Exercise 234Number theory › Divisibility and gcdHigh schoolAdded 3 Nov 2026

The HCF of consecutive integers

#coprime#consecutive integers#gcd

Statement

Show that $\gcd(n,n+1)=1$ and $\gcd(2n+1,3n+2)=1$ for all $n\ge1$ .

Hint

For the first, note that a common divisor divides $(n+1)-n=1$ . For the second, find a linear combination that gives $1$ .

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First. If $d\mid n$ and $d\mid n+1$ , then $d\mid1$ . So $\gcd(n,n+1)=1$ .

Second. Compute $3(2n+1)-2(3n+2)=6n+3-6n-4=-1$ . So any common divisor divides $1$ .

Conclusion. Both pairs are coprime for all $n$ .