A number theory problem by Christopher Boo

Intuition behind $\gcd(a,b)\text{lcm}(a,b) = ab$ is that $a$ and $b$ are expressed as products of their prime factors with corresponding exponents and in $\gcd(a,b)$ we take the smaller exponent between $a$ and $b$ for every prime and in $\text{lcm}(a,b)$ the larger.

This means that the left side contains a product of primes with exponents being the smaller exponents multiplied by a product of primes with exponents being the larger exponents, but on the right side we have them shuffled between $a$ and $b$ .

Since, multiplication is commutative, the both sides of the equation have the same value.

However, for $3$ positive integers, there is no similar formula because by taking the smallest exponents with $\gcd$ and largest exponents with $\text{lcm}$ we have no way of taking the middle exponent into account.

This means that the formula above only works if the middle exponents are all $0$ , meaning that the smallest exponents are also $0$ and largest are whatever we want.

Hence, $\gcd(a,b,c)\text{lcm}(a,b,c) = abc$ is true iff two of the positive integers are $1$ .

To prove that the answer is $\textbf{No}$ we just need a counter-example

Just pick $a=b=c\neq 1$ :

$\gcd(a,b,c)=\gcd(a,a,a)=a$

$lcm(a,b,c)=lcm(a,a,a)=a$

$\gcd(a,b,c)\cdot lcm(a,b,c)=a\cdot a=a^2\neq a^3=a\cdot b\cdot c$