Find all integer value of $a$ ... 
Find all integer values of $a$ such that the quadratic expression
$(x+a)(x+1991)+1$
can be factored as a product $(x+b)(x+c)$ where $b$ and $c$ are integers.

I tried expanding the brackets and equating it with $(x+b)(x+c)$ :
$x^2 + 1991x + ax+1991a + 1 = x^2 + cx + bx + bc$
$=>(1991+a)x + (1991a+1)  = (c+b)x + bc$
If I equate corresponding coefficients, I get two equations with three unknowns.
What is the "proper" approach to solve such problems ?
EDIT : This problem has been taken from a math contest "RMO" held in the year 1991. So I think this problem can be generalized for the expression $(x+a)(x+\lambda)+1$. How to solve it in that case ?
 A: Your assumption is that the polynomial
$$
(x+a)(x+1991)+1
$$
has integer roots $-b, -c$.
Now if $\xi$ is an integer such that
$$
(\xi + a) (\xi + 1991) = -1,
$$
then there's not much choice, the two factors must be $1$ and $-1$.
You get two possibilities. One is
$$
\begin{cases}
\xi + a = 1\\
\xi + 1991 = -1\\
\end{cases}
$$
whence $a - 1991 = 2$ and $a = 1993$; the other is
$$
\begin{cases}
\xi + a = -1\\
\xi + 1991 = 1\\
\end{cases}
$$
whence $a - 1991 = -2$ and $a = 1989$.

The generalization to $(x + \lambda) (x + \mu) + 1$, where $\lambda, \mu$ are integers, is straightforward. You get either
$$
\begin{cases}
\xi + \lambda = 1\\
\xi + \mu = -1\\
\end{cases}
$$
whence $\lambda = \mu + 2$, or
$$
\begin{cases}
\xi + \lambda = -1\\
\xi + \mu = 1\\
\end{cases}
$$
whence $\lambda = \mu - 2$.
A: we have $$(x+a)(x+1991)+1=x^2+x(a+1991)+1991a+1$$ this should equal to $$x^2+x(b+c)+bc$$ and from here we get $$bc=1991a+1$$ and $$b+c=a+1991$$
solving both equations for $a$ we get
$$bc-1=1991b+1991c-1991^2$$
frm here we get
$$c=\frac{1991b+1-1991^2}{b-1991}$$
can you proceed?
A: It is most convenient to represent the quadratic equation as:
$$n\,{x}^{2}+2\,n\,p\,x+{p}^{2}-m$$
then equathion $\left( x+b\right) \,\left( x+c\right)$ write as: ${x}^{2}+2\,p\,x+{p}^{2}-{m}^{2}$
The generalization to $\left( x+a\right) \,\left( x+t\right) +k$, solving the system of equations get:
$$a=t\pm2\,\sqrt{{m}^{2}+k}$$
In our case $t=1991, k=1$, then $m=0$, $a=1993, a=1989$
