How to find the number $P$ of integers $(n,m)$ such that $\operatorname{lcm}(n,m) = k$? Only $k$ is given.
I only find the number of $n$ such that $\operatorname{lcm}(n,k) = k$.
Can anyone help me solve this problem?
Thanks.
How to find the number $P$ of integers $(n,m)$ such that $\operatorname{lcm}(n,m) = k$? Only $k$ is given.
I only find the number of $n$ such that $\operatorname{lcm}(n,k) = k$.
Can anyone help me solve this problem?
Thanks.
One can consider the decomposition $k=\prod\limits_pp^{v_p(k)}$ of $k$ as a product of powers of distinct primes and wonder what the decompositions $n=\prod\limits_pp^{v_p(n)}$ and $m=\prod\limits_pp^{v_p(m)}$ of $n$ and $m$ can be.
The hypothesis that $k=\operatorname{lcm}(n,m)$ translates in terms of each triplet $(v_p(k),v_p(n),v_p(m))$ as $v_p(k)=\underline{\qquad\qquad}$. Hence, for a given value of $v_p(k)$, there are exactly $\underline{\qquad\qquad}$ couples $(v_p(n),v_p(m))$ available.
Taking into account every prime factor $p$ of $k$, one gets $P=\prod\limits_p\underline{\qquad\qquad}$.