Have you seen this prime number equation?

  • Thread starter ArcanaNoir
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In summary: Anyway, the proof goes like this. Suppose we have two prime numbers p and q. We can form the product pq by multiplying p and q together. This product is always a prime number, because if it wasn't then we could take pq and divide it by p or q to get a number that isn't a prime number. However, we can't divide pq by p or q because doing so would give us an integer that isn't a prime number. This means that there is some number r such that pq = r^2 for some integer r. r is a prime number because if it wasn't then we could take pq and divide it by p or q to
  • #1
ArcanaNoir
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One of my professors told us about a prime number equation he came up with back in his own college days in Russia. It was published in a Russian journal in the problems section years and years ago and has not been translated to English. The thing is, his professor thought this equation must have already been presented somewhere. My professor could not find it anywhere. I can not find it. But it does seem like the kind of thing that should already have been said. So I'm wondering, has anyone seen this equation before?

an integer p is prime iff there exists a unique m and a unique n (m and n are natural numbers) such that: 1/p = 1/m - 1/n
 
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  • #2
I'm not familiar with it.

To unpack it a little,

[tex]\frac{1}{p}=\frac{1}{m}-\frac{1}{n}\quad\to\quad p=\frac{n\,m}{n-m}[/tex]

Added in Edit:

I did notice the following:

[tex]\frac{1}{p-1}-\frac{1}{p}=\frac{1}{p^2-p}[/tex]

So for any prime p, it's easy enough to find m & n:

m = p - 1

and

n = m p = p(p-1).

Are those (m & n) unique for all primes?
 
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  • #3
SammyS said:
Are those (m & n) unique for all primes?

Well, m and n depend on each other, so I can say at least that it is a unique pair. And m can be expressed depending only on p, so for each p a unique m. The same for n, it can be expressed in terms of p, so for each p a unique n. So, yes, if I understand the question correctly.
 
  • #4
The fact is that for any natural number k > 1, it's true that:

[tex]\frac{1}{k-1}-\frac{1}{k}=\frac{1}{k^2-k}\quad\to\quad\frac{1}{k}=\frac{1}{k-1}-\frac{1}{k^2-k}[/tex]

So that m = k - 1 and n = k(k-1). According to your professor's conjecture, if k is prime, then n and m are unique. Also, according to the conjecture, if k is not prime, then there must be at least on other pair of natural numbers which give 1/k in the prescribed way.

It turns out to be easy to show that if k is not prime, then the m, n pair are not unique.
 
  • #5
EDIT: Sorry for the latexfail, my post didn't seem to be compiling right and I was too lazy to figure out why so I just deleted all the latex code.

Here's a proof that m and n are unique when k is prime:

****Spoiler Alert****

Let m and n be natural numbers such that (1/m) - (1/n) = (1/k), and suppose that k is prime. Rearranging gives

(1) mn = (n - m)k.

Let d = gcd(m,n), and suppose that m = da, n = db, where gcd(a,b) = 1. It is well-known that the identity gcd(m,n) * lcm(m,n) = mn holds for all integers m,n; thus, since lcm(m,n) = dab, we can divide the above relation by d to get

(2) dab = (b - a)k.

However, since gcd(a,b) = 1, we have gcd(a, b - a) = gcd(b, b - a) = 1. In general, if an integer x divides a product of integers yz with gcd(x,y) = 1, then necessarily x divides z . In this case, it follows that a and b divide k . Since k is assumed prime, we have three cases: Either a = b = k , in which case the right-hand side of (2) vanishes (contradiction); or a = b = 1, which has the same problem; or a = 1, b = k (the sign is wrong if a = k, b = 1 ). Plugging this into (2) gives d = k - 1. Putting things together, we see that m = da = k - 1 and n = db = k(k - 1) = k^2 - k, which is the solution that has already been pointed out. This completes the proof.
 
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  • #6
I came up with a slightly different proof (really, I think, just different formulation) of the proof that m,n are unique for p prime.

First, note a few obvious facts from the problem statement (m,n natural numbers):

n > m
0 < m < p
0 < p-m < p
p = nm / (n-m)
n = pm / (p - m )

Considering the last of these formulas, if p-m=1 we have the universal solution m=p-1, n=p(p-1). So we inquire about a case where p-m > 1, thus 1 < p-m < p, and 0 < m < p-1. Since we assume p prime, we must have p-m a factor of m (from last formula). Thus:

m = k (p-m) [1]

which gives:

p = (k+1)m/k

k and k+1 can only have 1 as common factor (suppose qr=k; smallest possible way to use q as factor of k+1 is q(r+1) = qr + q = k+q > k+1 unless q=1). Therefore k must be a divisor of m (including 1 as possibility). But if m/k > 1, we have a factorization of p. Thus k=m. But this leads, from [1] above, to 1 = p-m, which contradicts our assumption that p-m > 1. QED.

---

Just for completeness, I note that for any p, if ab=p,

m = p-a
n = b(p-a)

is a solution, allowing at least a=1,b=p plus some other solution for p not prime.

----

This is a really cute prime property which is new to me (but then, I never studied any number theory).
 
  • #7
Is there anyway to verify that it hasn't been published elsewhere, or at least be reasonably confident? I mean other than scouring actual paper between bindings. Like, a way to google it or something? My google attempts weren't getting the message across to the system. Perhaps I don't know what syntax would be clear to it.
 
  • #8
This theorem (or conjecture) doesn't look all that useful to me. After all, for any integer, k>1, letting m = k-1, and n = k(k-1) gives [itex]\displaystyle \frac{1}{k}=\frac{1}{m}-\frac{1}{n}\,.[/itex] Determining whether these are the only values of m & n which give k in this way, seems to be at least as difficult as any other method for check the primality of k.
 
  • #9
I don't think the point is to use it for a prime check. You're right, that would be less than efficient. I think it's just an exploration of primes and falls more toward number theory.
 
  • #10
Well, yes, it is a nice proof to work out in a number theory class.
 

Related to Have you seen this prime number equation?

1. What is a prime number equation?

A prime number equation is a mathematical expression that represents a relationship between two or more prime numbers. It can be in the form of a formula or an equation that helps to identify or generate prime numbers.

2. How do you determine if a number is prime using an equation?

There are several equations that can be used to determine if a number is prime. One of the most well-known is the Sieve of Eratosthenes, which involves creating a list of all numbers up to the given number and then crossing out all the multiples of each prime number until only the prime numbers are left.

3. Can an equation be used to find all prime numbers?

No, there is no single equation that can generate all prime numbers. However, there are various equations, algorithms, and methods that can be used to find and generate prime numbers up to a certain limit.

4. Are there any famous prime number equations?

Yes, there are several famous prime number equations, including the Sieve of Eratosthenes, the Sieve of Sundaram, and the Sieve of Atkin. Additionally, the Goldbach Conjecture, which states that every even number can be expressed as the sum of two prime numbers, is a famous unsolved mathematical problem.

5. What is the importance of prime number equations?

Prime number equations are essential in cryptography, which is the science of encoding and decoding secret messages. They are also used in various fields of mathematics and computer science, such as number theory, data encryption, and data compression. Prime numbers also have real-world applications, such as in creating secure communication networks and generating random numbers.

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