Can two consecutive odd primes sum to a product of three integers?

[Andromache]

Prove that the sum of any two consecutive odd prime numbers can always be written as the product of three integers, all greater than 1.

I'm sure this is simpler than it looks. Any help?

 

[robert Ihnot]

7x11 =77, which is composed only of 7 and 11, so the conjecture is false.

 

[Andromache]

7x11 =77, which is composed only of 7 and 11, so the conjecture is false.

I think you misread the question. It's the sum not product.

 

[CRGreathouse]

Suppose p and q are consecutive odd primes. Can (p+q)/2 be prime?

 

[Andrea2]

if p and q are consecutive there isn't any prime number that is "between" the consecutive prime numbers p and q, so (p+q)/2 isn't a prime number.

 

[Andrea2]

p+q is an even number, so we have that there are two factors: 2 and a (even or odd) number. (p+q)/2 could be even or odd. If it's even we have that the sum of two consecutive prime numbers can be written as (2)x(2)x(number). If it's odd we have that it isn't a prime numbers because there isn't any prime number between two CONSECUTIVE prime numbers, so we can write the odd number (p+q)/2 as (odd number)x(another odd number) and the sum of the 2 consecutive prime numbers as (2)x(odd number)x(another odd number). In both cases we demonstrate that the sum of two consecutive prime numbers can always be written as the product of three integers, all greater than 1.

 

[dodo]

I think it's more straightforward starting from CRGreathouse's suggestion directly.

(p+q)/2 is not prime because, as Andrea said, it's between p and q, which are consecutive primes.

Now, if (p+q)/2 is composite, there are two integers a,b > 1 such that ab = (p+q)/2, and therefore 2ab = p + q. QED.