Existence of pairs

Question

Let $p,p_1,p_2$ be positive integers such that $lmc(p_1,p_2)=p$. Is $\min(p_1+p_2)$ unique?

Answer 1

If $p_1, p_2$ are two integers such that $\text{lcm}(p_1, p_2)=p$, the only way that $p_1+p_2$ could be minimum is if $p_1$ and $p_2$ had no common factors. If they had a common factor $q$ with $q$ being a prime, then either $p_1$ or $p_2$ could be divided by $q$ and $\text{lcm}(p_1, p_2)$ would still equal $p$, while $p_1+p_2$ would be smaller. So $\gcd(p_1, p_2)$ must equal $1$. Together with $\text{lcm}(p_1, p_2)=p$, that implies $p_1p_2=p$.

Now it remains to show that if $p_1\le p_2$ and $q_1\le q_2$ are integers such that $p_1p_2=p=q_1q_2$ and $p_1+p_2=q_1+q_2$, then $p_1=q_1, p_2=q_2$ (you don't even need the fact that $\gcd(p_1,p_2)=1=\gcd(q_1,q_2)$). This follows from the fact that $p_1p_2=p$ and $p_1+p_2=n$ completely constrains what $p_1$ and $p_2$ can be. The only solution would be $p_1=\frac{n-\sqrt{n^2-4p}}{2}, p_2=\frac{n+\sqrt{n^2-4p}}{2}$.

Intuitively, you can think of this as meaning that if $p_1p_2=p$ and $p_1\le p_2$, then decreasing $p_1$ would increase $p_2$ more than $p_1$ decreases.