Register to reply 
Find A CounterExample to Thisby Atran
Tags: counterexample 
Share this thread: 
#1
Jun1214, 06:44 PM

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 11^{2}, therefore it's a prime number. I'm not a python programmer, but this code yields prime numbers using the above formula:
Thanks for help. 


#2
Jun1214, 09:58 PM

P: 2,796

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



#3
Jun1214, 11:56 PM

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. 


#4
Jun1314, 05:36 AM

P: 2,796

Find A CounterExample 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.



#5
Jun1314, 06:30 AM

Mentor
P: 18,038

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 [tex](p(x+1)/p(x))^2\leq pq\leq p(x)2i < (p(x+1)/p(x))^2[/tex] which is a contradiction. 


Register to reply 
Related Discussions  
Where can I find an Ion counter?  Atomic, Solid State, Comp. Physics  0  
Find a counter example/predicate logic  Calculus & Beyond Homework  8  
Where can I find counterclaims to David Cole's documentary?  General Discussion  7  
Geiger counter, ionization counter with a gain of unity  Advanced Physics Homework  0  
A Conjecture : can anyone find a counterexample or otherwise disprove ?  General Math  2 