# Find A Counter-Example to This

1. ### Atran

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:

Code (Text):
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.

### Staff: Mentor

What is your question? to find a counter example? or to prove your conjecture?

3. ### Atran

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.

### Staff: Mentor

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.

5. ### micromass

19,190
Staff Emeritus
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
$$(p(x+1)/p(x))^2\leq pq\leq p(x)-2i < (p(x+1)/p(x))^2$$