Find Integral values in a Quadratic equation

[addzy94]

Homework Statement

The number of integral values of 'a' for which the quadratic equation (x + a) (x + 1991) + 1 = 0
has integral roots are:

Homework Equations

D = b² - 4ac

The Attempt at a Solution

What I did was simplify the given equation and I got:

x² + (1991 + a)x + (1991a + 1) = 0

Now: (1991+a)² > 4(1991a + 1)

What do I do after this: Please help, I am not even sure that I am on the correct path.

 

[Saitama]

Homework Statement

The number of integral values of 'a' for which the quadratic equation (x + a) (x + 1991) + 1 = 0
has integral roots are:

Homework Equations

D = b² - 4ac

The Attempt at a Solution

What I did was simplify the given equation and I got:

x² + (1991 + a)x + (1991a + 1) = 0

Now: (1991+a)² > 4(1991a + 1)

What do I do after this: Please help, I am not even sure that I am on the correct path.

Assume that $b$ and $c$ are the integer roots. Then,
$$(x+a)(x+1991)+1=(x+b)(x+c)$$
Substitute $x=-b$, you should be able to take it from here.

 

[addzy94]

I don't know how that would work. It's alright I got the answer. As both a and x are integers

(x + a)(x+1991) = -1

Thus (x + a) = 1 or -1

(x+1991) = -1 or 1

So we get two integral values for a. I would also like to know how to do it your way.

 

[HallsofIvy]

I think you are misunderstanding the question. Isn't it obvious that there will be an infinite number of such a?

Your "(1991+a)² > 4(1991a + 1)" is simply the condition that the roots be real numbers.

 

[pasmith]

I think you are misunderstanding the question. Isn't it obvious that there will be an infinite number of such a?

There are indeed only two integer values of $a$ which ensure that $x$ is an integer, but the OP's argument in post #3 doesn't actually prove that.

Your "(1991+a)² > 4(1991a + 1)" is simply the condition that the roots be real numbers.

Instead the condition the OP needs is that $(1991 + a)^2 - 4(1991 a + 1)$ must be the square of an integer.

 

[addzy94]

There are indeed only two integer values of $a$ which ensure that $x$ is an integer, but the OP's argument in post #3 doesn't actually prove that.

There can only be two integers whose multiplication can result in -1. That is 1 and -1.

In my previous post I substitute 1 and minus -1 alternatively for the two integer values (x+a) and (x+1991) which are integer values themselves, thus proving it.