Menu Close

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-

Tinku Tara 0 Comments
Pin




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.
$$\mathrm{Find}\:\mathrm{all}\:\mathrm{integer}\:\mathrm{values}\:\mathrm{of}\:{a}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{quadratic}\:\mathrm{expression} \\ $$$$\left({x}\:+\:{a}\right)\left({x}\:+\:\mathrm{1991}\right)\:+\:\mathrm{1}\:\mathrm{can}\:\mathrm{be}\:\mathrm{factored} \\ $$$$\mathrm{as}\:\mathrm{a}\:\mathrm{product}\:\left({x}\:+\:{b}\right)\left({x}\:+\:{c}\right)\:\mathrm{where}\:{b}\:\mathrm{and} \\ $$$${c}\:\mathrm{are}\:\mathrm{integers}. \\ $$
Commented by Tikufly last updated on 23/Sep/17
a=1993 or a=1989
$${a}=\mathrm{1993}\:{or}\:{a}=\mathrm{1989} \\ $$
Commented by Tikufly last updated on 23/Sep/17
Sol (x+b)(x+c)=(x+a)(x+1991)+1 =>x^2 +(b+c)x+bc=x^2 +(a+1991)x+1991a+1 then b+c=(a+1991) and bc=(1991a+1) we know that (b+c)^2 −4bc=(b−c)^2 =>(a+1991)^2 −4(1991a+1)=(b−c)^2 =>(a+1991)^2 −4×a×1991−4=(b−c)^2 =>(a−1991)^2 −4=(b−c)^2 =>(a−1991)^2 −(b−c)^2 =4 The diff of two squares of integers is 4. This is possible if and only if b−c=0 =>(a−1991)^2 −2^2 =0 =>(a−1991−2)(a−1991+2)=0 =>a=1993 or a=1989
$$\mathrm{Sol} \\ $$$$\left({x}+{b}\right)\left({x}+{c}\right)=\left({x}+{a}\right)\left({x}+\mathrm{1991}\right)+\mathrm{1} \\ $$$$=>{x}^{\mathrm{2}} +\left({b}+{c}\right){x}+{bc}={x}^{\mathrm{2}} +\left({a}+\mathrm{1991}\right){x}+\mathrm{1991}{a}+\mathrm{1} \\ $$$${then}\:{b}+{c}=\left({a}+\mathrm{1991}\right)\:{and}\:{bc}=\left(\mathrm{1991}{a}+\mathrm{1}\right) \\ $$$$ \\ $$$${we}\:{know}\:{that} \\ $$$$\left({b}+{c}\right)^{\mathrm{2}} −\mathrm{4}{bc}=\left({b}−{c}\right)^{\mathrm{2}} \\ $$$$=>\left({a}+\mathrm{1991}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1991}{a}+\mathrm{1}\right)=\left({b}−{c}\right)^{\mathrm{2}} \\ $$$$=>\left({a}+\mathrm{1991}\right)^{\mathrm{2}} −\mathrm{4}×{a}×\mathrm{1991}−\mathrm{4}=\left({b}−{c}\right)^{\mathrm{2}} \\ $$$$=>\left({a}−\mathrm{1991}\right)^{\mathrm{2}} −\mathrm{4}=\left({b}−{c}\right)^{\mathrm{2}} \\ $$$$=>\left({a}−\mathrm{1991}\right)^{\mathrm{2}} −\left({b}−{c}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$${The}\:{diff}\:{of}\:{two}\:{squares}\:{of}\:{integers}\:{is}\:\mathrm{4}. \\ $$$${This}\:{is}\:{possible}\:{if}\:{and}\:{only}\:{if}\:{b}−{c}=\mathrm{0} \\ $$$$=>\left({a}−\mathrm{1991}\right)^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} =\mathrm{0} \\ $$$$=>\left({a}−\mathrm{1991}−\mathrm{2}\right)\left({a}−\mathrm{1991}+\mathrm{2}\right)=\mathrm{0} \\ $$$$=>{a}=\mathrm{1993}\:{or}\:{a}=\mathrm{1989} \\ $$
Commented by Tinkutara last updated on 23/Sep/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
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+(1991a+1)=0 ⇒x=((−(a+1991)±(√((a+1991)^2 −4(1991a+1))))/2) =((−(a+1991)±(√((a−1991)^2 −4)))/2) case 1: a is odd, i.e. a=2i+1 (2i+1−1991)^2 −4=(2k)^2 ⇒(i−995)^2 −1=k^2 ⇒(i−995)^2 −k^2 =1 ⇒(i+k−995)(i−k−995)=1 ⇒i+k−995=±1 ⇒i−k−995=±1 ⇒i=±1+995=996 or 994, k=0 ⇒a=2×996+1=1993 or ⇒a=2×994+1=1989 case 2: a is even, i.e. a=2i (2i−1991)^2 −4=(2k+1)^2 (2i−1991)^2 −(2k+1)^2 =4 (2i−1991+2k+1)(2i−1991−2k−1)=4 (i+k−995)(i−k−996)=1 ⇒i+k−995=±1 ⇒i−k−996=±1 ⇒i=((±2+995+996)/2)≠integer ⇒no solution for case 2 ⇒a=1989 or 1993
