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Prime Numbers

Date: 12/15/2000 at 14:14:40
From: Joel Dick
Subject: Prime Numbers

How can I prove that if

     A = (P1P2+1)^2-1

where P1 and P2 are two distinct primes, then A is a multiple of four 
distinct primes?

Date: 12/15/2000 at 17:51:06
From: Doctor Roy
Subject: Re: Prime Numbers


I am not sure that you can prove this. Consider the case when P1 = 2 
and P2 = 3, and we obtain that:

     A = (P1*P2 + 1)^2 - 1
       = (2*3 + 1)^2 - 1
       = 7^2 - 1
       = 49 - 1
       = 48

On the other hand, we know that 48 equals  16 * 3, which is 2^4 * 3. 
So, we have a case where A satisfies the equation, yet is only the 
multiple of two distinct primes, so we have a case in which the 
statement is not true, and therefore it cannot be true in general. 
Perhaps there is a slightly different statement of this problem that 
can be proved.

I hope this helps, and please feel free to write back with any further 
questions you may have.

- Doctor Roy, The Math Forum   

Date: 12/17/2000 at 00:17:57
From: Joel Dick
Subject: Re: Prime Numbers

You're right! I got it wrong. I think it was:

     A = (P1*P2 + 1)^4 - 1

and A is a multiple of three distinct primes.

I already factored A down to:

     P1*P2*(P1*P2-2)*((P1*P2-1)^2 + 1)

So that takes care of two primes (P1 and P2) so now we have to prove 

     (P1*P2-2)*((P1*P2-1)^2 + 1)

is either, (1) prime and distinct from both P1 and P2 or, (2) a 
multiple of P1 and/or P2 and a third distinct prime. I've plugged in a 
few numbers for P1 and P2 and I've never found (1) to be true. So 
basically, it's narrowed down to proving that (1) is false and (2) is 
true, but I cannot factor P1 or P2 out of the above expression, so 
that's where I need your help.


Date: 12/18/2000 at 11:56:15
From: Doctor Rob
Subject: Re: Prime Numbers

Dr. Roy asked for help in answering your question.

     A = (P1*P2-1)^4 - 1
       = ((P1*P2-1)^2 - 1)*((P1*P2-1)^2 + 1)
       = P1*P2*(P1*P2-2)*(P1^2*P2^2-2*P1*P2+2)

There are two cases to be considered: when one of the primes is 2, and 
when neither is.

Case 1: P1 = 2, P2 >= 3.


     A = 2^3*P2*(P2-1)*(2*P2^2-2*P2+1)

Call the last factor Q. Now it is possible for P2-1 to be a power of 
2. In any case, Q = 1 (mod P2-1), so Q is relatively prime to 2, P2, 
and P2-1; so it must be divisible by at least one prime factor other 
than 2, P2, or any prime factor of P2-1. In this case, you may have 
exactly three distinct prime factors of A, such as when P2 = 3, and so 
A = 2^4*3*13.

Case 2: 3 <= P1 < P2.

Then all four factors of A odd. Furthermore, P1*P2-2 is relatively 
prime to P1 and P2, and is >= 13, so it must be divisible by a prime 
number other than P1 and P2. Furthermore, the last factor is also 
relatively prime to P1 and P2 and P1*P2-2, and is >= 197. This implies 
that A has at least four different prime factors. It may have just 
four, as when P1 = 3, P2 = 5, and A = 3*5*13*197; or it may have more, 
as when P1 = 3, P2 = 11, and A = 3*5^2*11*31*41; or P1 = 5, P2 = 7, 
and A = 3*5*7*11*13*89.

- Doctor Rob, The Math Forum   
Associated Topics:
High School Number Theory

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