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-

Question Number 21422 by Tinkutara last updated on 23/Sep/17

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 .

Commented by Tikufly last updated on 23/Sep/17

a = 1993 o r a = 1989

Commented by Tikufly last updated on 23/Sep/17

Sol

(x + b) (x + c) = (x + a) (x + 1991) + 1

=> x^{2} + (b + c) x + b c = x^{2} + (a + 1991) x + 1991 a + 1

t h e n b + c = (a + 1991) a n d b c = (1991 a + 1)

w e k n o w t h a t

{(b + c)}^{2} - 4 b c = {(b - c)}^{2}

=> {(a + 1991)}^{2} - 4 (1991 a + 1) = {(b - c)}^{2}

=> {(a + 1991)}^{2} - 4 \times a \times 1991 - 4 = {(b - c)}^{2}

=> {(a - 1991)}^{2} - 4 = {(b - c)}^{2}

=> {(a - 1991)}^{2} - {(b - c)}^{2} = 4

T h e d i f f o f t w o s q u a r e s o f i n t e g e r s i s 4 .

T h i s i s p o s s i b l e i f a n d o n l y i f b - c = 0

=> {(a - 1991)}^{2} - 2^{2} = 0

=> (a - 1991 - 2) (a - 1991 + 2) = 0

=> a = 1993 o r a = 1989

Commented by Tinkutara last updated on 23/Sep/17

Thank you very much Sir!

Answered by mrW1 last updated on 24/Sep/17

if (x + a) (x + 1991) + 1 can be factored

as (x + b) (x + c), it means the equation

(x + a) (x + 1991) + 1 = 0

has two integer solutions : - b and - c .

(x + a) (x + 1991) + 1 = 0

x^{2} + (a + 1991) x + (1991 a + 1) = 0

\Rightarrow x = \frac{- (a + 1991) \pm \sqrt{{(a + 1991)}^{2} - 4 (1991 a + 1)}}{2}

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

case 1 : a is odd, i . e . a = 2 i + 1

{(2 i + 1 - 1991)}^{2} - 4 = {(2 k)}^{2}

\Rightarrow {(i - 995)}^{2} - 1 = k^{2}

\Rightarrow {(i - 995)}^{2} - k^{2} = 1

\Rightarrow (i + k - 995) (i - k - 995) = 1

\Rightarrow i + k - 995 = \pm 1

\Rightarrow i - k - 995 = \pm 1

\Rightarrow i = \pm 1 + 995 = 996 or 994, k = 0

\Rightarrow a = 2 \times 996 + 1 = 1993 or

\Rightarrow a = 2 \times 994 + 1 = 1989

case 2 : a is even, i . e . a = 2 i

{(2 i - 1991)}^{2} - 4 = {(2 k + 1)}^{2}

{(2 i - 1991)}^{2} - {(2 k + 1)}^{2} = 4

(2 i - 1991 + 2 k + 1) (2 i - 1991 - 2 k - 1) = 4

(i + k - 995) (i - k - 996) = 1

\Rightarrow i + k - 995 = \pm 1

\Rightarrow i - k - 996 = \pm 1

\Rightarrow i = \frac{\pm 2 + 995 + 996}{2} \neq integer

\Rightarrow no solution for case 2

\Rightarrow a = 1989 or 1993

Commented by Tinkutara last updated on 24/Sep/17

Thank you very much Sir!

More specific and proving!