HCF-of-n-2-10-n-1-2-10-n-N-

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^2 +10, b=(n+1)^2 +10=n^2 +10+2n+1 gcd(a,b)∣(a−b=)2n+1∣n(2n+1) gcd(a,b)∣2n^2 +n−2(n^2 +10)=n−20 ⇒gcd(a,b)∣2(n−20)−2n−1=−41 ⇒gcd(n^2 +10,n^2 +10+2n+1)=1 or 41 For example: when n=1, gcd(11,14)=1 when n=20, gcd(410,451)=41

$${Let}\:{a}={n}^{\mathrm{2}} +\mathrm{10},\:{b}=\left({n}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{10}={n}^{\mathrm{2}} +\mathrm{10}+\mathrm{2}{n}+\mathrm{1} \\ $$$${gcd}\left({a},{b}\right)\mid\left({a}−{b}=\right)\mathrm{2}{n}+\mathrm{1}\mid{n}\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$${gcd}\left({a},{b}\right)\mid\mathrm{2}{n}^{\mathrm{2}} +{n}−\mathrm{2}\left({n}^{\mathrm{2}} +\mathrm{10}\right)={n}−\mathrm{20} \\ $$$$\Rightarrow{gcd}\left({a},{b}\right)\mid\mathrm{2}\left({n}−\mathrm{20}\right)−\mathrm{2}{n}−\mathrm{1}=−\mathrm{41} \\ $$$$\Rightarrow{gcd}\left({n}^{\mathrm{2}} +\mathrm{10},{n}^{\mathrm{2}} +\mathrm{10}+\mathrm{2}{n}+\mathrm{1}\right)=\mathrm{1}\:{or}\:\mathrm{41} \\ $$$${For}\:{example}:\:{when}\:{n}=\mathrm{1},\:{gcd}\left(\mathrm{11},\mathrm{14}\right)=\mathrm{1} \\ $$$${when}\:{n}=\mathrm{20},\:{gcd}\left(\mathrm{410},\mathrm{451}\right)=\mathrm{41} \\ $$

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HCF-of-n-2-10-n-1-2-10-n-N-

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