2008 iTest Problems/Problem 84

Problem

Let $S$ be the sum of all integers $b$ for which the polynomial $x^2+bx+2008b$ can be factored over the integers. Compute $|S|$.

Solutions

Solution 1 (credit to official solution)

Let the roots of the quadratic be $r$ and $s$. By Vieta's Formulas, $r+s = -b$ and $rs$ = $2008b$.


We know that one of the possible values of $b$ is 0 because $x^2$ has integer roots. However, adding or removing 0 does not affect the value of $S$, so we can divide both sides by $-b$. Doing so results in \begin{align*} \frac{rs}{r+s} &= -2008 \\ rs &= -2008r - 2008s \\ rs + 2008r + 2008s &= 0 \\ (r+2008)(s+2008) &= 2008^2. \end{align*} WLOG, let $|a| \le 2008$ be a factor of $2008^2$, so $r+2008 = a$ and $s+2008 = \tfrac{2008^2}{a}$. Thus, \[-r-s = b = -a - \tfrac{2008^2}{a} + 4016.\] Since $a$ can be positive or negative, the positive values cancel with the negative values. The prime factorization of $2008^2$ is $2^6 \cdot 251^2$, so there are $\frac{21+2}{2} = 11$ positive factors that are less than $2008$. Thus, there are a total of $22$ values of $a$, so the absolute value of the sum of all values of $b$ equals $4016 \cdot 22 = \boxed{88352}$.

Solution 2

The discriminant of the function is $b^2 - 8032b$. Since all roots are integers and leading term is 1, the discriminant must equal $n^2$, where $n$ is an integer.


Thus, we know that \begin{align*} b^2 - 8032b =& n^2 \\ (b-4016)^2 - n^2 &= 4016^2 \\ (b-4016+n)(b-4016-n) &= 4016^2. \end{align*} Let $x$ be a factor of $4016^2$, so \begin{align*} b - 4016 + n &= x \\ b - 4016 - n &= \frac{4016^2}{x} \\ b &= 4016 + \frac{x}{2} + \frac{4016^2}{2x} \end{align*} Note that if $x$ is odd, then $\tfrac{4016^2}{x}$ is even, so $b$ can not be an integer. Thus, $x$ must be even. Let $x = 2y$, so $b = 4016 + y + \frac{2008^2}{y}$.


If $y_0 = \frac{2008^2}{y}$, then $y + \tfrac{2008^2}{y} = y_0 + \tfrac{2008^2}{y_0}$. Also, since $y$ can be positive or negative, the positive values cancel with the negative values. So WLOG, let $|y| \le 2008$.


The prime factorization of $2008^2$ is $2^6 \cdot 251^2$, so there are $\frac{21+2}{2} = 11$ positive factors that are less than $2008$. Thus, there are a total of $22$ values of $y$, so the absolute value of the sum of all values of $b$ equals $4016 \cdot 22 = \boxed{88352}$.

See Also

2008 iTest (Problems)
Preceded by:
Problem 83
Followed by:
Problem 85
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Invalid username
Login to AoPS