2010 AIME II Problems/Problem 6

Problem

Find the smallest positive integer n with the property that the polynomial $x^4 - nx + 63$ can be written as a product of two nonconstant polynomials with integer coefficients.

Solution

There are 2 ways for a monic fourth degree polynomial to be factored, into a cubic and a linear equation, or 2 quadratics.

Case 1) cubic and linear

Let $(x-r_1)$ be the linear equation (it must contain one root of the quatic)

and $x^3+ax^2+bx+c$ be the cubic.

By rational roots theorem, $r_1=1,3,7, 9$ , or $63$

$(x^3+ax^2+bx+c)(x-r_1)=x^4+(a-r_1)x^3+(b-ar_1)x^2+(c-br_1)x-cr_1$

So $a-r_1=0 \rightarrow a=r_1$ , $b-ar_1=0\rightarrow b=a^2$ ,

$c-br_1=-n\rightarrow n=a^3-c$ , and $-cr_1=63 \rightarrow c=\frac{-63}{a}$ .

So $n=a^3+\frac{63}{a}$ , $r_1=1,3,7, 9$ , or $63$ , which reach minimum when $r_1=3$ , where $n=48$

Case 2) 2 quadratic

Let $x^2+ax+b$ and $x^2+cx+d$ be the two quadratics,

$(x^2 + ax + b )(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd$

Therefore, we have

$a + c = 0\rightarrow a=-c$ , $b + d + ac = 0\rightarrow b+d=a^2$ , $ad + bc = - n$

and $bd = 63$ .

$b+d=a^2$ ,hence the only possible values for (b,d) are (1,63) and (7,9). From this we find that the possible values for n are $\pm 8 * 62$ and $\pm 4 * 2$ . Therefore, the answer is $\boxed{8}$ .

2010 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2010 AIME II Problems/Problem 6

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