Abundant numbres proof

• Mar 23rd 2008, 11:09 AM
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 $\displaystyle \sigma (n) = \sigma (945) \sigma (k) = \sigma (3^3) \sigma (5) \sigma (7) \sigma (k) = \frac {3^4-1}{3-1} \frac {5^2-1}{5-1} \frac {7^2-1}{7-1} \sigma (k) = 1920 \sigma (k)$

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

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

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

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

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

Thanks.
• Mar 24th 2008, 09:17 AM
ThePerfectHacker
Quote:

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