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

$$\mathrm{Find}\mathrm{all}\mathrm{integer}\mathrm{values}\mathrm{of}a\mathrm{such}\mathrm{that}$$$$\mathrm{the}\mathrm{quadratic}\mathrm{expression}$$$$(x+a)(x+1991)+1\mathrm{can}\mathrm{be}\mathrm{factored}$$$$\mathrm{as}\mathrm{a}\mathrm{product}(x+b)(x+c)\mathrm{where}b\mathrm{and}$$$$c\mathrm{are}\mathrm{integers}.$$

Commented by Tikufly last updated on 23/Sep/17

$$a=1993ora=1989$$

Commented by Tikufly last updated on 23/Sep/17

$$\mathrm{Sol}$$$$(x+b)(x+c)=(x+a)(x+1991)+1$$$$=>x^2+(b+c)x+bc=x^2+(a+1991)x+1991a+1$$$$thenb+c=(a+1991)andbc=(1991a+1)$$$$$$$$weknowthat$$$${(b+c)}^2-4bc={(b-c)}^2$$$$=>{(a+1991)}^2-4(1991a+1)={(b-c)}^2$$$$=>{(a+1991)}^2-4\times a\times 1991-4={(b-c)}^2$$$$=>{(a-1991)}^2-4={(b-c)}^2$$$$=>{(a-1991)}^2-{(b-c)}^2=4$$$$Thediffoftwosquaresofintegersis4.$$$$Thisispossibleifandonlyifb-c=0$$$$=>{(a-1991)}^2-2^2=0$$$$=>(a-1991-2)(a-1991+2)=0$$$$=>a=1993ora=1989$$

Commented by Tinkutara last updated on 23/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Answered by mrW1 last updated on 24/Sep/17

$$\mathrm{if}(\mathrm{x}+\mathrm{a})(\mathrm{x}+1991)+1\mathrm{can}\mathrm{be}\mathrm{factored}$$$$\mathrm{as}(\mathrm{x}+\mathrm{b})(\mathrm{x}+\mathrm{c}),\mathrm{it}\mathrm{means}\mathrm{the}\mathrm{equation}$$$$(\mathrm{x}+\mathrm{a})(\mathrm{x}+1991)+1=0$$$$\mathrm{has}\mathrm{two}\mathrm{integer}\mathrm{solutions}:-\mathrm{b}\mathrm{and}-\mathrm{c}.$$$$$$$$(\mathrm{x}+\mathrm{a})(\mathrm{x}+1991)+1=0$$$${\mathrm{x}}^2+(\mathrm{a}+1991)\mathrm{x}+(1991\mathrm{a}+1)=0$$$$\Rightarrow \mathrm{x}=\frac{-(\mathrm{a}+1991)\pm \sqrt{{(\mathrm{a}+1991)}^2-4(1991\mathrm{a}+1)}}{2}$$$$=\frac{-(\mathrm{a}+1991)\pm \sqrt{{(\mathrm{a}-1991)}^2-4}}{2}$$$$$$$$\mathrm{case}1:\mathrm{a}\mathrm{is}\mathrm{odd},\mathrm{i}.\mathrm{e}.\mathrm{a}=2\mathrm{i}+1$$$${(2\mathrm{i}+1-1991)}^2-4={\left(2\mathrm{k}\right)}^2$$$$\Rightarrow {(\mathrm{i}-995)}^2-1={\mathrm{k}}^2$$$$\Rightarrow {(\mathrm{i}-995)}^2-{\mathrm{k}}^2=1$$$$\Rightarrow (\mathrm{i}+\mathrm{k}-995)(\mathrm{i}-\mathrm{k}-995)=1$$$$\Rightarrow \mathrm{i}+\mathrm{k}-995=\pm 1$$$$\Rightarrow \mathrm{i}-\mathrm{k}-995=\pm 1$$$$\Rightarrow \mathrm{i}=\pm 1+995=996\mathrm{or}994,\mathrm{k}=0$$$$\Rightarrow \mathrm{a}=2\times 996+1=1993\mathrm{or}$$$$\Rightarrow \mathrm{a}=2\times 994+1=1989$$$$$$$$\mathrm{case}2:\mathrm{a}\mathrm{is}\mathrm{even},\mathrm{i}.\mathrm{e}.\mathrm{a}=2\mathrm{i}$$$${(2\mathrm{i}-1991)}^2-4={(2\mathrm{k}+1)}^2$$$${(2\mathrm{i}-1991)}^2-{(2\mathrm{k}+1)}^2=4$$$$(2\mathrm{i}-1991+2\mathrm{k}+1)(2\mathrm{i}-1991-2\mathrm{k}-1)=4$$$$(\mathrm{i}+\mathrm{k}-995)(\mathrm{i}-\mathrm{k}-996)=1$$$$\Rightarrow \mathrm{i}+\mathrm{k}-995=\pm 1$$$$\Rightarrow \mathrm{i}-\mathrm{k}-996=\pm 1$$$$\Rightarrow \mathrm{i}=\frac{\pm 2+995+996}{2}\ne \mathrm{integer}$$$$\Rightarrow \mathrm{no}\mathrm{solution}\mathrm{for}\mathrm{case}2$$$$$$$$\Rightarrow \mathrm{a}=1989\mathrm{or}1993$$

Commented by Tinkutara last updated on 24/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$$$\mathrm{More}\mathrm{specific}\mathrm{and}\mathrm{proving}!$$

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