# Find A Counter-Example to This

by Atran
Tags: counterexample
 P: 84 Hi! Say p(x) is the product of the first x odd prime numbers (e.g. p(4)=3*5*7*11) and i is at least one. Then consider: 1 < p(x) ± 2*i < (p(x+1)/p(x))2 My hypothesis is that the above formula, obeying the restrictions, always produces a prime number. For example if x=3 and i=13, then p(3)-2*13=79. 79 is bigger than 1 and smaller than 112, therefore it's a prime number. I'm not a python programmer, but this code yields prime numbers using the above formula: from math import ceil def is_prime(number): if(number < 2): return False i = 2 while i <= ceil(number**0.5): if(number%i == 0): return False i = i + 1 return True def f(number_of_primes): flag = True prime_product = 1 prime_array = [] n, i = 3, 0 while i < number_of_primes: if(is_prime(n) == True): prime_product = prime_product * n prime_array.append(n) i = i + 1 n = n + 1 while True: if(is_prime(n) == True): largest_prime = n break n = n + 1 i = 1 while True: for x in range(len(prime_array)): if(i%prime_array[x] == 0): flag = False break if flag == False: flag = True i = i + 1 continue n = prime_product - 2*i if n <= 1: break if n < largest_prime**2: if is_prime(n) == False: print("NOT : ", n, " : ", i) break else: print("NMB : ", n, " : ", i) i = i + 1 def g(number_of_primes): flag = True prime_product = 1 prime_array = [] n, i = 3, 0 while i < number_of_primes: if(is_prime(n) == True): prime_product = prime_product * n prime_array.append(n) i = i + 1 n = n + 1 while True: if(is_prime(n) == True): largest_prime = n break n = n + 1 i = 1 while True: for x in range(len(prime_array)): if(i%prime_array[x] == 0): flag = False break if flag == False: flag = True i = i + 1 continue n = prime_product + 2*i if n < largest_prime**2: if is_prime(n) == False: print("NOT : ", n, " : ", i) break else: print("NMB : ", n, " : ", i) else: break i = i + 1 x = 4 # Enter a non-negative integer f(x) print("- - - - -") g(x) Scroll down and assign a non-negative integer to x in the code and run it. If the program detects a non-prime then it should output "NOT" followed by the generated number. If the generated number is a prime, then it outputs "NMB". The second number after the second colon is the value of i. So the above example would be displayed as: "NMB : 79 : 13" (excluding the double quotes) Thanks for help.
 P: 2,969 What is your question? to find a counter example? or to prove your conjecture?
 P: 84 I want somebody to prove this wrong, by finding a counter example. And I'm really sorry, I forgot to mention that i must not be divisible by any of the primes found in the product p(x). So again, 1 < p(x) ± 2*i < (p(x+1)/p(x))2 where p(x) is the product of the first x odd prime numbers, and i is a positive integer not divisible by any prime in the product p(x). Example: p(4) - 2*514 = 127. 514 is positive and not divisible by 3, 5, 7, or 11. And 127 is less than 169, so according to the conjecture 127 should be a prime, and it is.
 P: 2,969 Find A Counter-Example to This Your best strategy here would be to let the program find the counter example by finding a list of primes for input and then running it thru its paces. Also if you could find a related program that can check for primeness of a number you could incorporate it in to the mix.
Mentor
P: 18,277
 Quote by Atran I want somebody to prove this wrong, by finding a counter example. And I'm really sorry, I forgot to mention that i must not be divisible by any of the primes found in the product p(x). So again, 1 < p(x) ± 2*i < (p(x+1)/p(x))2 where p(x) is the product of the first x odd prime numbers, and i is a positive integer not divisible by any prime in the product p(x). Example: p(4) - 2*514 = 127. 514 is positive and not divisible by 3, 5, 7, or 11. And 127 is less than 169, so according to the conjecture 127 should be a prime, and it is.
It's a true conjecture. Here's a proof:

First, we note that since ##p(x)## is not divisible by ##2## and ##2i## is divisible by ##2##, then ##p(x) - 2i## is not divisible by ##2##.
Second, if ##p## is an odd prime occuring in the product of ##p(x)##, then ##p## divides ##p(x)## and ##p## does not divide ##2i##, thus ##p## does not divide ##p(x) - 2i##.

So we deduce that if ##p## is any prime dividing ##p(x)-2i##, then ##p## cannot occur in ##p(x)##. Thus ##p\geq p(x+1)/p(x)##.

So, assume that ##p(x) - 2i## is composite, then there are there are two prime numbers ##p## and ##q## that divide ##p(x) - 2i##. Thus we have ##pq\leq p(x)-2i##. But we also have that ##p\geq p(x+1)/p(x)## and ##q\geq p(x+1)/p(x)##. Thus ##pq\geq (p(x+1)/p(x))^2##. So we see that
$$(p(x+1)/p(x))^2\leq pq\leq p(x)-2i < (p(x+1)/p(x))^2$$