AdrienMarie Legendre (17521833), known for important concepts such as the Legendre polynomials and Legendre transformation, states that given an integer n > 0, there exists a prime number, p, between and , or alternatively, . This conjecture constitutes one of Edmund Landau’s
four “unattackable” problems mentioned at the 1912 Fifth Congress of Mathematicians in Cambridge.
four “unattackable” problems mentioned at the 1912 Fifth Congress of Mathematicians in Cambridge.
A table of the results of the primes
that exist between and can be seen below.
n

Condition:

Prime
numbers, p 
Number of primes between and

1

1
< p < 4 
2, 3

2

2

4
< p < 9 
5, 7

2

3

9
< p < 16 
11, 13

2

4

16
< p < 25 
17, 19, 23

3

5

25
< p < 36 
29, 31

2

6

36
< p < 49 
37, 41, 43, 47

4

7

49
< p < 64 
53, 59, 61

3

8

64
< p < 81 
67, 71, 73, 79

4

We can see from the above table that indeed, the Legendre Conjecture holds true for the cases considered, with at least two prime numbers, p, fulfilling the inequality, . Furthermore, the positive integers, and are referred to as square numbers. But, how would one go about proving the conjecture true for any n?The best effort so far, it seems, was given by Chinese mathematician, Chen Jingrun (陳景潤) (19331996), who proved a slightly weaker version of Legendre’s Conjecture: there is either a prime, p in the interval, or a semiprime, pq, in the interval, , where q is a prime not equal to p. A semiprime is a composite number > 1 (not prime) that is the product of two prime numbers (could even possibly be equal primes). Examples of the initial semiprime numbers are 4, 6, 9, 10, 14, 15, 21, 22, etc. Examples of the first several semiprime numbers whose factors are distinct and for which Jingrun proved a slightly weaker version of Legendre’s Conjecture are: 6, 10, 14, 15, 21, 22, 26, 33, 34, etc. If the Legendre Conjecture turns out to be true, then the gap between any two primes, and would be , where p is a prime number, and O constitutes the big O
notation. Understanding the big O notation, let’s look at the table above for the case n = 2, for which it is found that the prime numbers 5 and 7 fall within the interval, ; therefore the gap between 5 and 7 should be which is about 2.236. Accordingly, the gap between prime numbers in big O notation has to be adjusted to become more representative of the actual gap between prime numbers if the Legendre Conjecture is true.This problem definitely seems hard to tackle,
but time will tell whether it will be proven, whether rigorously in analytic sense or with the aid of computers in a numerical fashion.What are your thoughts on approaching Legendre’s Conjecture?
notation. Understanding the big O notation, let’s look at the table above for the case n = 2, for which it is found that the prime numbers 5 and 7 fall within the interval, ; therefore the gap between 5 and 7 should be which is about 2.236. Accordingly, the gap between prime numbers in big O notation has to be adjusted to become more representative of the actual gap between prime numbers if the Legendre Conjecture is true.This problem definitely seems hard to tackle,
but time will tell whether it will be proven, whether rigorously in analytic sense or with the aid of computers in a numerical fashion.What are your thoughts on approaching Legendre’s Conjecture?
I am not sure if it is relevant, but I have something I call silly prime tricks, where I could see the pattern of primes geometrically, and tap them out musically. y=2x+3 will intersect every positive odd number, from which you can measure the hypotenuse distance between any odd or odd prime number using the Pythagorean theorem. All odd primes Ive plugged in work. The distance between 5 and 7 is the square root of 5, which is very close to the value of the gap indicated.
I’m an iranian student have been working for 7 months on this problem…ì acchived good results…this is not as hard as what you think…only a simple creative solution will solve it…
mahdokht: Do you have it?
It’s obviously correct, but that knowledge doesn’t prove anything. For example, you only need to multiply 1/2 * 1/3 * 1/5 * 1/7… (the prime factors) to the largest prime less than half a quadratic interval to see there must be an odd (prime) deficit. That’s an application of the Euler product formula. Mathematicians know about that, and it doesn’t provide a deterministic proof to infinity as far as they’re concerned.
After obsessing about this conjecture for a yearandahalf, I have a solution that I’m satisfied with: https://www.quora.com/profile/MichaelMRoss/Mathrodite/LinearParityProvesLegendresConjecture
I’m guessing you’re an enthusiast – not a trained mathematician – like me. Wish you the best of luck.
I meant “to add 1/2 + 1/3 + 1/5 + 1/7…”
Updated URL: https://www.quora.com/q/mathrodite/LegendresConjecture
My thoughts on the conjecture are stated in my blog entry on the subject:
https://justmathstuff.wordpress.com/2017/11/21/atreatiseontheprimenumbersincludingproofsofbrocardslegendresandandricasconjectures/
FYI: Proof of Legendre Cojecture via application of PNT
Key Result:
π((n+1)^2) – π (n^2) → π(n) = n / log(n) as n →∞.
Relevant Reference Link:
https://www.quora.com/Canwedisprovethatforanypositiveintegernthereareatleasttwoprimenumbersbetweenn2andn12/answer/DavidCole146
Oops! …Conjecture…