Prove that there exist $2015$ consecutive abundant numbers

Question

A positive integer $N$ is called abundant if the sum of its divisors is greater than $N$: $\delta (N) >N$.

My question is:

Prove that there exists an integer: $k\in\mathbb N\setminus\{0\}$ such that the numbers $k+1,k+2,k+3,......,k+2015$ are all abundant.

Answer 1

Key idea: Construct $2015$ pairwise coprime abundant numbers to conclude the existence of $2015$ consecutive abundant numbers using the Chinese Remainder Theorem.

Let $I(n)=\frac{\sigma(n)}n$ denote the abundancy index of $n$; if $n=\prod p_k^{a_k}$ we have $I(n)=\prod\frac{p_k-p_k^{-a_k}}{p_k-1}$.

Let $p_n$ denote the $n$th prime and $P_n=\prod_{k\leq n} p_k$, with $I(P_n)=\prod_{k\leq n}\frac{p_k-p_k^{-1}}{p_k-1}$. Observe $\frac{p_k-p_k^{-1}}{p_k-1}=1+\frac{1-p_k^{-1}}{p_k-1}\geq1+\frac1{2(p_k-1)}$.

Because $\sum_{k\geq0}\frac 1{p_k}$ diverges, so does $\sum\frac 1{p_k-1}$ and hence $I(P_n)\to\infty$. This means we can find $n_1,n_2,\ldots$ such that $I(\frac{P_{n_{k+1}}}{P_{n_k}})>2$ for all $k\geq1$. Because the numbers $Q_k=\frac{P_{n_{k+1}}}{P_{n_k}}$ are pairwise coprime, we can find $N$ such that $$\left\{\begin{array}l N\equiv-1\pmod{Q_1}\\ N\equiv-2\pmod{Q_2}\\ \;\;\;\;\;\vdots\\ N\equiv-2015\pmod{Q_{2015}} \end{array}\right.$$

by CRT.

Because a multiple of an abundant number is itself abundant (in general we have $I(a)<I(b)$ if $a\mid b$), this means that $N+1,\ldots,N+2015$ are all abundant.