find the no. of integers values of such that the quadratic expression (x+a)(x+1991) + 1can be factored as (x+b)(x+c). where b and c are integers

To determine the number of integer values for \( a \) such that the quadratic expression \( (x + a)(x + 1991) + 1 \) can be factored as \( (x + b)(x + c) \), where \( b \) and \( c \) are also integers, we need to analyze the expression step by step.

Expanding the Expression

First, let's expand the original expression:

• Starting with \( (x + a)(x + 1991) \), we have:
• \( x^2 + (a + 1991)x + a \cdot 1991 \)

Now, adding 1 to this gives:

• \( x^2 + (a + 1991)x + (a \cdot 1991 + 1) \)

Setting Up the Factorization

We want this quadratic to be factored as \( (x + b)(x + c) \). Expanding this gives:

• \( x^2 + (b + c)x + bc \)

For these two expressions to be equal, their coefficients must match:

• \( b + c = a + 1991 \)
• \( bc = a \cdot 1991 + 1 \)

Finding Relationships Between Variables

From the first equation, we can express \( c \) in terms of \( b \) and \( a \):

• \( c = a + 1991 - b \)

Substituting this into the second equation gives:

• \( b(a + 1991 - b) = a \cdot 1991 + 1 \)

Expanding this results in:

• \( ab + 1991b - b^2 = a \cdot 1991 + 1 \)

Rearranging terms leads to a quadratic in \( b \):

• \( b^2 - (1991 + a)b + (a \cdot 1991 + 1) = 0 \)

Applying the Discriminant Condition

For \( b \) to be an integer, the discriminant of this quadratic must be a perfect square. The discriminant \( D \) is given by:

• \( D = (1991 + a)^2 - 4(a \cdot 1991 + 1) \)

Expanding this gives:

• \( D = 1991^2 + 2 \cdot 1991a + a^2 - 4a \cdot 1991 - 4 \)
• \( D = a^2 - 2 \cdot 1991a + 1991^2 - 4 \)

This can be rewritten as:

• \( D = (a - 1991)^2 - 4 \)

Finding Integer Solutions

For \( D \) to be a perfect square, we set:

• \( (a - 1991)^2 - 4 = k^2 \) for some integer \( k \)

This leads to the equation:

• \( (a - 1991 - k)(a - 1991 + k) = 4 \)

The integer pairs that multiply to 4 are:

• \( (1, 4), (2, 2), (-1, -4), (-2, -2) \)

From each pair, we can solve for \( a \):

• For \( (1, 4) \): \( a - 1991 - k = 1 \) and \( a - 1991 + k = 4 \) leads to \( a = 1994 \)
• For \( (2, 2) \): \( a - 1991 - k = 2 \) and \( a - 1991 + k = 2 \) leads to \( a = 1993 \)
• For \( (-1, -4) \): \( a - 1991 - k = -1 \) and \( a - 1991 + k = -4 \) leads to \( a = 1988 \)
• For \( (-2, -2) \): \( a - 1991 - k = -2 \) and \( a - 1991 + k = -2 \) leads to \( a = 1989 \)

Counting the Solutions

Thus, the integer values of \( a \) that satisfy the conditions are \( 1994, 1993, 1988, \) and \( 1989 \). This gives us a total of four integer solutions.

In summary, the number of integer values for \( a \) such that the quadratic expression can be factored as required is 4.