Question Number 212043 by RojaTaniya last updated on 28/Sep/24
$$\:{HCF}\:{of}\:\left\{\left({n}^{\mathrm{2}} +\mathrm{10}\right),\:\left({n}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{10}\right\}=? \\ $$$$\:\:{n}\in{N} \\ $$
Answered by A5T last updated on 28/Sep/24
$${Let}\:{a}={n}^{\mathrm{2}} +\mathrm{10},\:{b}=\left({n}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{10}={n}^{\mathrm{2}} +\mathrm{10}+\mathrm{2}{n}+\mathrm{1} \\ $$$${gcd}\left({a},{b}\right)\mid\left({a}−{b}=\right)\mathrm{2}{n}+\mathrm{1}\mid{n}\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$${gcd}\left({a},{b}\right)\mid\mathrm{2}{n}^{\mathrm{2}} +{n}−\mathrm{2}\left({n}^{\mathrm{2}} +\mathrm{10}\right)={n}−\mathrm{20} \\ $$$$\Rightarrow{gcd}\left({a},{b}\right)\mid\mathrm{2}\left({n}−\mathrm{20}\right)−\mathrm{2}{n}−\mathrm{1}=−\mathrm{41} \\ $$$$\Rightarrow{gcd}\left({n}^{\mathrm{2}} +\mathrm{10},{n}^{\mathrm{2}} +\mathrm{10}+\mathrm{2}{n}+\mathrm{1}\right)=\mathrm{1}\:{or}\:\mathrm{41} \\ $$$${For}\:{example}:\:{when}\:{n}=\mathrm{1},\:{gcd}\left(\mathrm{11},\mathrm{14}\right)=\mathrm{1} \\ $$$${when}\:{n}=\mathrm{20},\:{gcd}\left(\mathrm{410},\mathrm{451}\right)=\mathrm{41} \\ $$