Just playing with lcm and gcd

Let's try simplifying the expression $lcm(a,b) \times gcd(a,b)$ first to find $b$ , the square root of their product.

Say $gcd(a,b)=c$ . We can write $a$ , $b$ as $c \times \frac{a}{c}$ , $c \times \frac{b}{c}$ . Thus, the least common multiple would be $c \times \frac{a}{c} \times \frac{b}{c}$ . Multiplying lcm and gcd together, we get $c \times \frac{a}{c} \times \frac{b}{c} \times c = ab.$

Now we could see that $b= \sqrt{ab}$ , which makes $a=b$ .

Relevant wiki: | Geometric Progression || gcd(a,b)*lcm(a,b)=ab | $$ Cause $lcm(a,b), a, gcd(a,b)$ are in a geometric progression, $lcm(a,b)=a*r,a=gcd(a,b)*r \iff lcm(a,b)=a*r,gcd(a,b)=a*\frac{1}{r}$ with $r$ is the ratio.So $ab=gcd(a,b)*lcm(a,b)=a*r*a*\frac{1}{r}=a^2*(r*\frac{1}{r})=a^2$ $\iff a^2=ab \iff \boxed{a=b}(\text{cause a, b are positive integers})$