Solutions to Quiz # 8 for Dr. Z.’s Number Theory Course for Nov. 14, 2013

1. (i) (2 pts.) Define σ_{2}(n) (ii) (2 pts.) State the formula for σ_{2}(n) in terms of the expression of n as a product of prime powers (iii) (2 pts.) Verify it for n = 15 by using the definition and the formula

Sol. to 1:

(i)

σ_{2}(n) := ∑_{d|n} d^{2} .

(ii) If

n = ∏_{i=1}^{k} p_{i}^{α_{i}}

is the factorization of n into products of prime powers guaranteed by the Fundamental Theorem of Arithmetic, then

σ_{2}(n) = ∏_{i=1}^{k}(1 + p_{i}^{2} + p_{i}^{4} + … + p_{i}^{2α_{i}})

Another way (summing the geometric series) is

σ_{2}(n) = ∏_{i=1}^{k} \frac{p_{i}^{2α_{i}+2} − 1)}{p_{i}^{2} − 1} .

(iii) Since

Div(15) = {1, 3, 5, 15} ,

From the definition

σ_{2}(15) = 1^{2} + 3^{2} + 5^{2} + 15^{2} = 1 + 9 + 25 + 225 = 260 .

From the formula, since 15 = 3^{1} · 5^{1},

σ_{2}(15) = (1 + 3^{2}) · (1 + 5^{2}) = 10 · 26 = 260 .

Yea!

2. (4 pts.) Prove that if p is a prime, and 2^{p} − 1 is also a prime, then

2^{p−1} · (2^{p} − 1)

is a perfect number.

Sol. of 2: Since both 2 and 2^{p} − 1 are prime, it follows from the formula for σ(n), that

σ(2^{p−1} · (2^{p} − 1)) = (1 + 2 + … + 2^{p−1}) · (1 + (2^{p} − 1)) .

By the famous geometric series formula

1 + q + … + q^{N} = \frac{q^{N+1} − 1}{q − 1} ,

with q = 2 and N = p − 1 we have

1 + 2 + … + 2^{p−1} = \frac{2^{p−1+1} − 1}{2 − 1} = 2^{p} − 1 .

By the famous theorem

1 + (N − 1) = N ,

applied to N = 2^{p} − 1, we have

1 + (2^{p} − 1) = 2^{p} .

Going back above we have

σ(2^{p−1} · (2^{p} − 1)) = (2^{p} − 1) · (2^{p}) = 2 · (2^{p−1}(2^{p} − 1)) .

Hence n = 2^{p−1}(2^{p} − 1) satisfies the condition

σ(n) = 2n ,

that this means that n is perfect. QED.