With σ(n) we denote the sum of natural divisors of the natural number n. Prove that, if n is the product of different prime numbers of the form 2k−1 for k∈N(Mersenne′s prime numbers) , than σ(n)=2m, for some m∈N. Is the inverse statement true?
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Tags: algebra, polynomial, function, inequalities, number theory, prime numbers, number theory unsolved
Thjch Ph4 Trjnh
22.05.2010 23:22
ridgers wrote: With σ(n) we denote the sum of natural divisors of the natural number n. Prove that, if n is the product of different prime numbers of the form 2k−1 for k∈N(Mersenne′s prime numbers) , than σ(n)=2m, for some m∈N. Is the inverse statement true? Assume n=∏ki=1(2αi−1). We have σ(n)=∏ki=1σ(2αi−1)=∏ki=12αi=2∑ki=1αi.
Joao Pedro Santos
02.09.2010 18:29
Since σ(n) is a multiplicative function, if σ(n) is a power of 2 whose exponent is a positive integer, then σ(pk−1) also is a power of 2 whose exponent is a positive integer, where p is a prime and k an integer larger than 1, such that pk−1∥n, since σ(pk−1)∣σ(n) and σ(n)=20⇔n=1.
Note that σ(pk−1)=2m, where m is a positive integer, is equivalent to pk−1p−1=2m, which is equivalent to ∏1<d∣kΦd(p)=2m.
Consider the following three cases:
Case 1: k has at least an odd prime divisor.
Let q be that odd prime divisor. Then Φq(p)=∑q−1i=0pi is larger than 1 and is a power of 2, therefore 2∣Φq(p).
But we have Φq(p)≡2∑q−1i=0pi≡21+∑q−1i=1p≡2p(q−1)+1≡21, a contradiction.
Case 2: k=2q, where q is an integer larger than 1.
Since 2∣2q and 4∣2q, then Φ2(p)=p+1 and Φ4(p)=p2+1 are larger than 1 and are powers of 2. Since p2+1>p+1, then p+1∣p2+1. Since p+1∣p2−1 and (p2+1)−(p2−1)=2, then p+1∣2, therefore p+1=2, which means p=1, which is not a prime, a contradiction.
Case 3: k=2.
In this case pk−1 is a prime number and we have p+1=2m, which means p is a Mersenne prime. We can conclude if σ(n) is a power of base 2 whose exponent is a positive integer, then n is the product of distinct Mersenne primes.
Jorge Miranda
02.09.2010 19:55
Clearly the first implication is trivial (because σ is multiplicative and the sum of the divisors of a Mersenne prime is a power of two), so suppose that σ(n)=2b, and WLOG let n=pa−1 be a power of a prime (just factor n and observe that the only divisors of a power of two are powers of two). Clearly p≠2 (or else σ(n)=1+p+⋯+pa−1≡1mod), so suppose that p\geq 3, and a\geq 2.
Case 1:p\equiv 1 \bmod{4}.
Suppose that 2^r\|p-1 and 2\|p+1, so that 2^{r+1-1}\|\frac{p^2-1}{2}. Using the Lifting lemma we conclude that 2^{r+v_2(a)}\|p^a-1.
But p^a-1=2^b(p-1), and this implies that 2^{b+r}\|p^a-1. We conclude that r+v_2(a)=r+b\Rightarrow b=v_2(a), and 2^b=1+p+\cdots+p^{a-1}>p^{a-1}\geq p^{2^b-1}>2^{2^b-1}. But this is a contradiction, as desired (because 2^b-1\geq b for all b\geq 1).
Case 2:p\equiv 3 \bmod{4}
Suppose that 2^r\|p+1 and 2\|p-1, so that 2^{r+1-1}\|\frac{p^2-1}{2}. Using the Lifting lemma we conclude that 2^{r+v_2(a)}\|p^a-1.
But p^a-1=2^b(p-1), and this implies that 2^{b+1}\|p^a-1. We conclude that r+v_2(a)=b+1\Rightarrow 2^b=\frac{1}{2}2^r\times 2^{v_2(a)}\leq 2^{r-1}a.
Now suppose that 2^r\neq p+1, so that 2^r\leq \frac{p+1}{2}, and p^{a-1}<1+p+\cdots+p^{a-1}=2^b\leq \frac{1}{4}(p+1)a. But using Bernoulli's inequality we get that p^{a-1}\geq 1+(p-1)(a-1)>(p-1)(a-1), and this is a contradiction, because \frac{p+1}{2}\leq p-1 for p\geq 3 and \frac{a}{2}\leq a-1 for a\geq 2, so that \frac{1}{4}(p+1)a=\frac{p+1}{2}\times \frac{a}{2}\leq (p-1)(a-1).
We conclude that 2^r=p+1\Leftrightarrow p=2^r-1, so that p is a Mersenne prime, and this implies the problem statement.
dattranquoc101201
15.06.2016 06:08
I think the inverse statement is not correct. Consider number 7. \sigma(7)=1+7=8=2^3 but 7 itself is a prime, which can't be a product of any two primes.