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 two ways for a monic fourth degree polynomial to be factored into two non-constant polynomials with real coefficients: into a cubic and a linear equation, or 2 quadratics.

Case 1: The factors are cubic and linear.

Let $x-r_1$ be the linear root, where $r_1$ is a root of the given quartic, and let $x^3+ax^2+bx+c$ be the cubic.

By the Rational Roots Theorem, then $r_1=1,3,7, 9$ , or $63$ . Observe that

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

Setting coefficients equal, we have $a-r_1=0 \Longrightarrow a=r_1$ , $b-ar_1=0\Longrightarrow b=a^2$ , and $c-br_1=-n \Longrightarrow n=a^3-c$ , and $-cr_1=63 \Longrightarrow c=\frac{-63}{a}$ .

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

Case 2: The factors are quadratics.

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

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

Therefore, again setting coefficients equal, $a + c = 0\Longrightarrow a=-c$ , $b + d + ac = 0\Longrightarrow b+d=a^2$ , $ad + bc = - n$ , and so $bd = 63$ .

Since $b+d=a^2$ , the only possible values for $(b,d)$ are $(1,63)$ and $(7,9)$ . From this we find that the possible values for $n$ are $\pm 8 \cdot 62$ and $\pm 4 \cdot 2$ . Therefore, the answer is $\boxed{008}$ .

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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