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

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Question Number 212043 by RojaTaniya last updated on 28/Sep/24
HCF of {(n^2 +10), (n+1)^2 +10}=? n∈N
HCFof{(n2+10),(n+1)2+10}=?n∈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
Leta=n2+10,b=(n+1)2+10=n2+10+2n+1gcd(a,b)∣(a−b=)2n+1∣n(2n+1)gcd(a,b)∣2n2+n−2(n2+10)=n−20⇒gcd(a,b)∣2(n−20)−2n−1=−41⇒gcd(n2+10,n2+10+2n+1)=1or41Forexample:whenn=1,gcd(11,14)=1whenn=20,gcd(410,451)=41

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