20210812, 06:14  #1 
Aug 2020
79*6581e4;3*2539e3
601 Posts 
Help with exercise questions from Elementary Number Theory
I'm struggling with this problem:
Show that the sum of the reciprocal of all primes is never an integer. I know there are many solutions online, but at a quick glance they often give away the answer immediately or seem to take a different path, so I'd rather like a hint if my attempt is worthwhile: I came up with the general form of the fraction which has primorial pn# as denominator and the numerator is the sum of the products of all possible combinations of n1 primes, i.e. "n choose n1" addends. For example: A sum will never have a prime factor that isn't a prime factor of all addends. Since for all primes <= pn there is a addend that doesn't contain it, the sum, i.e. the numerator, will not be divisible by any of p1 ... pn. So it doesn't share any factor with the denominator. Thus, the fraction cannot be reduced to an integer. Is that reasoning correct? And if so, I guess the proof would have to contain a proof for why the fraction is of that form? (how do you produce a # in Tex here? \# or \text{#} didn't work...) Last fiddled with by LaurV on 20210826 at 07:33 
20210812, 06:53  #2  
Jun 2003
3^{4}×67 Posts 
Quote:
Rest of the argument is fine, I think. You can't simplify that fraction any further (since none of the prime factors of the denominator appears in the numerator, by the above argument). 

20210812, 10:33  #3  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2^{2}×11×83 Posts 
Quote:


20210812, 12:43  #4 
Feb 2017
Nowhere
31·197 Posts 
The fact that, for each prime p, only *one* of the fractions has a denominator divisible by p is indeed the key here.
A slightly more sophisticated argument shows that for n > 1, the "harmonic numbers" all have denominators divisible by 2  and, in fact, the power of 2 dividing the denominator never goes down, and increases every time n is a power of 2. The sum of fractions with a common factor in the denominator can be a fraction without that factor in the denominator, e.g. 1/3 + 1/6 = 1/2. In fact, the common factor can show up in the numerator of the sum! We have the following well known result: If p > 3 is prime, then H_{p1} has a numerator divisible by p^{2} ["Wolstenholme's Theorem"]. It follows that if p > 3 is prime, then 1/p + 1/(2*p) + ... + 1/((p1)*p) has a numerator divisible by p. 
20210812, 13:11  #5  
Apr 2020
1553_{8} Posts 
Quote:
Like this: \(\#\) Or like this: \[\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}=\frac{p_1p_2+p_1p_3+p_2p_3}{p_3\#}\] 

20210812, 14:04  #6 
"Matthew Anderson"
Dec 2010
Oregon, USA
3^{2}×131 Posts 
harmonic series
I seem to remember a Mathlogger Youtube video about infinite harmonic series.
see https://www.youtube.com/watch?v=iJ8pnCO0nTY Hope this Helps. Matt 
20210813, 13:56  #7  
Aug 2020
79*6581e4;3*2539e3
1001011001_{2} Posts 
Thanks for all the helpful replies!
One question though: Quote:


20210813, 14:09  #8 
Apr 2020
5^{3}·7 Posts 
Indeed. Maybe Sweety meant "the denominator when the fraction is written in lowest terms". But there was no attempt at a proof.

20210814, 01:22  #9 
"Καλός"
May 2018
2^{3}×5×11 Posts 
The proof can be further simplified by noticing that the primorial (a term coined by Harvey Dubner) in the denominator is an even integer while the numerator is an odd integer.
By the way, one could possibly use the term 'Mersennerial', n#_{M}, for the product of the first n Mersenne prime exponents. For instance, 13#_{M} = 2 × 3 × 5 × 7 × 13. Also, π_{M}(n) could be the Mersenneprimeexponentcounting function, and π_{M}(13) = 5. 
20210826, 06:30  #10 
Aug 2020
79*6581e4;3*2539e3
601 Posts 
Could a mod change the thread titel to something like "Help with exercise questions from Elementary Number Theory"? So I don't have to open a new threads several times. Thanks.
Here's the next one where I'm struggling: Let p and p^2+8 be prime. Prove that p^3+4 is also prime. As an initial remark: This is true for p=3 and then I couldn't find any other examples. Which is due to: (3k+1)^2+8 = 9k^2+6k+9 = 3(3k^2+2k+3) (3k+2)^2+8 = 9k^2+12k+12 = 3(3k^2+4k+4) So only if p=3 we can have both p and p^2+8 be prime. A bit weird way to phrase the question in that case, but the proof should still be possible, I guess? I thought about algebraic factors of p^3+4 or p^2+8, came as far as (p+2)(p2)+12 = p^2+8. Other than that, no idea. Or is it really just about showing that it only holds for p=3 and since 3^3+4=31 is prime, that's the proof? Last fiddled with by bur on 20210826 at 06:30 
20210826, 07:33  #11  
Romulan Interpreter
"name field"
Jun 2011
Thailand
2×47×109 Posts 
Quote:
Related to your question, you just proved it, which is correct. What other proof do you want? (yes, the question is deliberately formulated in that tricky way, similar to "prove that if n1 is a square and n+1 is a cube, then n3, n+3, and n+5 are all prime", or the (in)famous "new mersenne conjecture", mathematicians have the right to joke too ) Last fiddled with by LaurV on 20210826 at 07:40 

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