Always multiple (Posted on 2008-09-07)

	Submitted by pcbouhid
Rating: 4.0000 (2 votes)
Solution:	(Hide)
Nice work, KS. Exactly what I had. 8640 = 2^6 * 3^3 * 5. So we need to show that the given poly is divisible by 2^6(=64), 3^3(=27), and 5. We have: x^9 - 6x^7 + 9x^5 - 4x^3 = = x^3(x^6 - 6x^4 + 9x^2 -4) = = x^3 * (x^2 - 4) * (x^4 - 2x^2 +1) = = x^3 * (x^2 - 4) * (x^2 - 1)^2 = = x^3 * (x+2) * (x-2) * (x+1)^2 * (x-1)^2 = = (x-2) * (x-1)^2 * x^3 * (x+1)^2 * (x+2) = = (x-2) * (x-1) *(x-1) * x * x * x * (x+1) * (x+1) * (x+2) 5 is a factor of (x-2)(x-1)x(x+1)(x+2) since those 5 factors contain all residues mod 5. 3 is a factor of (x-2)(x-1)x and of (x-1)x(x+1) and of x(x+1)(x+2) and therefore 3^3=27 is a factor of the polynomial. 8 is a factor of (x-2)(x-1)x(x+1) and (x-1)x(x+1)(x+2), because in both sets of factors we have a set of all residues mod 4, and therefore one factor is divisible by 2 and one of the others by 4, and therefore 8^2 = 2^6 is a factor of the polynomial. So, the polynomial is divisible by 5, 27 and 64 and therefore by 5 * 27 * 64 = 8640 for any integer x.

	Subject	Author	Date
	Solution	K Sengupta	2008-09-07 13:14:50