The way to prove that sigma (a)=3^k has no solution?

Alfredeim

Alfredeim

Answered question

2022-09-07

The way to prove that σ ( a ) = 3 k has no solution?
σ ( n ) = sum of divisors of n is a divisors function.
How to prove there are no such a and k 2 satisfy σ ( a ) = 3 k .
This proplem can be simplify to the case when a is a power of prime ( a = p α ) because
if a = p 0 α 0 p 1 α 1 . . . p n α n , then
σ ( a ) = σ ( p 0 α 0 ) σ ( p 1 α 1 ) . . . σ ( p n α n ) = i = 0 n p i α i + 1 1 p i 1

Answer & Explanation

faliryr

faliryr

Beginner2022-09-08Added 15 answers

Step 1
Let (p,q) be primes. Suppose we want to prove whether or not there exists at least one pair (r,k) of positive integers with σ ( p r ) = q k .
Assume that at least one solution exists, and let (r,k) be a solution with r minimal. Let d > 1 divide r + 1, then
q k = σ ( p r ) = p r + 1 1 p 1 = p r + 1 1 p d 1 p d 1 p 1 .
Hence, there exists a positive integer ℓ such that (d,ℓ) is also a solution. By minimality, d = r + 1 and r + 1 is prime.
Next, let j := ord q ( p 1 ). Note that
ord q ( p r + 1 1 ) = ord q ( ( p 1 ) q k ) = j + k j + 1..
Hence, p 1 ( mod q j + 1 ) while p r + 1 1 ( mod q j + 1 ). This means that r + 1 is not coprime to the order of the multiplicative group ( Z / q j + 1 Z ) × , which is q j ( q 1 ). Therefore, r + 1 is not coprime to q ( q 1 ). Because r + 1 is prime, we find that
r + 1 q ( q 1 ) .
Step 2
Now let's apply this knowledge to the case q = 3. We find that r { 1 , 2 }. If r = 1, then p + 1 = 3 k , whence p is even, whence p = 2 and k = 1.
If r = 2, then p 2 + p + 1 = 3 k , so p 2 + p + ( 1 3 k ) = 0. As a quadratic polynomial in p, this has discriminant
1 4 ( 1 3 k ) = 4 3 k 3 = 3 ( 4 3 k 1 1 ) ..
In order for this to be a perfect square, we need 3 4 3 k 1 1, which is only the case for k = 1. However, this gives p = 1, which is a contradiction.
Therefore, all solutions have p = 2. In this case, we may rewrite σ ( 2 r ) = 3 k as 2 r + 1 1 = 3 r , which has only the solution ( r , k ) = ( 1 , 1 ), by Mihăilescu's theorem. We are done.
Raven Mosley

Raven Mosley

Beginner2022-09-09Added 14 answers

Step 1
As you stated, since σ ( ) is a multiplicative function, we only need to check the prime powers, i.e., for prime p, e 1 and j 1, that
(1) σ ( p e ) = i = 0 e p i = p e + 1 1 p 1 = 3 j
First, p = 2 and e = 1 gives σ ( 2 ) = 2 + 1 = 3, so j = 1 works. For e > 1, then
(2) σ ( 2 e ) = 2 e + 1 1 = 3 j 2 e + 1 3 j = 1
Here, j > 1, but Mihăilescu's theorem shows there's no such solution (also, more simply, with 2 e + 1 = 3 j + 1, then e + 1 3 means 3 j 7 ( mod 8 ), but 3 j 1 ( mod 8 ) or 3 j 3 ( mod 8 ) only). This means a has at most 1 factor of 2 and there must be odd prime factors. Since p = 3 gives σ ( 3 e ) 1 ( mod 3 ), this means p 5.
With the summation in (1), as there are e + 1 terms, getting an odd sum requires e + 1 to be odd, i.e., e + 1 = 2 r + 1 for some integer r 1. Using the fraction part in (1), if p 2 ( mod 3 ), then p e + 1 1 ( 2 2 ) r ( 2 ) 1 ( 1 ) r ( 2 ) 1 1 ( mod 3 ) and p 1 1 ( mod 3 ), so the result would be 1 ( mod 3 ). Thus, this means p 1 ( mod 3 ).
Step 2
Once again from the summation in (1), as each term is 1 ( mod 3 ), there must be a multiple of 3 terms, i.e., e + 1 = 3 m for some integer m 1. The fractional part of (1) becomes
(3) p e + 1 1 p 1 = p 3 m 1 p 1 = ( p m 1 p 1 ) ( p 2 m + p m + 1 )
Note that p m 1 p 1 = i = 0 m 1 p is an integer. Also, since p 1 ( mod 3 ), then p m = 3 q + 1 for some integer q 1. Thus,
(4) p 2 m + p m + 1 = ( 3 q + 1 ) 2 + ( 3 q + 1 ) + 1 = ( 9 q 2 + 6 q + 1 ) + ( 3 q + 1 ) + 1 = 9 q 2 + 9 q + 3 = 3 ( 3 q 3 + 3 q + 1 )
However, 3 q 3 + 3 q + 1 > 1 and is not a power of 3, so (1) cannot hold. This therefore proves what is requested, i.e., there's no integer a such that σ ( a ) = 3 k for k 2.

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