$$\mathrm{if}\:\left(\mathrm{x}+\mathrm{a}\right)\left(\mathrm{x}+\mathrm{1991}\right)+\mathrm{1}\:\mathrm{can}\:\mathrm{be}\:\mathrm{factored} \\ $$$$\mathrm{as}\:\left(\mathrm{x}+\mathrm{b}\right)\left(\mathrm{x}+\mathrm{c}\right),\:\mathrm{it}\:\mathrm{means}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left(\mathrm{x}+\mathrm{a}\right)\left(\mathrm{x}+\mathrm{1991}\right)+\mathrm{1}=\mathrm{0}\: \\ $$$$\mathrm{has}\:\mathrm{two}\:\mathrm{integer}\:\mathrm{solutions}:\:−\mathrm{b}\:\mathrm{and}\:−\mathrm{c}. \\ $$$$ \\ $$$$\left(\mathrm{x}+\mathrm{a}\right)\left(\mathrm{x}+\mathrm{1991}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} +\left(\mathrm{a}+\mathrm{1991}\right)\mathrm{x}+\left(\mathrm{1991a}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}=\frac{−\left(\mathrm{a}+\mathrm{1991}\right)\pm\sqrt{\left(\mathrm{a}+\mathrm{1991}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1991a}+\mathrm{1}\right)}}{\mathrm{2}} \\ $$$$=\frac{−\left(\mathrm{a}+\mathrm{1991}\right)\pm\sqrt{\left(\mathrm{a}−\mathrm{1991}\right)^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{1}:\:\:\mathrm{a}\:\mathrm{is}\:\mathrm{odd},\:\mathrm{i}.\mathrm{e}.\:\mathrm{a}=\mathrm{2i}+\mathrm{1} \\ $$$$\left(\mathrm{2i}+\mathrm{1}−\mathrm{1991}\right)^{\mathrm{2}} −\mathrm{4}=\left(\mathrm{2k}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{i}−\mathrm{995}\right)^{\mathrm{2}} −\mathrm{1}=\mathrm{k}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{i}−\mathrm{995}\right)^{\mathrm{2}} −\mathrm{k}^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\left(\mathrm{i}+\mathrm{k}−\mathrm{995}\right)\left(\mathrm{i}−\mathrm{k}−\mathrm{995}\right)=\mathrm{1} \\ $$$$\Rightarrow\mathrm{i}+\mathrm{k}−\mathrm{995}=\pm\mathrm{1} \\ $$$$\Rightarrow\mathrm{i}−\mathrm{k}−\mathrm{995}=\pm\mathrm{1} \\ $$$$\Rightarrow\mathrm{i}=\pm\mathrm{1}+\mathrm{995}=\mathrm{996}\:\mathrm{or}\:\mathrm{994},\:\mathrm{k}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{a}=\mathrm{2}×\mathrm{996}+\mathrm{1}=\mathrm{1993}\:\mathrm{or} \\ $$$$\Rightarrow\mathrm{a}=\mathrm{2}×\mathrm{994}+\mathrm{1}=\mathrm{1989}\: \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{2}:\:\:\mathrm{a}\:\mathrm{is}\:\mathrm{even},\:\mathrm{i}.\mathrm{e}.\:\mathrm{a}=\mathrm{2i} \\ $$$$\left(\mathrm{2i}−\mathrm{1991}\right)^{\mathrm{2}} −\mathrm{4}=\left(\mathrm{2k}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{2i}−\mathrm{1991}\right)^{\mathrm{2}} −\left(\mathrm{2k}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$$\left(\mathrm{2i}−\mathrm{1991}+\mathrm{2k}+\mathrm{1}\right)\left(\mathrm{2i}−\mathrm{1991}−\mathrm{2k}−\mathrm{1}\right)=\mathrm{4} \\ $$$$\left(\mathrm{i}+\mathrm{k}−\mathrm{995}\right)\left(\mathrm{i}−\mathrm{k}−\mathrm{996}\right)=\mathrm{1} \\ $$$$\Rightarrow\mathrm{i}+\mathrm{k}−\mathrm{995}=\pm\mathrm{1} \\ $$$$\Rightarrow\mathrm{i}−\mathrm{k}−\mathrm{996}=\pm\mathrm{1} \\ $$$$\Rightarrow\mathrm{i}=\frac{\pm\mathrm{2}+\mathrm{995}+\mathrm{996}}{\mathrm{2}}\neq\mathrm{integer} \\ $$$$\Rightarrow\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:\mathrm{case}\:\mathrm{2} \\ $$$$ \\ $$$$\Rightarrow\mathrm{a}=\mathrm{1989}\:\mathrm{or}\:\mathrm{1993} \\ $$
Commented by Tinkutara last updated on 24/Sep/17
Thank you very much Sir! More specific and proving!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$$$\mathrm{More}\:\mathrm{specific}\:\mathrm{and}\:\mathrm{proving}! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *