Home > Number theory, Problem Solving > IMC 2014 Day 1 Problem 4

IMC 2014 Day 1 Problem 4


Let {n > 6} be a perfect number and {n = p_1^{e_1}...p_k^{e_k}} be its prime factorization with {1<p_1<...<p_k}. Prove that {e_1} is an even number.

A number {n} is perfect if {s(n)=2n}, where {s(n)} is the sum of the divisors of {n}.

IMC 2014 Day 1 Problem 4

Advertisements
  1. IMC2010
    August 5, 2014 at 8:48 pm

    If \p_1=2, then assuming \e_1 is odd yields that \p_2 = 3. However, if n is divisible by 6, then it is easy to see that \sigma(n)>2n, by using simple inequality argument.
    In the other case, assuming e_1 odd would imply that n has a prime divisor that divides p_1+1, which contradicts the fact that p_1 is the smallest.

  2. IMC2010
    August 5, 2014 at 8:52 pm

    Correction to the above parsing errors:
    If the smallest prime divisor of n is 2 and if e_1 is odd, then it would imply that n is also divisible by 6. However, it is easy to see that no perfect number is divisible by 6 except 6 itself.

  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: