How to find the number of roots for a constant in a quadratic equation when the independent term is unknown?

Question

I'm going in circles with this question. As I don't know how to deal with the fact that the independent term is unknown. Typically when solving a quadratic equation you know the terms or they can be expressed as letters. But here I don't know how to find the number requested. Can somebody help me?

The problem is as follows:

If $n$ is a positive integer less than $100$, how many values can $n$ have so that the equation $x^2+x-n=0$ has two roots which are integers and different?

The alternatives found in my book are as follows:

$\begin{array}{ll} 1.&8\\ 2.&10\\ 3.&12\\ 4.&6\\ 5.&9\\ \end{array}$

In my attempt to solve the problem I tried what it was obvious and was calling the solution of the quadratic equation using the discriminant as follows:

Given the function:

$F(x)=x^2+x-n=0$

$x_{1,2}=\frac{-1\pm\sqrt{1^2+4n}}{2}$

However I'm stuck at the part where it mentions an integer less than $100$.

For this to happen the value on the numerator must be divisible by $2$. So the sum has to end in either $2,4,6,8,0$.

Since it is -1 when summed the square root result must end with those numbers and when subtracted from the square root must also end with those numbers.

It just happens that positive uneven numbers when summed to $-1$ will return an even number hence becoming divisible by $2$ and when subtracted to $-1$ will increase to an even number.

Therefore the result of the square root must be an uneven number. After I found this fact, I had to consider what values less than $100$ would also yield the same. $99\times 4 +1 = 397$ which is a number with no integer as square root, but gives us the idea that the number I'm looking for is of three digits and less than the square root of $397$.

The greatest uneven number which can happen before that number could be $19$ as $19^{2}=367$ but I found that there is no way to attain that number as $18$ is not divisible by $4$. So I kept searching. The next one $17$ yields $16$ so this can be a number, the next one $15$ doesn't, $13$ checks, $9$ also checks, $5$ is the final choice.

So from this I've found that the possible numbers would be:

$17^{2}=1+4n$

$n_{1}=\frac{288}{4}=68$

$13^{2}=1+4n$ $n_{2}=\frac{168}{4}=42$

$9^{2}=1+4n$ $n_{3}=\frac{80}{4}=20$

$5^{2}=1+4n$ $n_{4}=\frac{24}{4}=6$

So there are only four possibilities. These produces different roots and integers.

$x_{1}=\frac{-1\pm \sqrt{289}}{2}=\frac{-1+17}{2}=8$

$x_{2}=\frac{-1-17}{2}=-9$

This would also happen with the rest. However this answer is not within the alternatives. Could it be that I overlooked any possible solutions?. Can somebody help me with this problem?.

Answer 1

As you said, $$x_{1,2}=\frac{-1\pm\sqrt{1+4n}}{2}$$ must be integers and $n<100$. This means ${4n+1}$ must be an odd square $(9,25,49,...).$

Therefore there are $9$ possibilities for $n<100$: $2,6,12,20,30,42,56,72,$ and $ 90.$

Answer 2

Write $x^2+x-n=(x-r)(x-s)$ for some integers $r,s$.

Then $r+s=-1$ and $rs=-n$. The second expression implies that $r$ and $s$ have opposite signs, so without loss of generality assume $r>0$ and $s<0$. Observe that $n=r(-s)=r(r+1)$.

There are only $9$ ways that an integer between $0$ and $100$ can be factored as $r$ and $r+1$, namely, $1\cdot 2$, $2\cdot 3$, $\ldots,$ $8\cdot 9$, $9\cdot 10$.