Indivisibility by 529 (Posted on 2016-07-25)

	Submitted by Brian Smith
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First note that 529 is the square of the prime number 23. Rearrange the quadratic x^2+7x+18 into (x+15)*(x-8) + 23*6. x+15 and x-8 are either both multiples of 23 or neither are. Then (x+15)*(x-8) is either a multiple of or coprime to 529. 23*6 is a multiple of 23 but not 529. If (x+15)*(x-8) is a multiple of 529 then the total is a multiple of 23 but not 529. Also if (x+15)*(x-8) is not a multiple of 529 then the total is coprime to 23 and 529. In either case x^2+7x+18 is not a multiple of 529.

	Subject	Author	Date
	re: completing the square	Ady TZIDON	2016-07-25 23:34:57
	completing the square	xdog	2016-07-25 17:47:27
	Brute force.	broll	2016-07-25 14:38:59