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

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Therefore, the answer is <math>4 \cdot 2 = \boxed{008}</math>.
 
Therefore, the answer is <math>4 \cdot 2 = \boxed{008}</math>.
 
== Solution 2 ==
 
== Solution 2 ==
 +
<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)(x^2-ax+\frac{63}{b})</math>. Expanding gives us <math>x^4-nx+63=x^4-(a^2+b+\frac{63}{b})x^2+(\frac{63a}{b}-ab)x+63</math>
  
 
== See also ==
 
== See also ==

Revision as of 16:26, 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

$x^4-nx+63=(x^2+ax+b)(x^2+cx+d)$. 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 $x^4-nx+63=(x^2+ax+b)(x^2-ax+\frac{63}{b})$. Expanding gives us $x^4-nx+63=x^4-(a^2+b+\frac{63}{b})x^2+(\frac{63a}{b}-ab)x+63$

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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