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

1. Find σ(200) Sol. to 1: First we factorize 200

200 = 2^{3} · 5^{2}

Now

σ(200) = (1 + 2 + 2^{2} + 2^{3}) · (1 + 5 + 5^{2}) = 15 · 31 = 465 .

Answer to 1: σ(200) = 465 2 (a) Define a perfect number Sol to 2: A positive integer is perfect if it equals to the sum of its proper divisors, or equivalenly the sum of the divisors is twice that number. In symbols: σ(2n) = 2σ(n).

(b) Prove that if both p and 2^{p} − 1 are primes, then 2^{p−1} · (2^{p} − 1)

is a perfect number. Sol. of 2(b): 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.