Very very perfect
How many perfect numbers are there which are perfect squares?
The answer is 0.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
EDIT: This proof only works for even perfect squares.
By the Euclid-Euler Theorem , we know that every even perfect number ≥ 6 can be written in the form 2 p − 1 ( 2 p − 1 ) whenever ( 2 p − 1 ) is a Mersenne prime and p ≥ 2 is a prime.
Now, since ( 2 p − 1 ) is a prime, it follows that 2 p − 1 = 2 p − 1 for the number to be a perfect square. Hence,
2 p − 1 = 2 p − 1 ⟹ 2 p − 2 p − 1 = 1 ⟹ 2 p − 1 ( 2 − 1 ) = 1 ⟹ 2 p − 1 = 1 ⟹ p − 1 = 0 ⟹ p = 1
This contradicts the fact that p ≥ 2 is a prime and hence there are no even perfect numbers which are also perfect squares.
Log in to reply
Every even perfect number can be written in that form. Doesn't say anything about the odd perfect numbers, which are precisely the numbers we want to attack.
Let n 2 is a perfect number n 2 = p 1 2 a 1 ∗ p 2 2 a 2 ∗ … ∗ p n 2 a n
Then σ ( n 2 ) = σ ( p 1 2 a 1 ) ∗ σ ( p 2 2 a 2 ) ∗ … ∗ σ ( p n 2 a n )
σ ( n 2 ) = 2 ∗ p 1 2 a 1 ∗ p 2 2 a 2 ∗ … ∗ p n 2 a n is an even number.
For a prime number p of an even power 2 a , the number of its divisors = 2 a + 1 , which implies that when we disregard 1 , there will be 2 a even number of divisors. Without losing the generality, we will make p and a represent all p i a n d a i
When p = 2 , ( 1 + p + p 2 + … + p 2 a ) = 1 + even + even + … + even = odd. So the product of the prime divisors of m is the product of odd numbers so m is odd.
When p is a prime number other than 2 , ( 1 + p + p 2 + … + p 2 a = 1 + odd + odd + … + odd = odd. Hence σ ( n 2 is product of odd numbers. So it is also odd. Thus it cannot be equal to 2 ∗ p 1 2 a 1 ∗ p 2 2 a 2 ∗ … ∗ p n 2 a n as it is even
So the answer is 0