# Math Help - Abundant numbres proof

1. ## Abundant numbres proof

Prove that there exist infinitely many odd abundant numbers.

Proof.

Consider the integers n = 945k, where k is any positive integers not divisible by 2,3,5, or 7.

Since 945 = (3^3)(5)(7), so gcd(945,k)=1, implies that $\sigma (n) = \sigma (945) \sigma (k) = \sigma (3^3) \sigma (5) \sigma (7) \sigma (k) = \frac {3^4-1}{3-1} \frac {5^2-1}{5-1} \frac {7^2-1}{7-1} \sigma (k) = 1920 \sigma (k)$

Therefore I have $\sigma (945k) = 1920 \sigma (k)$

Now, I claim that $\sigma (945k) = 1920 \sigma (k) > (2)945k = 1890k \ \ \ \ \ \forall k \in \mathbb {N}$, and I'm using induction to prove it.

Certainly, $1920 \sigma (1) = 1920 > 1890$, so the claim is true when k=1.

Suppose the claim is true for k=t, we then have $1920 \sigma (t) > 1890t$

Now, I want to show that $(1920) \sigma (t+1) = 1890 (t+1)$, how would I do that?

Thanks.

Since 945 = (3^3)(5)(7), so gcd(945,k)=1, implies that $\sigma (n) = \sigma (945) \sigma (k) = \sigma (3^3) \sigma (5) \sigma (7) \sigma (k) = \frac {3^4-1}{3-1} \frac {5^2-1}{5-1} \frac {7^2-1}{7-1} \sigma (k) = 1920 \sigma (k)$
All you need to show is that $\sigma (n) > 2n$. Now $\sigma(n) = 1920 \sigma(k)$ and $2n = 1890k$. Since $\sigma(k) > k$ it immediatelly completes the proof.