# Difference between revisions of "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
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