Integer coefficients (Posted on 2007-03-13)

	Submitted by K Sengupta
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Clearly, the only possible integer roots of the polynomial x^5 + 9x +1 are +/- 1 and these are not zeroes of this polynomial, it follows that it has no linear factors. Accordingly, we express x^5 + 9x +1 = p(x)*q(x); where p(x) is of degree 2 and q(x) is of degree 3. Since the coefficient of x^0 in the given polynomial is 1, we can assume without loss of generality p(x) = x^2 + a*x +b and q(x) = x^3 + c*x^2 + d*x +e. Accordingly, be=1, so that : (b,e) =(1,1) or (-1,-1). Case I: b=e=1 Comparing the coefficients of x, we obtain ae + bd=9; that is, a+d=9-------(1) Also, comparing the coefficients of x^4 and x^3, we obtain: a+c=0 giving a=-c and, b+ac+d = 0; or, b+d=a^2; or a^2 = d+1.----------(2) From (1) and (2), we obtain: a^2 -1 + a = 9; giving, a^2 + a -10 = 0. This equation does not admit of any integer roots, and this is accordingly, a contradiction. Case II: b=e=-1. Hence, a+d=-9; b+d = a^2 and a^2 = d-1 Therefore, a + a^2 + 1 = -9; giving, a^2 + a + 10 = 0, which does not possess any integer roots. This is a contradiction. . Consequently, there cannot exist polynomials p(x) and q(x) each having integer coefficients and of degree greater than or equal to 1 such that: p(x)*q(x)= x^5 + 9x +1.

	Subject	Author	Date
	No Linear Factors	Brian Smith	2007-03-14 21:09:54
	Outline of solution	Jer	2007-03-13 09:34:49
	cheating (WIMS Factoris)	Charlie	2007-03-13 08:31:49