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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 ?

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    Hint: $1991 = 11\cdot 181$ – 2017-01-14
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    Two equations with three unknowns, but with additional constraint that all are integers. That means factorization can be used and the number options is greatly reduced (and enumerable). – 2017-01-14
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    @Peter Thanks, I just forgot to type that, edited accordingly... – 2017-01-14

3 Answers 3

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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 $\lambda = \mu - 2$.

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    @mvw, please re-read the question. – 2017-01-14
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    And if $(x+a)(x+1991)+1 = (x+b)(x+c)$, then the roots of the polynomial $(x+a)(x+1991)+1$ are...? – 2017-01-14
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    $-b$ and $-c$. That is a good deduction. – 2017-01-14
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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?

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    I did the same till here , I actually do not know "how to proceed?" – 2017-01-14
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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 $m=0$, $a=1993, a=1989$