Exercise 919Number theory › Divisibility and gcdL1Added 3 Nov 2026

GCD of powers

#gcd#powers#prime factorization

Statement

Show that $\gcd(a^m,b^m)=\gcd(a,b)^m$ for all $m\ge1$ .

Hint

Use prime factorizations: the exponent of $p$ in $a^m$ is $m\cdot v_p(a)$ .

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Prime factorization. For each prime $p$ , $v_p(\gcd(a^m,b^m))=\min(m v_p(a),m v_p(b))=m\min(v_p(a),v_p(b))$ .

GCD. $m\min(v_p(a),v_p(b))=m v_p(\gcd(a,b))=v_p(\gcd(a,b)^m)$ .

Conclusion. The prime factorizations agree, so $\gcd(a^m,b^m)=\gcd(a,b)^m$ .