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Exercise 920Number theory › Divisibility and gcdL1Added 3 Nov 2026

A coprimality proof

#coprime#proof#linear combination

Statement

Show that for any integer nn, 2n+12n+1 and 3n+23n+2 are coprime.

Hint

Find a linear combination equal to 11.

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Solution

Combination. 3(2n+1)−2(3n+2)=−13(2n+1)-2(3n+2)=-1.

Conclusion. So 2(3n+2)−3(2n+1)=12(3n+2)-3(2n+1)=1, giving a Bézout relation. Hence gcd⁡(2n+1,3n+2)=1\gcd(2n+1,3n+2)=1.

Application. The fraction 2n+13n+2\frac{2n+1}{3n+2} is irreducible for all nn.