For what values of a in mathbb{Z},the quadrat | vedclass

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(A) Given $(x+a)(x+1991)+1 = (x+b)(x+c)$.Expanding the left side: $x^2 + (a+1991)x + 1991a + 1 = x^2 + (b+c)x + bc$.Comparing coefficients,we have $b+

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$1989, 1993$

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(A) Given $(x+a)(x+1991)+1 = (x+b)(x+c)$.Expanding the left side: $x^2 + (a+1991)x + 1991a + 1 = x^2 + (b+c)x + bc$.Comparing coefficients,we have $b+c = a+1991$ and $bc = 1991a+1$.Since $b$ and $c$ are roots of the quadratic equation $x^2 - (b+c)x + bc = 0$,the discriminant $D$ must be a perfect square,say $m^2$.$D = (b+c)^2 - 4bc = (a+1991)^2 - 4(1991a+1) = m^2$.$(a+1991)^2 - 4(1991a) - 4 = m^2$.$a^2 + 2(1991)a + 1991^2 - 4(1991)a - 4 = m^2$.$a^2 - 2(1991)a + 1991^2 - 4 = m^2$.$(a-1991)^2 - 4 = m^2$.$(a-1991)^2 - m^2 = 4$.$(a-1991-m)(a-1991+m) = 4$.Let $X = a-1991-m$ and $Y = a-1991+m$. Then $XY = 4$.Since $Y-X = 2m$,$X$ and $Y$ must have the same parity. Since their product is $4$ (even),both must be even.The possible pairs $(X, Y)$ are $(2, 2)$ and $(-2, -2)$.Case $1$: $a-1991-m = 2$ and $a-1991+m = 2$. Adding gives $2(a-1991) = 4$ $\Rightarrow a-1991 = 2$ $\Rightarrow a = 1993$.Case $2$: $a-1991-m = -2$ and $a-1991+m = -2$. Adding gives $2(a-1991) = -4$ $\Rightarrow a-1991 = -2$ $\Rightarrow a = 1989$.Thus,$a \in \{1989, 1993\}$.

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 $b, c \in \mathbb{Z}$?

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