re(3): when the base is odd...	Richard	2005-05-30 23:37:47
	re(2): when the base is odd...	pcbouhid	2005-05-30 22:14:18
	re: when the base is odd...	Richard	2005-05-30 22:05:55
	when the base is odd...	pcbouhid	2005-05-30 20:10:36