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Difference between revisions of "2010 AIME II Problems/Problem 6"
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Let <math>x-r_1</math> be the linear root, where <math>r_1</math> is a root of the given quartic, and let <math>x^3+ax^2+bx+c</math> be the cubic.  
 
Let <math>x-r_1</math> be the linear root, where <math>r_1</math> is a root of the given quartic, and let <math>x^3+ax^2+bx+c</math> be the cubic.  
  
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By the [[rational roots theorem]], then <math>r_1=1,3,7, 9</math>, or <math>63</math>. Observe that
+
By the [[Rational Roots Theorem]], then <math>r_1=1,3,7, 9</math>, or <math>63</math>. Observe that
  
 
<cmath>(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.</cmath>
 
<cmath>(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.</cmath>

Revision as of 04:00, 28 December 2014

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 $\boxed{008}$.

See also

2010 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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