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

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==Problem==
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== Problem ==
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Find the smallest positive integer <math>n</math> with the property that the [[polynomial]] <math>x^4 - nx + 63</math> can be written as a product of two nonconstant polynomials with integer coefficients.
  
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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 integer coefficients.  
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==Solution==
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You can factor the polynomial into two quadratic factors or a linear and a cubic factor.
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For two quadratic factors, let <math>x^2+ax+b</math> and <math>x^2+cx+d</math> be the two quadratics, so that
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<cmath>(x^2 + ax + b )(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd.</cmath>
  
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==Solution==
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Therefore, again setting coefficients equal, <math>a + c = 0\Longrightarrow a=-c</math>, <math>b + d + ac = 0\Longrightarrow b+d=a^2</math> , <math>ad + bc = - n</math>, and so <math>bd = 63</math>.
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Since <math>b+d=a^2</math>, the only possible values for <math>(b,d)</math> are <math>(1,63)</math> and <math>(7,9)</math>. From this we find that the possible values for <math>n</math> are <math>\pm 8 \cdot 62</math> and <math>\pm 4 \cdot 2</math>.
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For the case of one linear and one cubic factor, doing a similar expansion and matching of the coefficients gives the smallest <math>n</math> in that case to be <math>48</math>.
  
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Will be up right now, typing^v^
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Therefore, the answer is <math>4 \cdot 2 = \boxed{008}</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}}
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Revision as of 23:42, 28 May 2021

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

You can factor the polynomial into two quadratic factors or a linear and a cubic factor.

For two quadratic factors, 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 $\pm 4 \cdot 2$.

For the case of one linear and one cubic factor, doing a similar expansion and matching of the coefficients gives the smallest $n$ in that case to be $48$.

Therefore, the answer is $4 \cdot 2 = \boxed{008}$.

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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