Thread: 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.

Originally Posted by tttcomrader

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) $

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.