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

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== Solution 2 ==
 
== Solution 2 ==
 
Let <math>x^4-nx+63=(x^2+ax+b)(x^2+cx+d)</math>. From this, we get that <math>bd=63\implies d=\frac{63}{b}</math> and <math>a+c=0\implies c=-a</math>. Plugging this back into the equation, we get <math>x^4-nx+63=(x^2+ax+b)\left(x^2-ax+\frac{63}{b}\right)</math>. Expanding gives us <math>x^4-nx+63=x^4+\left(-a^2+b+\frac{63}{b}\right)x^2+\left(\frac{63a}{b}-ab\right)x+63</math>. Therefore <math>-a^2+b+\frac{63}{b}=0</math>. Simplifying gets us <math>b(a^2-b)=63</math>. Since <math>a</math> and <math>b</math> must be integers, we can use guess and check for values of <math>b</math> because <math>b</math> must be a factor of <math>63</math>. Note that <math>b</math> cannot be negative because <math>a</math> would be imaginary. After guessing and checking, we find that the possible values of <math>(a,b)</math> are <math>(\pm 8, 1), (\pm 4, 7), (\pm 4, 9),</math> and <math>(\pm 8, 63)</math>. We have that <math>n=ab-\frac{63a}{b}</math>. Plugging in our values for <math>a</math> and <math>b</math>, we get that the smallest positive integer value <math>n</math> can be is <math>\boxed{008}</math>.
 
Let <math>x^4-nx+63=(x^2+ax+b)(x^2+cx+d)</math>. From this, we get that <math>bd=63\implies d=\frac{63}{b}</math> and <math>a+c=0\implies c=-a</math>. Plugging this back into the equation, we get <math>x^4-nx+63=(x^2+ax+b)\left(x^2-ax+\frac{63}{b}\right)</math>. Expanding gives us <math>x^4-nx+63=x^4+\left(-a^2+b+\frac{63}{b}\right)x^2+\left(\frac{63a}{b}-ab\right)x+63</math>. Therefore <math>-a^2+b+\frac{63}{b}=0</math>. Simplifying gets us <math>b(a^2-b)=63</math>. Since <math>a</math> and <math>b</math> must be integers, we can use guess and check for values of <math>b</math> because <math>b</math> must be a factor of <math>63</math>. Note that <math>b</math> cannot be negative because <math>a</math> would be imaginary. After guessing and checking, we find that the possible values of <math>(a,b)</math> are <math>(\pm 8, 1), (\pm 4, 7), (\pm 4, 9),</math> and <math>(\pm 8, 63)</math>. We have that <math>n=ab-\frac{63a}{b}</math>. Plugging in our values for <math>a</math> and <math>b</math>, we get that the smallest positive integer value <math>n</math> can be is <math>\boxed{008}</math>.
 +
-Heavytoothpaste
  
 
== See also ==
 
== See also ==

Latest revision as of 16:47, 29 June 2022

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 1

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 $4 \cdot 2 = \boxed{008}$.

Solution 2

Let $x^4-nx+63=(x^2+ax+b)(x^2+cx+d)$. From this, we get that $bd=63\implies d=\frac{63}{b}$ and $a+c=0\implies c=-a$. Plugging this back into the equation, we get $x^4-nx+63=(x^2+ax+b)\left(x^2-ax+\frac{63}{b}\right)$. Expanding gives us $x^4-nx+63=x^4+\left(-a^2+b+\frac{63}{b}\right)x^2+\left(\frac{63a}{b}-ab\right)x+63$. Therefore $-a^2+b+\frac{63}{b}=0$. Simplifying gets us $b(a^2-b)=63$. Since $a$ and $b$ must be integers, we can use guess and check for values of $b$ because $b$ must be a factor of $63$. Note that $b$ cannot be negative because $a$ would be imaginary. After guessing and checking, we find that the possible values of $(a,b)$ are $(\pm 8, 1), (\pm 4, 7), (\pm 4, 9),$ and $(\pm 8, 63)$. We have that $n=ab-\frac{63a}{b}$. Plugging in our values for $a$ and $b$, we get that the smallest positive integer value $n$ can be is $\boxed{008}$. -Heavytoothpaste

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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