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Difference between revisions of "2010 AIME II Problems/Problem 6"

m (Created page with '==Problem== Find the smallest positive integer n with the property that the polynomial <math>x^4 - nx + 63</math> can be written as a product of two nonconstant polynomials with…')
 
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==Solution==
 
==Solution==
  
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There are 2 ways for a monic fourth degree polynomial to be factored, into a cubic and a linear equation, or 2 quadratics.
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<br/>
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<b>Case 1)</b> cubic and linear
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Let <math>(x-r_1)</math> be the linear equation (it must contain one root of the quatic)
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and <math>x^3+ax^2+bx+c</math> be the cubic.
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By rational roots theorem, <math>r_1=1,3,7, 9</math>, or <math>63</math>
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<math>(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</math>
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So <math>a-r_1=0 \rightarrow a=r_1</math>, <math>b-ar_1=0\rightarrow b=a^2</math>,
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<math>c-br_1=-n\rightarrow n=a^3-c</math>, and <math>-cr_1=63 \rightarrow c=\frac{-63}{a}</math>.
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<br/>
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So <math>n=a^3+\frac{63}{a}</math>, <math>r_1=1,3,7, 9</math>, or <math>63</math>, which reach minimum when <math>r_1=3</math>, where <math>n=48</math>
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<br/>
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<br/>
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<b>Case 2)</b> 2 quadratic
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Let <math>x^2+ax+b</math> and <math>x^2+cx+d</math> be the two quadratics,
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<math>(x^2 + ax + b )(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd</math>
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Therefore, we have
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<math>a + c = 0\rightarrow a=-c</math>, <math>b + d + ac = 0\rightarrow b+d=a^2</math> , <math>ad + bc = - n</math>
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and <math>bd = 63</math>.
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<math>b+d=a^2</math> ,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 <math>\pm 8 * 62</math> and <math>\pm 4 * 2</math>. Therefore, the answer is <math>\boxed{8}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2010|num-b=5|num-a=7|n=II}}
 
{{AIME box|year=2010|num-b=5|num-a=7|n=II}}

Revision as of 17:09, 3 April 2010

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 $\boxed{8}$.

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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