Difference between revisions of "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 Root 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 $\boxed{008}$.