(See question towards the end)
Suppose we take the first four consecutive primes, $$2, 3, 5, 7$$ Since these are prime numbers, the greatest common divisor will be 1. In other words, they will be co-prime. Knowing this, this also means their lowest common multiple will be the product of the primes multiplied. $2*3*5*7$ will result in $210$.
$210$ obviously will not be prime as it is divisible by $2, 3, 5,$ and $7$, but this must mean $210$'s only factors are $2, 3, 5,$ and $7$ because every number has only one prime factorization which is also unique (ignore composite factors, they will not matter in my method).
Following this logic, $23$, equal to the expression $(2*3*5) - 7$, cannot be divisible by $2, 3, 5,$ or $7$.
$23$'s not divisible by $7$ because we know the first term in the expression, $(2*3*5)$, wasn't a multiple of 7 (since it didn't have 7 in its prime factorization), so subtracting 7 from it won't change that. Also, since $7$ is co-prime with the other three primes, subtracting $7$ from $(2*3*5)$ makes $23$ not divisible by $2, 3,$ or $5$ as well.
$23$ then is not divisible by $2, 3, 5,$ or $7$. Due to how factors work, to check if any positive integer, c, is prime, you can take the square root of c and find primes below that value. If no primes less than the square root of c divide evenly into c, c is prime.
If we apply this to $23$, we get $\lfloor{\sqrt23}\rfloor = 4$. We showed earlier $23$ is not divisible by primes up to $7$, so it's not divisible by primes up to $4$ either. Therefore $23$ is prime.
Generalizing this, we can take the first n consecutive primes ( $p_1, p_2, p_3, ... , p_{n-1}, p_n$) and organize these primes into two groups however you'd like. Then, take the products of the primes within each group. Let's name the larger product, a, and the smaller product, b.
Take the difference of $a - b$. This difference between a and b will always be prime as long as the statements: $$\sqrt{a - b} \leq p_n$$ and $${a - b} >1$$
is true where $p_n$ is the nth prime.
For example, $227$ is a prime which I found using this method. We take the first 8 consecutive primes, $$2, 3, 5, 7, 11, 13, 17, 19$$ and split them into any two groups we'd like, in this case:
$2, 5, 17, 19$
$3, 7, 11, 13$
Taking the products of each group we get:
$2*5*17*19 = 3,230$
$3*7*11*13 = 3,003$
Have the larger product be a and the the smaller product be b. Afterwards, take the difference of $a - b$ : $$3,230 - 3,003 = 227$$
$\lfloor{\sqrt227}\rfloor = 15$.
$15 < 19$ and $227 > 1$, so $227$ is prime.
My Question:
Is this method for finding primes valid? If so, would it be effective to use when trying to find large primes?
(This question was really hard to articulate, and I'm aware I likely didn't do a good job. Edits, suggestions, and clarification is very much welcome!)
from Hot Weekly Questions - Mathematics Stack Exchange
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