Factorize this!!

Expanding the expression, we have $f(x) = x^{2} + (1991 + a)x + (1991a + 1)$ .

Now $f(x) = 0$ when

$x = \dfrac{-(1991 + a) \pm \sqrt{(1991 + a)^{2} - 4*(1991a + 1)}}{2} =$

$\dfrac{-(1991 + a) \pm \sqrt{(1991 - a)^{2} - 4}}{2} =$

$\dfrac{-(1991 + a) \pm \sqrt{((1991 - a) + 2)((1991 - a) - 2)}}{2} =$

$\dfrac{-(1991 + a) \pm \sqrt{(1993 - a)(1989 - a)}}{2}$ .

In order to have integer roots, we will require that $(1993 - a)(1989 - a)$ be a perfect square. Now let $m = 1989 - a$ . Then

$(1993 - a)(1989 + a) = (m + 4)m = (m^{2} + 4m + 4) - 4 = (m + 2)^{2} - 4$ .

Now since the only pair of perfect squares which differ by $4$ are $0$ and $4$ , we must have either $m = 0$ or $m + 4 = 0$ , i.e., $a = 1989$ or $a = 1993$ .

The desired sum is then $1989 + 1993 = \boxed{3982}$ .

Comment: Note that in either case we end up with a double root, as the discriminant is zero. With $a = 1989$ we get $b = c = 1990$ , and with $a = 1993$ we get $b = c = 1992$ .

Let $p(x) = (x+a)(x+1991)+1 = (x+b)(x+c)$ .

Then the roots of $p(x)$ are $-b$ and $-c$ .

$p(-b) = (-b+a)(-b+1991)+1 = 0$

$(a-b)(1991-b) = -1$

Since $(a-b)$ and $(1991-b)$ are integers,

and $-1 = (-1)(1) = (1)(-1)$ ;

we can conclude that either $a-b = -1$ and $1991-b = 1$ ,

or $a-b = 1$ and $1991-b = -1$ .

The first case implies that $b = 1990$ and $a = 1989$ .

The second case implies that $b = 1992$ and $a = 1993$ .

Therefore, the required sum is $1989+1993 = 3982$ .

Expand both of the equations to get $x^{ 2 }+(a+1991)x+(1991a+1)=0,\quad x^2+(b+c)x+bc=0$

Since both equations have to be the same... $\begin{cases} b+c=a+1991\longrightarrow 1991b+1991c=1991a+1991^{ 2 } \\ bc=1991a+1 \end{cases}$ Subtract the second equation from the first equation to get $1991b+1991c-bc=1991^{ 2 }-1\longrightarrow bc-1991b-1991c+1991^{ 2 }=1\\ (b-1991)(c-1991)=1$

Because b,c are both integers, this has two solutions.

$b-1991=1,\quad c-1991=1\rightarrow b=1992,\quad c=1992\\ b+c=a+1991\longrightarrow 1992+1992=a+1991\\ \therefore a=1993$

$b-1991=-1,\quad c-1991=-1\rightarrow b=1990,\quad c=1990\\ b+c=a+1991\longrightarrow 1990+1990=a+1991\\ \therefore a=1989$

Therefore, the sum of the values of a is 1993+1989=3982

this is simply completing the square $x^2 + (1991+a)x + 1991a + 1$ since the constant can be completed by dividing the middle term by 2 and then squaring it, we can equate it as such: ( $\frac{1991+a}{2}$ )^2 = 1991a + 1

Let $f(x) = x+b, g(x) = x+c$ , so $P(x) = (x+a)(x+1991) + 1 = (x+b)(x+c) = f(x)g(x)$ . Observe $\begin{aligned} f(-b)g(-b) & =1 \\ f(-c)g(-c) &= 1 \end{aligned}$ Since they are all integer polynomials, we have $f(-b) = g(-b), f(-c) = g(-c)$ . So $f(x) - g(x)$ has 2 roots but degree at most 1. Thus it must be zero, meaning $b = c$ .

We have $(x+b)^2 = (x+a)(x+1991) +1 \implies (x + b-1)(x + b + 1) = (x+a)(x+1991).$ Matching the roots gives us $a = 1989, 1993$ .