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Question Number 109029 by nimnim last updated on 20/Aug/20

$$Findthevalue\left(s\right)ofa,bandcif:$$$$(x+a)(x+2020)+1=(x+b)(x+c)$$$$wherea,bandcarenaturalnumbers.$$

Answered by floor(10²Eta[1]) last updated on 21/Aug/20

$${\mathrm{x}}^2+(2020+\mathrm{a})\mathrm{x}+2020\mathrm{a}+1={\mathrm{x}}^2+(\mathrm{b}+\mathrm{c})\mathrm{x}+\mathrm{bc}$$$$\Rightarrow 2020+\mathrm{a}=\mathrm{b}+\mathrm{c}$$$$\Rightarrow 2020\mathrm{a}+1=\mathrm{bc}$$$$\mathrm{a}=\mathrm{b}+\mathrm{c}-2020$$$$2020(\mathrm{b}+\mathrm{c}-2020)+1=\mathrm{bc}$$$$2020\mathrm{b}+2020\mathrm{c}-{2020}^2+1=\mathrm{bc}$$$$\mathrm{bc}-2020\mathrm{b}-2020\mathrm{c}+{2020}^2=1$$$$(\mathrm{b}-2020)(\mathrm{c}-2020)=1$$$$\mathrm{the}\mathrm{only}\mathrm{ways}\mathrm{we}\mathrm{can}\mathrm{factor}1\mathrm{are}$$$$1.1\mathrm{or}(-1)(-1)$$$$\mathrm{I}\mathrm{case}:$$$$\mathrm{b}-2020=1=\mathrm{c}-2020$$$$\Rightarrow \mathrm{b}=\mathrm{c}=2021\therefore \mathrm{a}=2022$$$$\mathrm{II}\mathrm{case}:$$$$\mathrm{b}-2020=-1=\mathrm{c}-2020$$$$\Rightarrow \mathrm{b}=\mathrm{c}=2019\therefore \mathrm{a}=2018$$$$$$$$\mathrm{so}\mathrm{all}\mathrm{solutions}\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{are}:$$$$(2022,2021,2021),(2018,2019,2019)$$

Commented by Rasheed.Sindhi last updated on 22/Aug/20

$$\underset{\mathrm{W}}{\stackrel{\mathrm{W}}{\mathrm{O}}}\underset{!}{\stackrel{!}{!}},great!...\mathcal{T}hanksfloor\mathcal{S}ir!$$

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