How to prove that if $LCM(a,b) = GCF(a,b)^2$, then $b = a^2$ ?
2026-02-24 11:58:04.1771934284
Prove that if $LCM(a,b) = GCF(a,b)^2$, then $b = a^2$
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First, the conclusion isn't exactly true. It should be $b=a^2$ OR $a=b^2$, by symmetry.
Anyway, it's false: take $a=12$, $b=18$. Then $\gcd(a,b)=6$, $\operatorname{lcm}(a,b)=36$, yet $b\ne a^2$.
The assertion would be true if $\gcd(a,b)$ were a prime power.