# 1989 AHSME Problems/Problem 8

## Problem

For how many integers $n$ between 1 and 100 does $x^2+x-n$ factor into the product of two linear factors with integer coefficients? $\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }$

## Solution 1

For $x^2+x-n$ to factor into a product of two linear factors, we must have $x^2+x-n = (x + a)(x + b)$, where $a$ and $b$ are integers.

By expansion of the product of the linear factors and comparison to the original quadratic, $ab = -n$ $a + b = 1$.

The only way for this to work if $n$ is a positive integer is if $a = -b +1$.

Here are the possible pairs: $a = -1, b = 2$ $a = -2, b = 3$ $\vdots$ $a = -9, b = 10$

This gives us 9 integers for $n$, $\boxed{\text{D}}$.

## Solution 2

For $x^2+x-n$ to factor into a product of two linear factors, the discriminant must be a perfect square. The discriminant equals $\sqrt{1^2-4(1)(-n)}$ or $\sqrt{4n+1}$.

Since $n$ must be a positive integer, $4n+1$ must be odd because if it were even, $4n$ would have to be both odd and divisible by 4, which is a contradiction. Therefore, $n$ must be even.

The maximum value of $4n+1$ is 401, and the minimum is 5. Therefore, we must find the number of odd perfect squares between 5 and 401. There are 9: $9,25,49,81,121,169,225,289,$and $361$. Therefore, the answer is $\boxed{D}$.

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