# Prove any integer multiple of a divisor rich number is divisor rich itself?

A positive integer n is divisor rich if the sum of its proper divisors is strictly greater than n. For instance, 10 is not divisor rich because its proper divisors are 1, 2, and 5, which add to 8. Also, squares don't have the same proper divisor twice, only once. Also, the number n itself doesn't count as a proper divisor, just like in the example with 10. Is there any algebraic proof to this?

Thanks!

• The standard terminology is abundant, not divisor rich. – Robert Israel Dec 30 '16 at 19:35

If $n$ has proper divisors $d_1, \ldots, d_k$ with $d_1 + \ldots + d_k > n$, then the proper divisors of $mn$ include $md_1, \ldots, m d_k$ with ...
Hint: If $d$ is a proper divisor of $n$, then $dk$ is a proper divisor of $nk$.
If $n=p_1^{a_1}p_2^{a_2}\dots p_r^{a_r}$ then $\frac{\sigma(n)}{n}=\prod\limits_{j=1}^r \frac{p^{a_j+1}-1}{p^j(p-1)}$.
Clearly each factor is an increasing function with respect to $a_j$ and every factor is greater than or equal to $1$. So increasing the number of factors or increasing the exponents only makes the fraction larger.