Question

For what values of \(a \in \mathbb{Z}\), the quadratic expression \[ (x+a)(x+1991)+1 \] can be factorised as \[ (x+b)(x+c), \] where \(b,c \in \mathbb{Z}\)?

• A. \(1990\)
• B. \(1989\)
• C. \(1991\)
• D. \(1992\)

Hint:

When comparing two quadratic expressions, equate coefficients of corresponding powers of \(x\). Completing factor-type rearrangements often helps simplify integer factorization problems.

Answer 1

The Correct Option is B

Step 1: Aim for a perfect square.
A neat way to make $(x+a)(x+1991)+1$ factor over integers is to make it a perfect square $(x+k)^2$.
Step 2: Expand the perfect square.
$(x+k)^2=x^2+2kx+k^2$.
Step 3: Expand the given expression.
$(x+a)(x+1991)+1=x^2+(a+1991)x+1991a+1$.
Step 4: Match the linear coefficients.
$a+1991=2k$.
Step 5: Match the constant terms.
$1991a+1=k^2$. Substituting $k=\dfrac{a+1991}{2}$ gives $1991a+1=\dfrac{(a+1991)^2}{4}$, i.e. $4(1991a+1)=(a+1991)^2$, which simplifies to $a^2-2\cdot1991a+1991^2-4=0$, so $(a-1991)^2=4$.
Step 6: Solve and pick the option.
Then $a-1991=\pm2$, giving $a=1993$ or $a=1989$. The value listed is $1989$, which is option (2).
\[ \boxed{1989} \]