Furlong's Conjecture

This past Saturday, I re-entered the "prime" of my life, I turned 53, by the grace of God! As everyone knows, a prime number is a whole number (integer, conventionally denoted as p, for "prime") greater than 1 that has 2 and only 2 distinct whole number divisors, whole numbers that divide into another whole number without leaving any non-zero remainder, namely 1 and itself, p. Indeed, p divided by 1 equals p, and p divided by p equals 1. The first prime number p_1 is 2, which is the unique even prime number (every even number greater than 2 is divisible by 2, in addition to being divisible by 1 and itself, so it cannot be prime)! The next prime number p_2 is 3, since 3 is not evenly divisible by 2. The first composite number, a number greater than 1 that is not prime, is 4, which is evenly divisible by 2 exactly twice. The third prime number p_3 is 5, the second composite number is 6 (which also happens to be the first perfect number, a number that is equal to the sum of all its proper or aliquot divisors, all its divisors that are strictly less than the number itself: the proper divisors of 6 are 1, 2, and 3, and 1+2+3=6), the fourth prime number p_4 is 7, the third composite number is 8, which is evenly divisible by 2 exactly thrice, the fourth composite number is 9, which is (oddly) divisible by 3 exactly twice, and the fifth composite number is 10, which is evenly divisible by 2 exactly once and by 5 exactly once. One can proceed similarly forever!

There happen to be 4 prime numbers less than 10, 25 prime numbers less than 100, 168 prime numbers less than 1000, 1229 prime numbers less than 10000, 9592 prime numbers less than 100000, and 78498 prime numbers less than 1000000, for example! The first 25 prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97. Note that there is one unique pair of neighboring prime numbers, 2 and 3. There are also a fair number of "twin primes" separated by one and only one composite number, such as 3 and 5, 5 and 7, and 11 and 13, for example. It is not known whether there are an unlimited number of twin primes. Euclid cleverly proved centuries ago that there are an infinite number of prime numbers. Assume to the contrary that there are only a finite number of prime numbers, p_1, p_2, p_3, ..., p_n, where p_n is the "largest known prime number." Multiply all of the known primes together and then add 1, forming N=p_n#+1, where p_n# is "p_n primorial", the product of all the prime numbers less than or equal to p_n. Since N is (obviously) greater tha 1, N either is prime or composite. If N is prime, then since N is evidently (much, much) larger than p_n, the so-called "largest known prime," we have demonstrated that there is always at least one more prime number than we thought there were! If N is composite and not prime then N must necessarily have one or more prime divisors that are less than N. However, clearly none of the "known" prime numbers, p_1, p_2, p_3, ..., p_n, are able to divide N without leaving a remainder of 1, so there must exist at least one more prime number p that is greater than the so-called "largest prime number p_n, implying once again that our original assumption that there are only a finite number of prime numbers is wrong and there must therefore be an infinite or unlimited number of prime numbers!

Brocard's conjecture is that there are at least 4 prime numbers between the squares of successive odd prime numbers. I'm pleased to make "Furlong's conjecture," which necessarily includes Brocard's conjecture, but is much more restrictive: there are at least 2 twin primes between the squares of successive odd prime numbers. I have absolutely no idea of how to go about proving my conjecture! I have been able to verify the validity of my conjecture, by judicious sampling, for prime numbers up into the billions (and beyond)! I haven't found a counter-example, yet, but I'll keep looking and be sure to let you all know the outcome!