Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2010 AIME II Problems/Problem 6

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_6&oldid=175480"